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Solutions toTopology

Chapter 3 - Connectedness and CompactnesJames Munkres

Solutions by positrón0802https://positron0802.wordpress.com

1 January 2021

Contents

3 Connectedness and Compactness 123 Connected Spaces . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 124 Connected Subspaces of the Real Line . . . . . . . . . . . . . . . . . . . . . . . . . 325 Components and Local Connectedness . . . . . . . . . . . . . . . . . . . . . . . . 726 Compact Spaces . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1227 Compact Subspaces of the Real Line . . . . . . . . . . . . . . . . . . . . . . . . . . 1628 Limit Point Compactness . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1829 Local Compactness . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 21Supplementary Exercises: Nets . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 24

3 Connectedness and Compactness

23 Connected Spaces

Exercise 23.1. If (-,� ′) is connected, then (-,�) is connected. The converse is not true ingeneral.

Exercise 23.2. Suppose that⋃= �= = � ∪�, where � and � are disjoint open subsets of

⋃= �= .

Since �1 is connected and a subset of � ∪ �, by Lemma 23.2 it lies entirely within either � or�. Without any loss of generality, we may assume �1 ⊂ �. Note that given =, if �= ⊂ � then�=+1 ⊂ �, for if �=+1 ⊂ � then �= ∩�=+1 ⊂ � ∩� = ∅, in contradiction with the assumption. Byinduction, �= ⊂ � for all = ∈ Z+, so that

⋃= �= ⊂ �. It follows that

⋃= �= is connected.

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23 Connected Spaces

Exercise 23.3. For each U we have � ∩ �U ≠ ∅, so each � ∪ �U is connected by Theorem 23.3.In turn {� ∪ �U }U is a collection of connected spaces that have a point in common (namely anypoint in �), so

⋃U (� ∪�U ) = � ∪ (

⋃U �U ) is connected by Theorem 23.3.

Exercise 23.4. Suppose that � is a non-empty subset of - that is both open and closed, i.e., �and - \ � are �nite or all of - . Since � is non-empty, - \ � is �nite. Thus � cannot be �nite as- \� is in�nite, so � is all of - . Therefore - is connected.

Exercise 23.5. Let - have the discrete topology and let . be a subspace of - containing at leasttwo di�erent points. Let ? ∈ . . Then {?} and. \{?} are non-empty disjoint open sets in. whoseunion is ., so . is not connected. It follows that a discrete space - is totally disconnected.

The converse does not hold: - = Q (with the standard topology) is totally disconnected(Example 4), but its topology is not the discrete topology.

Exercise 23.6. Suppose that�∩Bd� = �∩�∩- −� = ∅. Then�∩� and�∩(- \�) are a pair ofdisjoint non-empty sets whose union is all of�, neither of which contains a limit point of the other.Indeed, if�∩(-−�) contains a limit point G of�∩�, then G ∈ �∩(-−�)∩�′ ⊂ �∩�∩- −� = ∅,a contradiction, and similarly� ∩� does not contain a limit point of� ∩ (- −�). Then� ∩� and� ∩ (- −�) constitute a separation of� , contradicting the fact that� is connected (Lemma 23.1).

Exercise 23.7. The space Rℓ is not connected, as (−∞, 0) and [0, +∞) are disjoint non-emptyopen sets in Rℓ whose union is all of Rℓ .

Exercise 23.8. First, we show that Rl is not connected in the uniform topology. Let � and �denote the subsets of Rl consisting of all bounded and all unbounded sequences respectively.Then � ∪ � = Rl and � ∩ � = ∅. If a ∈ Rl , then the set

* (a, 1) = (01 − 1, 01 + 1) × · · · × (0= − 1, 0= + 1) × · · ·

contains the ball �d (a, Y) if Y < 1, and consists entirely of bounded sequences if a ∈ �, and ofunbounded sequences if a ∈ �. Thus� and � are open in Rl . Since they are non-empty, it followsthat Rl is not connected in the uniform topology.

Exercise 23.9. This is similar to the proof of Theorem 23.6. Take 2 × 3 ∈ (- \�) × (. \ �) . Foreach G ∈ - \�, the set

*G = (- × {3}) ∪ ({G} × . )

is connected since - × {3} and {G} × . are connected and have the common point G × 3. Then* =

⋃G ∈-\�*G is connected because it is the union of the connected spaces *G which have the

point 2 × 3 in common. Similarly, for each ~ ∈ . \ � the set

+~ = (- × {~}) ∪ ({2} × . )

is connected, so+ =⋃~∈.\� +~ is connected. Thus (- ×. ) \ (� × �) = * ∪+ is connected since

2 × 3 is a common point of* and + .

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24 Connected Subspaces of the Real Line

Exercise 23.10. (a) Let = {U1, . . . , U=} ⊂ � . Then the function 5 : - → -U1 × · · · × -U= givenby 5 (x) = (GU1, . . . , GU= ) is a homeomorphism. Since the latter is a �nite product of connectedspaces, it is connected, and therefore so is - .

(b) Since a is a point in common of the collection {- } of connected spaces - , for ⊂ �

�nite, it follows that . is connected.(c) Let x = (GU )U ∈ - \ . and let * be a (standard) basis element for

∏U -U containing x, so

that * =∏U *U where *U is open in -U for all U and *U = -U except for �nitely many indices,

say U1, . . . , U= . Let = {U1, . . . , U=}. Then is a �nite subset of � . As x ∉ ., there exists someindex V ∈ � \ such that GV ≠ 0V . Let y = (~U )U ∈

∏U -U be such that~U8 = GU8 for all 8 = 1, . . . , =,

and ~U = 0U for all other indices. Then y ≠ x, and y ∈ * ∩ - ⊂ * ∩ . . Since * was arbitrary,we have x ∈ . ′. It follows that - = . . We deduce that - is connected from (b) and Theorem 23.4.

Exercise 23.11. Suppose that * and + constitute a separation of - . If ~ ∈ ? (* ), then ~ = ? (G)for some G ∈ * , so that G ∈ ?−1({~}). Since ?−1({~}) is connected and G ∈ * ∩ ?−1({~}), wehave ?−1({~}) ⊂ * . Thus ?−1({~}) ⊂ * for all ~ ∈ ? (* ), so that ?−1(? (* )) ⊂ * . The inclusion* ⊂ ?−1(? (* )) if true for any subset and function, so we have the equality * = ?−1(? (* )) andtherefore * is saturated. Similarly, + is saturated. Since ? is a quotient map, ? (* ) and ? (+ ) aredisjoint non-empty open sets in . . But ? (* ) ∪ ? (+ ) = . as ? is surjective, so ? (* ) and ? (+ )constitute a separation of ., contradicting the fact that . is connected. We conclude that - isconnected.

Exercise 23.12. Suppose that � and � form a separation of . ∪ �. Since . ⊂ � ∪ � and . isconnected, . is entirely contained in either� or� ; suppose. ⊂ �, so that� ⊂ �. Since� is openand closed in - \., there exist* open in - and � closed in - such that � = (- \. ) ∩* = * \.and � = (- \. ) ∩ � = � \. . Note that � ⊂ � ⊂ * and � ⊂ � ⊂ � . Since � is open in . ∪� and* ⊂ (* − . ) ∪ . = . ∪ �, it follows that � is open in * . Similarly, � is closed in � . Thus � isa non-empty subset open and closed in -, a contradiction. Hence . ∪ � is connected. Similarly,. ∪ � is connected.


Exercise 24.1. (a) We follow the hint. If ℎ : [0, 1] → (0, 1] is a homeomorphism, then the restric-tion ℎ′ : (0, 1) → (0, 1] \ {ℎ(0), ℎ(1)} of ℎ is a homeomorphism between a connected space and adisconnected space, a contradiction. Similarly, (0, 1) is neither homeomorphic to [0, 1] nor (0, 1] .

(b) The function 5 : (0, 1) → (0, 1] given by 5 (G) = G/2 is continuous with image (0, 1/2),and the restriction 5 ′ : (0, 1) → (0, 1/2) is a homeomorphism, so 5 is an imbedding. Similarly,6 : (0, 1] → (0, 1) given by 6(G) = G/2 is an imbedding. But we know from (a) that (0, 1) and (0, 1]are not homeomorphic.

(c) If 5 : R= → R is a homeomorphism, the restriction 5 ′ : R= \ {0} → R \ {5 (0)} is a homeo-morphism, but R= \ {0} is connected (Example 4) while R \ {5 (0)} can’t be connected. So thereis no homeomorphism between R= and R if = > 1.



Exercise 24.2. Let 5 : (1 → R be continuous. Let G ∈ (1. If 5 (G) = 5 (−G) we are done, soassume 5 (G) ≠ 5 (−G) . De�ne 6 : (1 → R by setting 6(G) = 5 (G) − 5 (−G) . Then 6 is continuous.Suppose 5 (G) > 5 (−G), so that 6(G) > 0. Then −G ∈ (1 and 6(−G) < 0. By the intermediate valuetheorem, since (1 is connected and 6(−G) < 0 < 6(G), there exists ~ ∈ (1 such that 6(~) = 0. i.e,5 (~) = 5 (−~) . Similarly, if 5 (G) < 5 (−G), then 6(G) < 0 < 6(−G) and again the intermediatevalue theorem gives the result.

Exercise 24.3. If 5 (0) = 0 or 5 (1) = 1 we are done, so suppose 5 (0) > 0 and 5 (1) < 1. Let6 : [0, 1] → [0, 1] be given by 6(G) = 5 (G) −G . Then 6 is continuous, 6(0) > 0 and 6(1) < 0. Since[0, 1] is connected and 6(1) < 0 < 6(0), by the intermediate value theorem there exists G ∈ (0, 1)such that 6(G) = 0, that is, such that 5 (G) = G .

If- equals [0, 1) or (0, 1) this is no longer true. For example 5 : (0, 1) → (0, 1) given by 5 (G) =G/2 is continuous with no �xed points. Similarly 6 : [0, 1) → [0, 1) given by 6(G) = G/2 + 1/2 iscontinuous with no �xed points.

Exercise 24.4. First we prove that - has the least upper bound property. Let � be a non-emptysubset of - that is bounded above. Suppose that � does not have a supremum, so that the set �of all upper bounds for � does not have a smallest element. Consider the subsets

� =⋃0∈�(−∞, 0) and � =

⋃1∈�(1, +∞)

of- . Note� and� are non-empty since� does not have a supremum, and they are union of openrays, so they are open in - . They are also disjoint, for suppose G ∈ � ∩ �. Then G ∈ (1, +∞) forsome 1 ∈ �, so G is an upper bound for �. Also, G ∈ (−∞, 0) for some 0 ∈ �, so 2 < 0 ≤ 2, absurd.Thus � ∩ � = ∅. Finally, let G ∈ - . Then either G ∈ � or G ∈ - \ �. If G ∈ �, since � does nothave a smallest element, there exist 1 ∈ � such that 1 < 2, so 2 ∈ (1, +∞) ⊂ �. If G ∈ - \ �, thereexists 0 ∈ � such that G < 0, so G ∈ (−∞, 0) ⊂ �. It follows that � and � form a separation of -,contradicting the fact that - is connected. Hence, � has a supremum. Since � was arbitrary, -has the least upper bound property.

Now, if G < ~ and there is no element I such that G < I < ~, then the sets (−∞, ~) and (G, +∞)form a separation of -, a contradiction. We conclude that - is a linear continuum.

Exercise 24.5. (a) We prove that � ≔ Z+ × [0, 1) is a linear continuum. Let G × ~, I × F ∈ �and suppose that G × ~ < I × F. If G = I and ~ < F, then G × ~ < G × ~+F

2 < I × F. If G < I,

then G × ~ < G+I2 × 0 < I × F. Now, if ( ⊂ � is non-empty and bounded above, let c1 : � →

Z+ and c2 : � → [0, 1) denote the projections onto the �rst and second factor respectively; letG = maxc1(() and ~ = supc2(( ∩ ({G} × (0, 1])). If ~ ≠ 1, then G × ~ = sup (. If ~ = 1, then(G + 1) × 0 = sup (. Since ( was arbitrary, � have the least upper bound property. We concludethat � = Z+ × [0, 1) is a linear continuum.

(b) Since 0 × 1 < 0 × 2 but there is no element I in [0, 1) × Z+ such that 0 × 1 < I < 0 × 2, itfollows that [0, 1) × Z+ is not a linear continuum.



(c) We show that� ≔ [0, 1) × [0, 1] is a linear continuum. Similarly as in (a), for every G,~ ∈ �such that G < ~, there exists I ∈ � such that G < I < ~. Let ( ⊂ � be non-empty and boundedabove, and let c1 : � → [0, 1) and c2 : � → [0, 1] denote the projections. Let G = supc1(() . IfG ∈ c1((), let ~ = supc2(( ∩ ({G} × [0, 1])); then G × ~ = sup (. If G ∉ c1((), then G × 0 = sup (.Since ( was arbitrary, � have the least upper bound property, so � = [0, 1) × [0, 1] is a linearcontinuum.

(d) [0, 1] × [0, 1) is not a linear continuum, since for any G ∈ [0, 1), the subset ( = {G} × [0, 1)is bounded above but does not have a least upper bound.

Exercise 24.6. Let- be a well-ordered set. G ×~, I×F ∈ - × [0, 1); suppose that G ×~ < I×F. IfG = I and~ < F, then G×~ < G× ~+F2 < I×F. If G < I, then G×~ < G× ~+12 < I×F.We prove that- × [0, 1) have the greatest lower bound property, which is equivalent to having the least upperbound property. Let ( ⊂ - × [0, 1) be non-empty and bounded below; let c1 : - × [0, 1) → - andc2 : - × [0, 1) → [0, 1) denote the projections onto the �rst and second factor respectively. Since- is a well-ordered set, c1(() has a least element G . Then G × ~, where ~ = inf c2(( ∩ ({G} × ()),is a greatest lower bound for (. Since ( was arbitrary, - × [0, 1) have the greatest lower boundproperty. It follows that - × [0, 1) is a linear continuum.

Exercise 24.7. (a) Since 5 is order-preserving, it is injective, hence a bijection. Let (0, +∞) be anopen ray in ., G = 5 −1(0) . Then 5 −1((0, +∞)) = (G, +∞), so 5 −1((0, +∞)) is open in - . Similarly,5 −1((−∞, 1)) = (−∞, 5 −1(1)) is open in - for each 1 ∈ . . Since open rays constitute a subbasisfor the order topology on ., and their preimages are open in -, it follows that 5 is continuous.Similarly, 5 ((2, +∞)) = (5 (2), +∞) and 5 ((−∞, 3)) = (−∞, 5 (3)) are open in ., so the image ofevery open ray in - is open in . and hence 5 is an open map. Therefore, 5 a homeomorphism.

(b) Given ~ ∈ R+, the point ~1/= ∈ R+ satis�es 5 (~1/=) = ~, so 5 is surjective. If G,~ ∈ R+ andG < ~, then G2 < G~ < ~2. Inductively, G: < ~: for all : ∈ Z+, so 5 is order-preserving. By (a), 5is a homeomorphism, so its inverse, the =th root function, is continuous.

(c) Let G ∈ R. If G ≥ 0, then 5 (G) = G . If G < 0, then 5 (G − 1) = G . Thus 5 is surjective.Furthermore, 5 is clearly order-preserving. But 5 can’t be a homeomorphism, sinceR is connectedwhile (−∞,−1) ∪ [0, +∞) is not. (Recall from Theorem 16.4 that if - have the order topologyand . ⊂ - is convex, then the subspace topology on . equals the order topology on . . Since(−∞,−1) ∪ [0, +∞) is not convex, we do not expect that this two topologies are equal. Indeed,[0, +∞) is open in the subspace topology on (−∞,−1) ∪ [0, +∞) while is not open in the ordertopology on (−∞,−1) ∪ [0, +∞) .)

Exercise 24.8. (a) Let {-U }U ∈� be a collection of path-connected spaces and consider the productspace

∏U -U . Let x, y ∈ ∏

U -U . For each U, there exists a continuous function 5U : [0, 1] → -Usuch that 5U (0) = GU and 5U (1) = ~U . Let 5 : [0, 1] → ∏

-U be given by the equation 5 (C) =(5U (C))U ∈� . Then 5 is continuous by Theorem 19.6. Moreover, 5 (0) = x and 5 (1) = y, so x andy can be joined by a path in

∏U -U . Since x, y ∈ ∏

U -U were arbitrary, we conclude that theproduct space

∏U -U is path-connected.



(b) It’s not true that � path-connected implies � path-connected. For instance, see Example7: the subspace � = {G × sin(1/G) | 0 < G ≤ 1} of R2 is path-connected, but its closure ( =

( ∪ ({0} × [−1, 1]), the topologist’s sine curve, is not.(c) If 5 (G), 5 (~) ∈ 5 (- ), there exists a continuous function 6 : [0, 1] → - such that 6(0) = G

and 6(1) . Then 5 ◦6 : [0, 1] → 5 (- ) is continuous and (5 ◦6) (0) = 5 (G), (5 ◦6) (0) = 5 (~). Thus5 (- ) is path-connected.

(d) Let I ∈ ⋂U �U . Let G,~ ∈ ⋃

U �U . Then there exist indices V,W such that G ∈ �V and~ ∈ �W .Since �V is path-connected, there exist a continuous function 5 : [0, 1] → �V such that 5 (0) = Gand 5 (1) = I. Similarly, there is a continuous function 6 : [1, 2] → �W such that 6(1) = I and6(2) = ~. By Theorem 18.2(e), the functions 5 ′ : [0, 1] → �V ∪ �W and 6′ : [1, 2] → �V ∪ �Wobtained by expanding the range of 5 and 6 respectively are continuous. By the pasting lemma(Theorem 18.3), since [0, 1] and [1, 2] are closed in [0, 2] and 5 ′(1) = 6′(1), we can combine 5 ′and 6′ to obtain a continuous function ℎ : [0, 2] → �V ∪ �W by setting ℎ(G) = 5 ′(G) if G ∈ [0, 1]and ℎ(G) = 6′(G) if G ∈ [1, 2] . Applying Theorem 18.2(e) again, the function ℎ′ : [0, 2] → ⋃

U �Uobtained by expanding the range of ℎ is continuous, and furthermore, ℎ′(0) = G and ℎ′(2) = ~.Thus G and ~ can be joined by a path in

⋃U �U . It follows that

⋃U �U is path-connected.

Exercise 24.9. We use the hint. Let G,~ ∈ R2 \ �. There are uncountable many lines passingthrough G, and � is countable, so choose one of the lines that do not intersect �. Then choose aline passing though ~ that does not intersect � but intersects the line passing though G . Thesetwo lines are path-connected and have a point in common, so their union is path-connected byExercise 24.8(d). Thus G and ~ can be joined by a path in R2 \�. Since G,~ ∈ R2 \� were arbitrary,it follows that that R2 \� is path-connected.

Exercise 24.10. We follow the hint. Given G ∈ * , let�G be the set of points that can be joined toG by a path in * . If ~ ∈ �G , since * is open, there exists a ball �3 (~, A ) entirely contained within* . Then every point in �3 (~, A ) can be joined to ~ by a path in * , hence can be joined to G . So~ ∈ �3 (~, A ) ⊂ �G and �G is open in * . Similarly, if I ∈ * \ �G , let A ′ > 0 be so that the ball�3 (I, A ′) is entirely contained within * . Then none of the points in �3 (I, A ′) can be joined to Gby a path in * , for then I could be joined to G . So I ∈ �3 (I, A ′) ⊂ * \ �G and * \ �G is open in* . Thus �G is non-empty and both open and closed in* , which is connected, so �G = * . Since Gwas arbitrary, it follows that* is path-connected.

Exercise 24.11. Let �2 denote the closed unit ball in R2, and� denote the closed ball with centre2 × 0 and radius 1. Then 1 × 0 ∈ �2 ∩�, so � = �2 ∪� is connected. But Int� is not connected,since Int�2 and Int� form a separation of Int�. Furthermore, � = (0, 1) is connected whileBd� = {0, 1} is not.

The converse does not hold either: BdQ = R is connected while Q is not, and IntQ ∩ (0, 1) =(0, 1) is connected while Q ∩ (0, 1) is not.

Exercise 24.12. (a) First assume that [0, 2) has the order type of [0, 1) . There exists an order-preserving bijection 5 : [0, 2) → [0, 1), so its restriction to [0, 1) → [0, 5 (1)) is an order-preservingbijection. Thus [0, 1 (2)) has the order type of [0, 5 (1)), hence of [0, 1), and similarly for [1, 2) .


25 Components and Local Connectedness

Conversely, assume both [0, 1) and [1, 2) have the order type of [0, 1) . Then we have order-preserving bijections [0, 1) → [0, 1/2) and [1, 2) → [1/2, 1) which glue to an order-preservingbijection [0, 2) → [0, 1).

(b) First suppose that [G0, 1) has the order type of [0, 1) . Then, for given 8 we have G8 <

G8+1 < 1, so by part (a) we deduce that [G8 , G8+1) has the order type of (0, 1] . Conversely, supposethat each interval [G8 , G8+1) has the order type of [0, 1). Then for each 8 ∈ N there is an order-preserving bijection 58 : [G8 , G8+1) →

[1 − 1

28 , 1 −1

28+1

), and they all glue to an order-preserving

bijection [G0, 1) → [0, 1) .(c) We follows the hint. We prove the claim by trans�nite induction. The base case is when

0 is the successor of 00, and it is clear. Now let 0 have a immediate predecessor 1 (greater than00) in (Ω . Then [00 × 0, 1 × 0) has the order type of [0, 1) by induction hypothesis, and so does[1 × 0, 0 × 0] . It then follows from (a) that [00 × 0, 0 × 0) has the order type of [0, 1) . Now supposethat 0 is not a successor in (Ω . Then there is an increasing sequence 08 in (Ω with 0 = sup8{08}.By induction, each interval [00 × 0, 08 × 0) has the order type of (0, 1] . Then so does [00 × 0, 0 × 0)by part (b). We deduce by the principle of trans�nite induction that the interval [00 × 0, 0 × 0] of(Ω × [0, 1) has the order type of [0, 1) for all 0 in (Ω distinct from 00.

(d) Let 01 denote the successor of 00 in (Ω . Consider the point 01 × 0 ∈ !. As [00 × 0, 01 × 0)has the order type of [0, 1) by (c), any point in this interval can be connected to 01 × 0 by a pathin !. Given 1 ≥ 01, every point 1 × C for C ∈ [0, 1) can be connected to the point 1 × 0, and in turn1 × 0 can be connected to 01 × 0 by parts (a) and (b). Thus every point of ! can be joined by a pathto 01 × 0, so we deduce that ! is path-connected.

(e) This is clear for points of the form 00 × C, C ∈ [0, 1) . Now let 1 > 00 and C ∈ [0, 1) . Then(00 × 0, 1 × 0) ⊂ ! is homeomorphic to (0, 1) ⊂ R by part (c) and Exercise 24.7(a), and hence so is(00 × 0, 1 × C) .

(f) We follow the hint. Suppose that ! can be imbedded in R= for some =. In particular, ! issecond countable. Let . = {(08 , C8)}8∈N be any countable subset of - . Then there exists 1 ∈ (Ωsuch that 1 > 08 for all 8 ∈ N. Let 1 ′ = 1 + 1 denote the successor of 1. Then {1 ′} × (0, 1) isa neighbourhood of 1 ′ × 1

2 in ! which does not intersect . . This is a contradiction by Theorem30.3(b). It follows that ! cannot be imbedded in any R= .


Exercise 25.1. Let� be a non-empty connected subspace of Rℓ and let 0 ∈ �. Then [0, +∞)∩� isnon-empty and both open and closed in�, so [0, +∞)∩� = �. Thus, if 1 ∈ �, then 1 ∉ (−∞, 0), so1 ≥ 0. Similarly, 0 ≥ 1 for any 1 ∈ �, so � = {0}. Since each path component lies in a connectedcomponent, it follows that only the singletons {0} are path-connected in Rℓ .

If 5 : R→ Rℓ is continuous, its image 5 (R) must be connected in Rℓ , so must be a singleton.Hence the continuous maps 5 : R→ Rℓ are the constant maps (cf. Exercise 18.7(b)).

Exercise 25.2. (a) The product space Rl is connected by Example 7 of §13, so the whole space



is the only component. Since R is path-connected, the product space Rl is path-connected byExercise 24.8(a), so the whole space is the only path component.

(b) We use the hint. Note that given x ∈ Rl , the map 5 : Rl → Rl given by 5 (y) = x− y is ahomeomorphism. Therefore, x and y lie in the same component if and only if x−y lies in the samecomponent of 0. We prove that this occurs exactly when x − y is bounded. Recall from Exercise23.8 that the set � of all bounded sequences and the set � of all unbounded sequences constitutea separation of Rl . Thus, if x − y lies in the same component of 0, it must belong to �, i.e., x − ymust be bounded. Conversely, assume that x− y is bounded; we show that x− y lies in the samecomponent of 0. De�ne 5 : [0, 1] → Rl by 5 (C) = C (x− y) . Let " > 0 be such that |G= −~= | < "

for all = ∈ Z+. Given Y > 0, let X = Y/". Thus, if |C − C0 | < X, then d (5 (C), 5 (C0)) < X" = Y. Hence5 is a path in Rl from 0 to 5 (1) = x − y. So 0 and x − y lie in the same path component, andin particular in the same component. We conclude that x − y is bounded if and only if it lies thesame component of 0, and this occurs if and only if x and y lie in the same component.

(c) Let R∞ denote the subspace of Rl in the box topology consisting of all sequences whichare eventually zero. We have to prove that x and y lie in the same component of Rl in the boxtopology if and only if x − y ∈ R∞.

First assume x− y ∈ R∞. De�ne 5 : [0, 1] → Rl by 5 (C) = y + C (x− y) . Since x− y ∈ R∞, wehave G8−~8 = 0 for all but �nitely many indices 81, . . . , 8= . Let C ∈ R and let* =

∏8 *8 be a standard

basis element for Rl in the box topology containing 5 (C). For each 9 = 1, . . . , =, let Y 9 > 0 be suchthat the ball centred at the 8 9 -entry of 5 (C) and having radius Y 9 is entirely contained within *8 9 .Let Y = min{Y 9 }. Then + =, ∩ [0, 1], where, = (C − Y, C + Y) is a neighbourhood of C in [0, 1]and satis�es 5 (+ ) ⊂ * . Therefore 5 is continuous and hence a path in Rl joining x and y.

Conversely, assume that x − y ∉ R∞. Then there are in�nitely many indices 8 ∈ Z+ such thatG8 ≠ ~8 . De�ne ℎ(z) : Rl → Rl by the equation

ℎ(z) = (F1,F2, . . .), where F8 =

{I8 − G8 , if G8 = ~8 ,

8|G8−~8 | (I8 − G8), if G8 ≠ ~8 .

Then ℎ(x) = 0 is bounded. But ℎ(y) is not bounded since G8 ≠ ~8 for in�nitely many indices.Furthermore, ℎ is a homeomorphism of Rl onto itself by Exercise 19.8. Hence x and y do not liein the same component of R∞.

Exercise 25.3. Let � 2o denote the ordered square. If G × ~ ∈ � 2o and * is a neighbourhood ofG ×~, then* contains a basis interval. By Theorem 24.1, since � 2o is a linear continuum, this basisinterval is a connected neighbourhood of G × ~ contained in* . Hence � 2o is locally connected.

Now we prove that � 2o is not locally path-connected. Let 0 < ? < 1 and consider the point? × 1 ∈ � 2o . Let * be a neighbourhood of ? × 0; suppose that * contains a path-connected neigh-bourhood + of ? × 1. Then + contains a point @ × 0, where @ > ?. Similarly as in Example 6of §24, let 5 : [0, 1] → + be a path joining ? × 1 and @ × 0. By the intermediate value theorem.5 ( [0, 1]) contains every point G × ~ between ? × 1 and @ × 0. Thus, for each ? < G < @, theset *G = 5 −1({G} × (0, 1)) is non-empty and open in [0, 1], so we can choose a rational point



@G ∈ *G . Then the map G ↦→ @G is an injective mapping from (?, @) to Q, but the interval (?, @) isuncountable while Q is countable, a contradiction. Therefore, � 2o is not locally path-connected atthe points ? × 1 with 0 < ? < 1. Similarly, � 2o is not locally path-connected at the points @× 0 with0 < @ < 1. Hence the path components are the segments {G} × [0, 1] for G ∈ [0, 1] .

Exercise 25.4. Let* be a open connected set in - . By Theorem 25.4, each path component of*is open in -, hence open in * . Thus, each path component in * is both open and closed in * , somust be empty or all of* . It follows that* is path-connected.

Exercise 25.5. (a) Each line segment joining ? to a point of - is path-connected. Hence ) isunion of path-connected spaces with non-empty intersection, so) is path-connected by Exercise24.8(d).

The only connected open sets in - are singletons, so they are all path-connected.(b) In (a), ) is locally connected precisely at ?: if @ ∈ ) \ {?}, then any small neighbourhood

of @ intersects in�nitely many lines, and has in�nitely many components. We can extend theexample. Let . = Q∩ ([0, 1] × {1}). Let ' be the union of all line segments joining the point 1× 0to points in . . Similarly to (a), ' is path-connected. Then '∪) is path-connected as 1×0 ∈ '∩) .But ' ∪ ) is locally connected at none of its points, as now every small neighbourhood of anypoint, including ? and 1 × 0, has in�nitely many components.

Exercise 25.6. We use the hint. Let * be open in - and � be a component of * . If G ∈ �, thereis a connected subspace �G of - contained in * that contains a neighbourhood +G of G . Since�G is connected, it must lie entirely within the component � of * . Then G ∈ +G ⊂ �G ⊂ �, andtherefore � =

⋃G ∈� +G is open in - . Since � was arbitrary, each component of * is open in - ;

since* was arbitrary, it follows from Theorem 25.3 that - .

Exercise 25.7. We use the hint. We describe the set - . Without any loss of generality, we mayassume that ? is the origin of R2, ? = (0, 0) . There is a sequence {(08 , 0)} of points of R2 that liesin the G-axis, so that 08+1 < 08 for all 8 ∈ Z+, and such that (08 , 0) → ? = (0, 0) when 8 → ∞. Theline segment ! from ? to (01, 0) is in -, so - contains all points (08 , 0) . From each (08 , 0) there areemanating in�nite lines !8, 9 with endpoint (08+1, A 9 ) for some A 9 > 0, and such that A 9+1 < A 9 forall 9 and A 9 → 0 when 9 →∞. The space - is the union of all this lines !8, 9 and !.

First, we prove that - is not locally connected at ?. We show that any connected neighbour-hood, of ? must contain all the points (08 , 0) . Since, is open,, contains all but �nitely many(08 , 0) . Let # ∈ Z+ be such that (08 , 0) ∈ , for all 8 ≥ # and (0#−1, 0) ∉ , . Since, is openand contains (0# , 0), there is a ball � ≔ �3 ((0# , 0), Y) in R2 such that (0# , 0) ∈ � ∩ - ⊂ , .

Then � ∩ - intersects all but �nitely many of the lines !8, 9 . Each of these lines is connected andintersects,, so must lie entirely within, . But then (0#−1, 0) ∈,, contradicting the choice of# . Thus (08 , 0) ∈, for all 8 ∈ Z+. Therefore, if* is a neighbourhood of ? that does not contain all(08 , 0), then there is no connected neighbourhood of ? contained in* . For example, �3 (?, A ) ∩-where A < 3 (?, (01, 0)), is a neighbourhood of ? that contains no connected neighbourhood of ?.Hence - is not locally connected at ?.



Now we prove that - is weakly locally connected at ?. Let* be a neighbourhood of ?. Since* is open in -, there exists an open set + in R2 such that * = - ∩+ . Then + contains an openball � ≔ �3 (?, A ) for some A > 0. Since 08 → 0, � must contain all but �nitely many of the (08 , 0).Let # ∈ Z+ be such that (08 , 0) ∈ !8, 9 ⊂ � for all 8 ≥ # and all 9 ∈ Z+, and such that (0#−1, 0) ∉ �.Then

� =⋃8≥#

�8 , where �8 =⋃9 ∈Z+

(!8, 9 ∪ (! ∩* )

)for each 8,

is a contained in �, hence contained in * . Furthermore, each �8 is path-connected being a unionof path-connected spaces with the common point (08 , 0), so � is path-connected since ? ∈ �8 forall 8 . In particular,� is connected. Moreover,� contains the set �3 (?, X) ∩- for any X < 0# , so�contains a neighbourhood of ?. Since* was arbitrary, - is weakly locally connected at ?.

Exercise 25.8. We follow the hint. Let * be open in . and � be a component of * . We provethat � is open. If � is a component of ?−1(* ) that intersects ?−1(�), then ? (�) is connected andintersects �, so must lie entirely in �. Then � ⊂ ?−1(�) . This proves that ?−1(�) is union ofcomponents of ?−1(* ) . Since - is locally connected and ?−1(* ) is open in -, each component of?−1(* ) is open in - by Theorem 25.3, hence ?−1(�) is open in - . Since ? is an quotient map,� isopen in . . Therefore, each component of * is open in . . It follows from Theorem 25.3 that . islocally connected.

Exercise 25.9. We use the hint. By Exercise 4 of Supplementary Exercises of Chapter 2, givenG ∈ �, the maps ~ ↦→ G~ and ~ ↦→ ~G are homeomorphisms of � onto itself. Since � is acomponent, G� and �G are both components that contain G, so they are equal. Hence G� = �G

for all G ∈ �, so � is a normal subgroup of �.

Exercise 25.10. (a) The relation is clearly re�exive and symmetric. Now suppose G ∼ ~ and~ ∼ I. If G � I, then there exists a separation - = � ∪ � of - into disjoints open sets such thatG ∈ � and I ∈ �. If ~ ∈ �, then ~ � I, in contradicting with the assumption. If ~ ∈ �, then ~ � Gand again we have a contradiction. Thus G ∼ I and the relation is transitive, hence an equivalencerelation.

(b) Let � be a component of - . Let G,~ ∈ � and suppose G � ~. Then there is a separation- = � ∪ � of - into disjoint open sets such that G ∈ � and ~ ∈ �. Then � ∩ � and � ∩ � form aseparation of �, contradicting the fact that � is connected. Hence G ∼ ~ for all G,~ ∈ � and thus� lies within a quasicomponent of - .

Now suppose that - is locally connected. Let � be a component of - . If G ∈ �, there existsa connected neighbourhood * of G in - such that G ∈ * . Since * is connected and � ∩ * ≠ ∅,then G ∈ * ⊂ �. This proves that� is open in - . Since� is connected, we have� = � and hence� is also closed in - . Hence, the components of - are both open and closed in - . Now let & bea quasicomponent of - ; we show that & is connected. If G ∈ &, there exists a component � of -such that G ∈ �. This component must lie within a quasicomponent, so � ⊂ &. If ~ ∈ &, then ~lies in� as well, for if ~ ∈ - \�, then� and - \� would constitute a separation of - into disjointopen sets such that G ∈ � and ~ ∈ - \�. Thus ~ ∈ � for all ~ ∈ &, which proves & = �. Hence



& is connected. This proves that the components and quasicomponents of - are the same if - islocally connected.

(c) First consider � = ( × [0, 1]) ∪ {0 × 0} ∪ {0 × 1}. Let ) = {0 × 0} ∪ {0 × 1}. For each= ∈ Z+, let+= denote the vertical line+= = {1/=} × [0, 1] . We show that the path components andcomponents of � are {0 × 0}, {0 × 1} and each+= for = ∈ Z+. We show that the quasicomponentsof� are each+= and) . If G ∈ +=, ~ ∈ +< and = < <, let C ∈ [0, 1] be such that 1

<< C < 1

<−1 . Then� = � ∪�, where � = ((−∞, C) ×R) ∩� and � = ((C, +∞) ×R) ∩� are disjoint open sets in� suchthat G ∈ � and ~ ∈ �, so G � ~. Similarly, if G ∈ ) and ~ ∈ += for some =, choose 1

<< C < 1

<−1 ;then� = � ∪� where � = ((−∞, C) ×R) ∩� and � = ((C, +∞) ×R) ∩� are disjoint open sets in -such that G ∈ � and ~ ∈ �, so G � ~. So, each quasicomponent of � must lie within) or some+= .Since each+= is path-connected, is a path component, a component and a quasicomponent. Now) splits into two more components (hence path components): {0 × 0} and {0 × 1}. We prove that) is a quasicomponent. Suppose there is a separation � = � ∪ � of � into disjoint open sets in �such that 0 × 0 ∈ � and 0 × 1 ∈ �. Then � and � must intercept some common+=, so � ∩+= and� ∩ += is a separation of +=, contradicting the fact that += is connected. Hence there is no suchseparation, and ) is one single quasicomponent of �.

Now consider� = �∪([0, 1]×{0}). For each= ∈ Z+,+= = {1/=}×[0, 1] intersects ( [0, 1]×{0}).Since these two subspaces are path-connected, their union )= path-connected. Since � \ {0 × 1}equals the union of the )= and 0 × 0 ∈ )= for all =, then � \ {0 × 1} is path-connected, henceconnected. Since the closure of � \ {0 × 1} is �, it follows that � is connected, so it has onlyone component and one quasicomponent, namely itself. We now prove that there are two pathcomponents of �, namely � \ {0 × 1} and {0 × 1}. Suppose that 0 × 1 and 0 × 0 can be joinedby a path in �, so that there is a continuous function 5 : [0, 1] → � such that 5 (0) = 0 × 0 and5 (1) = 0 × 1. Let 51 = c ◦ 5 be the �rst coordinate function of 5 . Then 51 is continuous and51(0) = 0, 51(1) = 1, so there exists C0 ∈ (0, 1) such that 5 (C0) = 1/2 and 51(C) ≥ 1/2 for all C ≥ C0.Let � = � ∩ (R × [1/2, 1]) \ {0 × 1}. Then the restriction 5 ′ : [C0, 1] → � of 5 is continuous, so0 × 1 can be joined by a path in � to the point 5 (C0) ∈ � ⊂ � \ {0 × 1}, which is impossible since{0 × 1} is a path component of �. Therefore, there is no path between 0 × 0 and 0 × 1, and wededuce that � − {0 × 1} and {0 × 1} are the two path components of �.

Finally, consider� = ( ×[0, 1])∪(− ×[−1, 0])∪([0, 1]×− )∪([−1, 0]× ) . For each= ∈ Z+,let *= = [−1, 0] × {1/=}, += = {1/=} × [0, 1], ,= = [0, 1] × {−1/=, }, and /= = {−1/=} × [−1, 0] .Thus � =

⋃= (*= ∪+= ∪,= ∪ /=). The path components of � are each of the sets*=, +=, ,=, /= .

We prove that � is connected. Suppose it is not, so that there is a separation � = � ∪ � of �into two non-empty disjoint open sets. Fix # ∈ Z+ and consider the point 0 × (1/# ) ∈ *# . Wecan assume that 0 × (1/# ) ∈ �. Since *# is (path) connected, *# ⊂ �. Now, since 0 × (1/# ) isa limit point of

⋃=+= and � does not contain any limit point of �, all but �nitely many += are

contained in �. Let + denote the union of those += contained in �. Then each point 0 × (1/=)is a limit point of + , so 0 × (1/=) ∈ � for all =, and hence

⋃=*= ⊂ �. Similarly, each point

(−1/=) × 0 ∈ /= is a limit point of⋃=*=, so

⋃= /= ⊂ �. Continuing this argument, it follows

that � =⋃= (*= ∪ += ∪,= ∪ /=) ⊂ �, contradicting the fact that � is non-empty. Thus � is


26 Compact Spaces

connected, so it has only one component and one quasicomponent, namely itself.

26 Compact Spaces

Exercise 26.1. (a) If (-,� ′) is compact, then (-,�) is compact. The converse is not true ingeneral.

(b) Suppose that � ⊂ �′. Let � ∈ �

′. Then - \ � is closed in (-,� ′), so it is compact byTheorem 26.2. Since � ⊂ �

′, by (a) we have that - \ � is compact in (-,�), hence closed in(-,�) by Theorem 26.3. Thus � ∈ �, and hence � = �

′. This proves the claim.

Exercise 26.2. (a) Let � be a subspace of R in the �nite complement topology, and let � be acovering of � by open sets in R. Fix � ∈ �. Then R \� is �nite, say R \� = {21, . . . , 2=}. For each8 = 1, . . . , = such that 28 ∈ �, choose an element of � containing 28 . Then these points 28 and �constitute a �nite subcollection of � that also covers �, so � is compact by Lemma 26.1. Since �was arbitrary, it follows that every subspace of R in the �nite complement topology is compact.

(b) We show that [0, 1] is not compact in this topology. For each= ∈ Z+, let �= = {1/: | : ≥ =},and let �= = [0, 1] \ �= . Note that, for each =, �= ⊂ �=+1. The collection � = {�= | = ∈ Z+} is anopen covering of [0, 1]; suppose that � contains a �nite subcollection that covers [0, 1] . Let # bethe greatest positive number such that �# belongs to this subcollection. Then 1/# ∉ �# , so noset of this subcollection contains 1/#, contradicting the fact that this subcollection covers [0, 1] .Therefore � is an open covering of [0, 1] with no �nite subcovering, so [0, 1] is not compact inthis topology.

Exercise 26.3. Let .1, . . . , .= be compact subspaces of - . Let � be a covering of⋃=8=1 .8 by sets

open in - . For each 8 = 1, . . . , =,� is a covering of .8 by sets open in -, so by Theorem 26.1, thereis a �nite subcollection �8 ≔ {�1

8 , . . . , �=88} of � covering .8 . Then

⋃8 �8 is a �nite subcollection

of � covering⋃=8=1 .8 , so

⋃=8=1 .8 is compact by Theorem 26.1.

Exercise 26.4. Let . be a compact subspace of a metric space with metric 3. Since any metricspace is Hausdor�, . is closed by Theorem 26.3. Fix ~ ∈ . . For each = ∈ Z+, let �= ≔ �3 (~, =) bethe open ball centred at ~ having radius =. Then the collection � = {�=}=∈Z+ is an open coveringof ., so there is a �nite subcollection of � covering . . If # is the greatest positive number suchthat �# belongs to this subcollection, then . ⊂ �# , so . is bounded.

Now, let - be an in�nite space with the discrete topology. Then - is a metric space with thediscrete metric 3. Now - is closed in itself, and it is bounded since - ⊂ �3 (G, 2) for any G ∈ - .But - is not compact, for the collection of singletons {G}, G ∈ -, constitute an open coveringwith no �nite subcovering.

Exercise 26.5. For each G ∈ �, by Lemma 26.4 there exist disjoint open sets *G and +G of -containing G and �, respectively. Then {*G }G ∈� is an open covering of�, so there exist G1, . . . , G= ∈� such that {*G8 }=8=1 also covers �. The corresponding +G8 are open in - and contain �, so + =⋂=8=1+G8 is open in - and contains �. Let * =

⋃=8=1*G8 . Then * and + are open sets containing


26 Compact Spaces

� and � respectively. Furthermore, they are disjoint, for if 2 ∈ * ∩+ , then 2 ∈ *G8 ∩+G8 = ∅ forsome 8, a contradiction.

Exercise 26.6. Let � be closed in - . Then � is compact by Theorem 26.2, so 5 (�) is compact byTheorem 26.5. Thus 5 (�) is closed by Theorem 26.3, so 5 is a closed map.

Exercise 26.7. Let � be closed in - × . . Let G0 ∈ - \ c1(�) . Then the slice G0 × . of - × . iscontained in the open set (- ×. ) \� of- ×. . Thus, by the tube lemma (Lemma 26.8), (- ×. ) \�contains some tube, × . about G0 × ., where, is a neighbourhood of G0 in - . If G ∈,, thenG×~ ∈, ×. for all~ ∈ ., so G×~ ∉ � for all~ ∈ . and hence G ∉ c1(�). Then G0 ∈, ⊂ - \c1(�),so - \ c1(�) is open in - and c1(�) is closed in - . Since � was arbitrary, it follows that c1 is aclosed map.

Exercise 26.8. First suppose that 5 is continuous. We prove that (- × . ) \� 5 is open in - × . .Let G0 × ~0 ∈ (- × . ) \ � 5 , so that 5 (G0) ≠ ~0. Since . is Hausdor�, there exist disjoint opensets * and + in . such that 5 (G0) ∈ * and ~0 ∈ + . Then 5 −1(* ) × + is a neighbourhood ofG0 × ~0. Furthermore, it does not intercept � 5 , for if G × ~ ∈ (5 −1(* ) × + ) ∩ � 5 , then 5 (G) = ~and 5 (G) ×~ ∈ * ∩+ = ∅, absurd. Thus G0 ×~0 ∈ 5 −1(* ) ×+ ⊂ (- ×. ) \� 5 , so (- ×. ) \� 5 isopen in - × . and hence � 5 is closed in - × . .

Conversely, assume that � 5 is closed in - × . . We use the hint. Let G0 ∈ - and let + be aneighbourhood of 5 (G0) . Then � ≔ � 5 ∩ (- × (. \ + )) is closed in - × ., so by Exercise 26.7,since . is compact, c1(�) is closed in - . We claim that * ≔ - \ c1(�) is a neighbourhood of G0such that 5 (* ) ⊂ + . Indeed, G0 ∈ * since 5 (G0) ∉ . \+ . Let G ∈ * and suppose that 5 (G) ∉ + .Then G × 5 (G) ∈ � 5 ∩ (- × (. \+ )) = �, so c1(G × 5 (G)) = G ∈ c1(�), contradicting the fact thatG ∈ * . Therefore,* is a neighbourhood of G0 such that 5 (* ) ⊂ + , so 5 is continuous.

Exercise 26.9. Let 0 ∈ � and consider the slice 0 × �. Then 0 × � ⊂ # . Cover 0 × � by standardbasis elements, × / for - × . lying in # . Since 0 × � is compact, there are �nitely many suchbasis elements,1 × /1, . . . ,,= × /= that also cover 0 × �, such that,8 × /8 actually intersects0 × � for each 8 . Let

*0 =

=⋂8=1,8 and +0 =

=⋃8=1

/8 .

Then *0 is a neighbourhood of 0 and +0 is open in . and contains �. Furthermore, *0 ×+0 ⊂ # .Indeed, if G × ~ ∈ *0 × +0, then there exists 8 ∈ {1, . . . , =} such that ~ ∈ /8 , and hence G × ~ ∈,8×/8 ⊂ # . Now, each*0×+0 is open in- ×. , so the collection {*0×+0}0∈� is an open coveringof�×�. Since�×� is compact, a �nite subcollection also covers�×�, say*01×+01, . . . ,*0: ×+0: .Let

* =

:⋃8=1*0: and + =

:⋂8=1+0: ,

so that* is open in- and+ is open in. .We claim that�×� ⊂ * ×+ ⊂ # . Indeed, if 0×1 ∈ �×�,then 0 × 1 ∈ *08 ×+08 for some 8 = 1, . . . , :, so 0 ∈ * . Since each +0 contains �, we have � ⊂ + ,


26 Compact Spaces

so 1 ∈ + . Thus 0 × 1 ∈ * × + . Finally, if G × ~ ∈ * × + , then ~ ∈ +08 for some 8 = 1, . . . , :, soG × ~ ∈ *08 ×+08 ⊂ # .

Exercise 26.10. (a) Note that 5= (G) ≤ 5 (G) for all = ∈ Z+ and all G ∈ - . Let Y > 0. For each G ∈ -,there exists #G ∈ Z+ such that 5 (G) − 5= (G) < Y for all = ≥ #G . Since 5 and 5= are continuous,there exists a neighbourhood*G of G such that 5 (I) − 5= (I) < Y for all I ∈ *G . Now {*G }G ∈- is anopen covering of -, so there exist G1, . . . , G= ∈ - such that the �nite subcollection {*G8 }=8=1 alsocovers - . Let # = max{#G1, . . . , #G= }. Given G ∈ -, there exists 8 ∈ {1, . . . , =} such that G ∈ *G8 ;since # ≥ #G8 , we have 5 (G) − 5= (G) < Y for all = ≥ # . Thus 5 (G) − 5= (G) < Y for all = ≥ # andall G ∈ -, so 5= converges uniformly to 5 .

(b) In Exercise 21.9 we proved that the sequence of functions 5= : R→ R given by

5= (G) =1

=3 [G − (1/=)]2 + 1

converges to the zero function, but not uniformly as the sequence G= = 1/= converges to G = 0but 5= (G=) = 1 for all =. We can restrict its domain to any compact interval [0, 1], and the conver-gence is still not uniform. So the theorem fails if we delete the requirement that the sequence bemonotone.

Now consider 5= : R → R given by 5= (G) = arctan(G + =) . The sequence (5=)= convergespoint-wise to the constant (hence continuous) function 5 ≡ c/2, and the sequence is monotone.Since 5= (−=) = 0 for all =, the convergence is not uniform. So the theorem fails if we delete therequirement that - be compact.

Exercise 26.11. We use the hint. Since each � ∈ � is closed, . is closed. Suppose that � and �form a separation of . . Then � and � are closed in ., hence closed in - . Since - is compact, �and � are compact by Theorem 26.2. Since - is Hausdor�, by Exercise 26.5, there exist * and +open in - and disjoint containing � and �, respectively. We show that⋂

�∈�(� \ (* ∪+ ))

is not empty. Let {�1, . . . , �=} be a �nite subcollection of elements of �. We may assume that�8 ( �8+1 for all 8 = 1, . . . , = − 1. Then

=⋂8=1(�8 \ (* ∪+ )) = �1 \ (* ∪+ ) .

Suppose that �1 \ (* ∪+ ) = ∅. Then �1 ⊂ * ∪+ . Since �1 is connected and * ∩+ = ∅, �1 lieswithin either* or+ , say�1 ⊂ * . Then . ⊂ �1 ⊂ * , so that� = . ∩� ⊂ . ∩+ = ∅, contradictingthe fact that� and � form a separation of . . Hence,

⋂=8=1(�8 \ (* ∪+ )) is non-empty. Therefore,

the collection {� \ (* ∪+ ) | � ∈ �} has the �nite intersection property, so⋂�∈�(� \ (* ∪+ )) =

( ⋂�∈�

�

)\ (* ∪+ ) = . \ (* ∪+ )


26 Compact Spaces

is non-empty by Theorem 26.9. So there exists ~ ∈ . such that ~ ∉ * ∪+ ⊂� ∪ �, contradictingthe fact that� and � form a separation of . . We conclude that there is no such separation, so that. is connected.

Exercise 26.12. We use the hint. We �rst show that if* is an open set containing ?−1({~}), thenthere is a neighbourhood, of ~ such that ?−1(, ) is contained in* . Since - −* is closed in -,? (- − * ) is closed in . and does not contain ~, so, = . \ ? (- \ * ) is a neighbourhood of ~.Moreover, since - \* ⊂ ?−1(? (- \* )) (by elementary set theory), we have

?−1(, ) = ?−1(. \ ? (- \* )) = ?−1(. ) \ ?−1(? (- \* )) ⊂ - \ (- \* ) = * .

Now let � be an open covering of - . For each ~ ∈ ., let �~ be a subcollection of � such that

?−1({~}) ⊂⋃�∈�~

�.

Since ?−1({~}) is compact, there exists a �nite subcollection of �~ that also covers ?−1({~}), say{�1

~, . . . , �=~~ }. Thus

⋃=~

8=1�8~ is open and contains ?−1({~}), so there exists a neighbourhood,~

of ~ such that ?−1(,~) is contained in⋃=~

8=1�8~ . Then {,~}~∈. is an open covering of ., so there

exist ~1, . . . , ~: ∈ . such that {,~ 9 }:9=1 also covers . . Then

- = ?−1(. ) ⊂ ?−1( :⋃9=1,~ 9

)=

:⋃9=1?−1(,~ 9 ) ⊂

:⋃9=1

( =~9⋃8=1

�8~ 9

),

so {�8~ 9

}9=1,...,: .8=1,...,=~9 .

is a �nite subcollection of � that also covers - . Therefore, - is compact.

Exercise 26.13. (a) We follow the hint. Let 2 ∈ � not in � · �. Let 1 ∈ �. Then 2 · 1−1 ∉ �. BySupplementary Exercise 7(c) of Chapter 2, there exists a neighbourhood* of 2 ·1−1 disjoint from�. Then, = * · � is a neighbourhood of 2 disjoint from � · �. It follows that � · � is closed in�.

(b) Let� be closed in�. Let G ∈ � be such that G� ∉ ? (�) . Then G ∉ � ·�, which in closed in� by compactness of � and part (a), so there exists a neighbourhood * of G disjoint from � · �.By Supplementary Exercise 5(c) of Chapter 2, ? is an open map, so ? (* ) is a neighbourhood ofG� disjoint from ? (�). It follows that ? is a closed map.

(c) The projection ? : � → �/� is continuous and surjective, and it is closed by part (b). Forevery G ∈ �, the subspace ?−1(G� ) = G� is compact, being homeomorphic to � (SupplementaryExercise 4 of Chapter 2). We conclude by Exercise 26.12 that � is compact.


27 Compact Subspaces of the Real Line


Exercise 27.1. Let � be a subset of - bounded above. Then � is closed, hence compact byassumption. Applying Theorem 27.4 to the inclusion ] : �→ -, there exists 1 ∈ � such that 0 ≤ 1for all 0 ∈ �. We prove that 1 is the least upper bound of �. Indeed, since � ⊂ �, we have 0 ≤ 1for all 0 ∈ �. Suppose that there exists 2 ∈ - such that 2 < 1 and 0 ≤ 2 for all 0 ∈ �. Then 2 ∈ �,for if (2 − Y, 2 + Y) is a neighbourhood of 2 that does not intercept �, then the neighbourhood(1 − 2 − Y, 1 + Y) of 1 does not intercept � either, contradicting the fact that 1 ∈ �. Thus 2 ∈ �, so2 ≤ 1, contrary to the assumption. Hence 1 is the least upper bound of �. It follows that - hasthe least upper bound property.

Exercise 27.2. (a) If 3 (G,�) = X > 0, then �3 (G, X) ∩� = ∅, so G ∉ �. Conversely, if G ∉ �, thereexists X > 0 such that �3 (G, X) ∩� = ∅, and hence 3 (G,�) > X/2 > 0.

(b) The distance function 3 : � ×�→ R is continuous by Exercise 20.3(a). Thus, given G ∈ -,the function 3G : � → R given by 3G (0) = 3 (G, 0) is continuous by Exercise 19.11. Since � iscompact, by the extreme value theorem (Theorem 27.4), there exists 0 ∈ � such that 3G (0) =3 (G, 0) ≤ 3 (G, 0′) for all 0′ ∈ �. Then 3 (G,�) = 3 (G, 0) .

(c) Let ~ ∈ * (�, Y) . Then 3 (~,�) = inf{3 (~, 0) | 0 ∈ �} < Y, so there exists 0 ∈ � suchthat 3 (~, 0) < Y, so that ~ ∈ �3 (0, Y). Thus * (�, Y) ⊂ ⋃

0∈� �3 (0, Y) . Conversely, let 0 ∈ � andG ∈ �3 (0, Y) . Then 3 (G,�) ≤ 3 (G, 0) < Y, so G ∈ * (�, Y) . It follows that* (�, Y) = ⋃

0∈� �3 (0, Y) .(d) For each G ∈ �, there exists YG > 0 such that �3 (G, YG ) ⊂ * . Then {�3 (G, YG )}G ∈� is an open

covering of �, so there exist G1, . . . , G= ∈ � such that {�3 (G8 , YG8 )}=8=1 also covers �. Let X > 0 bethe Lebesgue number for this �nite covering, and let Y = X/2. Then for any 0 ∈ � the ball �3 (0, Y)has diameter less than X, so there exists 8 ∈ {1, . . . , =} such that �3 (0, Y) ⊂ �3 (G8 , YG8 ) . Therefore

* (�, Y) =⋃0∈�

�3 (0, Y) ⊂=⋃8=1

�3 (G8 , YG8 ) ⊂ * .

(e) Let � = {G × tanG | −c2 < G < c2 }. Then � is closed in - = R2, but is not compact (as not

bounded). Let * = (−c2 ,c2 ) × R. Then * is open in R2, but for any Y > 0, there exists 2 ∈ � such

that �3 (2, Y) ∩ (R2 −* ) ≠ ∅. So no Y-neighbourhood of � is contained in* .

Exercise 27.3. (a) The -topology and the standard topology on R are both Hausdor�. Further-more, [0, 1] is compact in the standard topology. If [0, 1] were compact in R , these topologieson [0, 1] would be equal by Exercise 26.1(b), but the -topology is strictly �ner than the standardtopology. Thus [0, 1] is not compact as a subspace of R .

(b) We use the hint. Clearly (−∞, 0) inherits its usual topology as subspace of R . Now, if+ is a standard basis element for (0, +∞) as a subspace of R , then + = * ∩ (0, +∞) for * astandard basis element for R . If * = (0, 1), then clearly + is open in (0, +∞) with the standardtopology. Let denote the closure of in the standard topology. If * = (0, 1) − , then + =

(0, 1) ∩ (0, +∞) − = (0, 1) ∩ (0, +∞) − , so + is open in (0, +∞) with the standard topology



as well. Thus (0, +∞) inherits its usual topology as a subspace of R . Hence (−∞, 0) and (0, +∞)are connected as subspaces of R , so their closures (−∞, 0] and [0, +∞) are connected and havethe common point 0. It follows from Theorem 23.3 that (−∞, 0] ∪ [0, +∞) = R is connected.

(c) Let R denote the real numbers with the standard topology. Suppose that there is a pathfrom 0 to 1 in R , that is, a continuous function 5 : [0, 1] → R such that 5 (0) = 0 and 5 (1) = 1.Since R is �ner than R, 5 is continuous when considered as a function to R. By the intermediatevalue theorem (Theorem 24.3), the interval [0, 1] is contained in 5 ( [0, 1]) . Since [0, 1] is compactin R and 5 is continuous, it follows that 5 ( [0, 1]) is compact in R . Thus [0, 1] is closed in R andcontained in the compact space 5 ( [0, 1]), so [0, 1] is compact in R , contradicting (a). We deducethat there is no path from 0 to 1 in R and hence R is not path-connected.

Exercise 27.4. The distance function 3 : - × - → R is continuous by Exercise 20.3(a), so givenG ∈ -, the function 3G : - → R given by 3G (~) = 3 (G,~) is continuous by Exercise 19.11. Since- is connected, the image 3G (- ) is a connected subspace of R, and contains 0 since 3G (G) = 0.Thus, if ~ ∈ - and ~ ≠ G, then 3G (- ) contains the set [0, X], where X = 3G (~) > 0. Therefore -must be uncountable.

Exercise 27.5. We follow the hint. Let * be any non-empty open set in - . We prove that * hasa point that is not in

⋃= �= . Since �1 has empty interior, we have* ⊄ �1. We show that there is

a non-empty open set +1 in - such that +1 ⊂ * \ �1. Indeed, let G ∈ * \ �1. Then �1 ∪ (- \* )is closed and does not contain G . Since - is compact, �1 ∪ (- \* ) is compact by Theorem 26.2.Thus, by Lemma 26.4, there exist disjoint open sets+1 and, of - containing G and �1 ∪ (- \* )respectively. Hence G ∈ +1 ⊂ +1 ⊂ - \ (�1 ∪ (- \ * )) = * \ �1. Similarly, +1 is a non-emptyopen set in - not contained in �2, so there exists a non-empty open set+2 such that+2 ⊂ +1 \�2.

Inductively we obtain a nested sequence

+1 ⊂+1 ⊂· · ·

of non-empty closed sets of - . Since - is compact,⋂∞8=1+8 is non-empty by Theorem 26.9. So

there exists G ∈ ⋂∞8=1+8 ⊂ +1 ⊂ * , and G ∉ �8 for all 8 ∈ Z+. Thus * has a point that is not in⋃

= �= . Since* was arbitrary, we deduce that⋃= �= has empty interior.

Exercise 27.6. (a) Let � be a connected subspace of �. Suppose that � contains two di�erentpoints G ≠ ~. By the Archimedean property, there exists = ∈ Z+ such that 1/3= < |G − ~ |. ThusG and ~ belong to di�erent closed intervals in �= . Let � ⊂ �= be the interval containing G . Then� ∩ � and � ∩ (�� ) form a separation of �, contradicting the fact that � is connected. It followsthat � is totally disconnected.

(b) Each �= is closed in [0, 1], so � is an intersection of closed sets, hence closed. Thus � is aclosed subspace of the compact space [0, 1], so � is compact by Theorem 26.2

(c) We use induction. Clearly �1 = [0, 13 ] ∪ [23 , 3] satis�es this property. At each step, each

closed interval of length 1/3= of �= is divided in three closed intervals of length 1/3=+1, and theinterior of the middle interval is removed, leaving in �=+1 two closed intervals of length 1/3=+1.


28 Limit Point Compactness

At each step only the interior of closed intervals are removed, so the endpoints of the intervalsare never removed and they lie in �.

(d) Let G ∈ �. Then G ∈ �= for each =, so there exist intervals �= ⊂ �= containing G . For eachinterval �=, we can choose an endpoint G= of the interval not equal to G . These endpoints G= arein � by (c), and G= → G . So G is not an isolated point of - . Since G was arbitrary, we deduce that� has no isolated points.

(e)� is non-empty compact Hausdor� with no isolated points, hence uncountable by Theorem27.7.


Exercise 28.1. Consider the subset . = {0, 1}l ⊂ [0, 1]l of sequences whose entries are 0’s and1’s. . is clearly in�nite. We show that. has no limit point in [0, 1]l . Let x ∈ [0, 1]l . If x ∉ ., thenthere exists 8 ∈ Z+ such that G8 ≠ 0, 1. Let X = min{1 − G8 , G8}. Then �d (x, X) is a neighbourhoodof x for which �d (x, X) ∩ . = ∅, so x ∉ . ′. Now, if x ∈ . and y ∈ . with x ≠ y, then thereexists 8 ∈ Z+ such that |G8 − ~8 | = 1. Thus, if x ∈ . and Y < 1, we have �d (x, X) ∩ . = {x} andhence x ∉ . ′. It follows that . = {0, 1}l is an in�nite subset of [0, 1]l that has no limit point, asclaimed.

Exercise 28.2. Let � = {1 − 1/= | = ∈ Z+}. Then � is in�nite. Clearly any point G ∈ [0, 1) is nota limit point of �: if 1 − 1/= ≤ G < 1 − 1/(= + 1), then [G, 1 − 1/(= + 1)) is a neighbourhood of Gwith no point in � \ {G}. Furthermore, if G = 1, then {1} = [1, 2) ∩ [0, 1] is a neighbourhood of Gwith no point in � \ {G}. Hence � is an in�nite subset of [0, 1] with no limit point, so [0, 1] is notlimit point compact as a subspace of Rℓ .

Exercise 28.3. (a) No. Consider- = Z+∪. of Example 1: . is a set of two points in the indiscretetopology. Then - is limit point compact and c1 : - → Z+ is continuous, but c1(- ) = Z+ is notlimit point compact.

(b) Yes. If� is closed, then� contains all its limit points. If � ⊂ � is in�nite, then � has a limitpoint G in - . Then G is also a limit point of �, so G ∈ �′ ⊂ �. Therefore, every in�nite subset of� has a limit point in �, so � is limit point compact.

(c) No. Consider (Ω in the order topology. It is Hausdor� by Exercise 17.10. By Example 2,(Ω is a limit point compact subspace of (Ω, but is is not closed since it does not contain its limitpoint Ω.

Exercise 28.4. First let - be a countable compact space. Note that if . is a closed subset of-, then . is countable compact as well, for if {*=}=∈Z+ is a countable open covering of ., then{*=}=∈Z+ ∪ (- \ . ) is a countable open covering of - ; there is a �nite subcovering of -, hencea �nite subcovering of . . Now let � be an in�nite subset. We show that � has a limit point. Let� be a countable in�nite subset of �. Suppose that � has no limit point, so that � is closed in - .Then � is countable compact. Since � has no limit point, for each 1 ∈ � there is a neighbourhood*1 of 1 that intersects � in the point 1 alone. Then {*1}1∈� is an open covering of � with no �nite



subcovering, contradicting the fact that � is countable compact. Hence � has a limit point, so that� has a limit point as well. Since � was arbitrary, we deduce that - is limit point compact. (Notethat the )1 property is not necessary in this direction.)

Now assume that - is a limit point compact )1 space. We show that - is countable compact.Suppose, on the contrary, that {*=}=∈Z+ is a countable open covering of - with no �nite sub-covering. For each =, take a point G= in - not in *1 ∪ · · · ∪ *= . By assumption, the in�nite set� = {G= | = ∈ Z+} has a limit point ~ ∈ - . Since {*=}=∈Z+ covers -, there exists # ∈ Z+ such that~ ∈ *1 ∪ · · · ∪*# . Now - is )1, so for each 8 = 1, . . . , # there exists a neighbourhood +8 of ~ thatdoes not contain G8 (see Exercise 17.15). Then

+ = (+1 ∩ · · · ∩+# ) ∩ (*1 ∪ · · · ∪*# )

is a neighbourhood of~ that does not contain any of the points G8 , contradicting the fact that~ is alimit point of�. It follows that every countable open covering of- must have a �nite subcovering,so - is countable compact.

Exercise 28.5. We could imitate the proof of Theorem 26.9, but we prove directly each direction.First let - be countable compact and let �1 ⊂�2 ⊂· · · be a nested sequence of closed non-

empty sets of - . For each = ∈ Z+, *= = - \ �= is open in - . Then {*=}=∈Z+ is a countablecollection of open sets with no �nite subcollection covering-, for if*81 ∪ · · · ∪*1= covers-, then�81 ∩ · · · ∩�8= is empty, contrary to the assumption. Hence {*=}=∈Z+ does not cover -, so thereexist G ∈ - \⋃=∈Z+ *= =

⋂=∈/+ (- \*=) =

⋂=∈/+ �= .

Conversely, assume that every nested sequence�1 ⊂�2 ⊂· · · of closed non-empty sets of -has a non-empty intersection and let {*=}=∈Z+ be a countable open covering of - . For each =,let += = *1 ∪ · · · ∪*= and �= = - \ += . Suppose that no �nite subcollection of {*=}=∈Z+ covers- . Then each �= is non-empty, so �1 ⊂�2 ⊂· · · is a nested sequence of non-empty closed setsand

⋂=∈Z+ �= is non-empty by assumption. Then there exists G ∈ ⋂

=∈Z+ �=, so that G ∉ += forall =, contradicting the fact that {*=}=∈Z+ covers - . It follows that there exists # ∈ Z+ such that�# = ∅, so that - = +# and hence some �nite subcollection of {*=}=∈Z+ covers - . We deducethat - is countable compact.

Exercise 28.6. We follow the hint. Note that 5 is an imbedding by Exercise 21.2. It remains toprove that 5 is surjective. Suppose it is not, and let 0 ∈ 5 (- ) . Since- is compact, 5 (- ) is compactand hence closed (every metric space is Hausdor�). Thus, there exists Y > 0 such that the Y-neighbourhood of 0 is contained in - \ 5 (- ) . Set G1 = 0, and inductively G=+1 = 5 (G=) for = ∈ Z+.We show that 3 (G=, G<) ≥ Y for = ≠<. Indeed, we may assume = < <. If = ≥ 1, then 3 (G=, G<) =3 (5 −1(G=), 5 −1(G<)) = 3 (G=−1, G<−1). By induction it follows that 3 (G=, G<) = 3 (G=−8 , G<−8) forall 8 ≥ 1, and hence 3 (G=, G<) = 3 (0, G<−=) = 3 (0, 5 (G<−=−1)) . Since 5 (G<−=−1) ∈ 5 (- ) and�(0, Y) ∩ 5 (- ) = ∅, we have 3 (G=, G<) ≥ Y, as claimed. Thus {G=}=∈Z+ is a sequence with noconvergent subsequence, so - is not sequentially compact. By Theorem 28.2, this contradicts thefact that - is compact. Therefore 5 is surjective and hence a homeomorphism.



Exercise 28.7. First we prove (b), which implies (a).(b) We use the hint. For each = ∈ Z+, let�= = 5 = (- ) . Since - is compact, each�= is compact,

hence closed. Thus � =⋂=∈Z+ �= is closed. Moreover, note that �=+1 ⊂ �= for all =, so � is

non-empty by Theorem 26.9. Now we prove that 5 (�) = �. Let ~ ∈ 5 (�) . There exists G ∈ �such that ~ = 5 (G) . Let = ∈ Z+ be such that G ∈ �= . If = = 1, then ~ ∈ 5 (- ) = �1 because G ∈ - .If = > 1, then ~ ∈ 5 = (- ) = �= because G ∈ �=−1. Thus ~ ∈ �= for all =, so that ~ ∈ � and hence5 (�) ⊂ �. Now let G ∈ �. Then G ∈ 5 = (- ) for all =, so there exists a sequence {G=}=∈Z+ suchthat 5 =+1(G=) = G for each =. Let ~= = 5 = (G=), = ∈ Z+. Since - is a compact metric space, it issequentially compact by Theorem 28.2, so {~=}=∈Z+ has a convergent subsequence, say {I=}=∈Z+with limit 0 ∈ - . Then I= → 0 and I= ∈ � for all =, so 0 ∈ � by the sequence lemma (Lemma21.2). Since � is closed, we have 0 ∈ �; since 5 is continuous and 5 (I=) = G for all =, we have5 (0) = G . Thus G ∈ 5 (�) and hence � ⊂ 5 (�) . Therefore � = 5 (�). Moreover, � has only onepoint. Indeed, suppose that� has more than one point. Since the distance function 3 : �×�→ R

is continuous (Exercise 20.3(a)), it follows from the extreme value theorem (Theorem 27.4) thatthere exists G ≠ ~ in � such that 3 (G ′, ~ ′) ≤ 3 (G,~) for all G ′, ~ ′ ∈ �. Let G = 5 (0) and ~ = 5 (1)with 0, 1 ∈ �. Then 3 (G,~) = 3 (5 (0), 5 (1)) < 3 (0, 1) ≤ 3 (G,~), a contradiction. Hence� containsonly one point, and is a �xed point of 5 . Now 5 has only one �xed point, for if G and ~ are twodi�erent �xed points of 5 , then 3 (G,~) = 3 (5 (G), 5 (~)) < 3 (G,~), absurd. Thus the point in � isthe only �xed point of 5 .

(c) Note that|5 (G) − 5 (~) | = |G − ~ |

��1 − G + ~2 ��for all G,~ ∈ - . Since |1 − G+~

2 | < 1 for all G,~ ∈ - with G ≠ ~, 5 is a shrinking map. We showthat 5 is not a contraction. Suppose there exists U < 1 such that

|5 (G) − 5 (~) | ≤ U |G − ~ |

for all G,~ ∈ - . Note that|5 (G) − 5 (0) | = (G − 0)

(1 − G

2

)for all G ∈ - . Thus, taking G ≠ 0 such that U < 1 − G

2 , we see that the condition for being acontraction is not satis�ed.

(d) Note that(G2 + 1)1/2 − (~2 + 1)1/2

G − ~ =G + ~

(G2 + 1)1/2 + (~2 + 1)1/2


29 Local Compactness

for all G,~ ∈ R, G ≠ ~. Thus

|5 (G) − 5 (~) | = |G − ~ |2

��1 + (G2 + 1)1/2 − (~2 + 1)1/2G − ~

��=|G − ~ |

2

��1 + G + ~(G2 + 1)1/2 + (~2 + 1)1/2

��<|G − ~ |

2+ |G − ~ |

2

��12 + 12

�� = |G − ~ |,for all G,~ ∈ R, G ≠ ~, so 5 is a shrinking map. Now suppose that there exists U < 1 such that��5 (G) − 1

2

�� = |5 (G) − 5 (0) | ≤ U |G |for all G ∈ R. Since 5 is strictly increasing, we have 5 (G) > 1/2 for all G > 0. Thus

5 (G) − 12= |5 (G) − 5 (0) | = |G |

2

��1 + G

(G2 + 1)1/2

�� = G

2

(1 + 1(1 + 1

G2)1/2

)for G > 0. So, if G > 0 satis�es

12

(1 + 1(1 + 1

G2)1/2

)> U,

then the condition for being a contraction is not satis�ed, and we can �nd such G since the aboveexpression tends to 1 as G → ∞. So 5 is not a contraction. Furthermore, it has no �xed point,since

5 (G) = G + (G2 + 1)1/22

>G + |G |

2> G

for all G ∈ R.


Exercise 29.1. First, we prove that each set Q ∩ [0, 1], where 0, 1 are irrational numbers, is notcompact. Indeed, sinceQ∩[0, 1] is countable, we can writeQ∩[0, 1] = {@1, @2, . . .}. Then {*8}8∈Z+,where*8 = Q∩ [0, @8) for each 8, is an open covering ofQ∩ [0, 1] with no �nite subcovering. Nowlet G ∈ Q and suppose thatQ is locally compact at G . Then there exists a compact set� containinga neighbourhood* of G . Then* contains a set Q∩ [0, 1] where 0, 1 are irrational numbers. Sincethis set is closed and contained in the compact�, it follows Q∩ [0, 1] is compact, a contradiction.Therefore, Q is not locally compact.

Exercise 29.2. (a) Recall that the projections cV :∏U -U → -V are continuous open maps.

Assume that∏U -U is locally compact. Let x ∈ ∏

U -U . Then there exist a compact subspace� of

∏U -U containing a neighbourhood of x. This neighbourhood contains a standard basis



element for∏U -U containing x, say* =

∏U *U where*U = -U for all but �nitely many indices,

say U1, . . . , U= . Let V ≠ U8 for all 8 = 1, . . . , =. Then cV (�) is compact and contains cV (* ) = -V ,

so cV (�) = -V and -V is compact. Hence -U is compact for all but �nitely many values of U.It remains to prove that -U8 is locally compact for 8 = 1, . . . , =. Let G ∈ -U8 . Let x ∈ ∏

U -U besuch that GU8 = G . There is a compact � containing a basis neighbourhood * of x. Then cU8 (�)is a compact subspace of -U8 containing the neighbourhood cU8 (* ) of G . Thus the spaces -U8 , for8 = 1, . . . , =, are indeed locally compact.

(b) Suppose that each -U is locally compact and -U is compact for all but �nitely many U. Letx ∈ ∏

U -U . For each U, there exists a compact subspace�U of -U containing a neighbourhood*Uof GU . By assumption, for all but �nitely many indices we may assume that�U = *U = -U . By theTychono� theorem,

∏U �U is compact, and it contains the neighbourhood

∏U *U of x. It follows

that∏U -U is locally compact.

Exercise 29.3. If 5 is continuous but not open, 5 (- ) is not necessarily locally compact. Indeed,consider Q3 , the rational numbers having the discrete topology, and let Q denote the rationalnumbers in the usual topology. Then Q3 is locally compact, for if G ∈ Q3 , then {G} is open andcompact. Let 8 : Q3 → Q be the identity map. Then 8 is continuous, but 8 (Q3 ) = Q is not locallycompact by Exercise 29.1.

Now, if 5 is continuous and open, then 5 (- ) is locally compact. Indeed, let ~ ∈ 5 (- ), so that~ = 5 (G) for some G ∈ - . Since - is locally compact, there exists a compact subspace � of -containing a neighbourhood * of G . Since 5 is continuous, 5 (�) is compact and since 5 is open,5 (* ) is open. Thus 5 (�) is compact and contains the neighbourhood 5 (* ) of~, so 5 (- ) is locallycompact.

Exercise 29.4. Consider 0 ∈ [0, 1]l and suppose that [0, 1]l is locally compact at 0. Then thereexists a compact � containing an open ball � = �d (0, Y) ⊂ [0, 1]l . Note that � = [0, Y]l . Then[0, Y]l is closed and contained in the compact�, so it is compact. But [0, Y]l is homeomorphic to[0, 1]l ,which is not compact by Exercise 28.1. This contradiction proves that [0, 1]l is not locallycompact in the uniform topology.

Exercise 29.5. Let .1 = -1 ∪ {∞1} and .2 = -2 ∪ {∞2} be the respective one-point compacti�c-ations. De�ne 6 : .1 → .2 by 6|-1 = 5 and 6(∞1) = ∞2. Then 6 is bijective. By symmetry of theproblem, it su�ces to prove that 6 is open. Let * be an open set of .1. If * is open in -1, then6(* ) = 5 (* ) is open in -1, hence in .1. If * = -1 \�, where � is a compact subspace of -, then6(* ) = 6(.1) \ 6(�) = .2 \ 6(�) . Since 5 is continuous and � ⊂ -, it follows 6(�) = 5 (�) iscompact, so .2 \ 6(�) is open in .2. Thus 6 is an open map, hence a homeomorphism.

Exercise 29.6. The function 5 : R→ (0, 1) given by

5 (G) = 11 + 2−G

is a homeomorphism, and the map 6 : (0, 1) → (1 \ {?}, where ? = 1 × 0, de�ned by

6(G) = cos(2cG) × sin(2cG)



is also a homeomorphism (as it is a bijective local homeomorphism). Thus R is homeomorphicto (1 \ {?} via ℎ ≔ 6 ◦ 5 . Furthermore, since (1 is compact Hausdor� and the one-point com-pacti�cation is uniquely determined up to homeomorphism, the one-point compacti�cation of(1 \ {?} is precisely (1. By Exercise 29.5, ℎ extends to a homeomorphism between the one-pointcompacti�cation of R and (1.

Exercise 29.7. Since the one-point compacti�cation is uniquely determined up to homeomorph-ism, it su�ces to prove that (Ω is compact Hausdor�. It is Hausdor� since it has the order topology(see Exercise 17.10). Now let � be an open covering of (Ω . Let� ∈ � contain Ω. Then� containsan interval (0, +∞) . Let 00 be the least element of (Ω . Since (Ω is well-ordered, [00, 0] is compactby Theorem 27.1. Thus �nitely many elements of � cover [00, 0], and these elements togetherwith � cover (Ω, so (Ω is compact. We conclude that the one-point compacti�cation of (Ω ishomeomorphic with (Ω .

Exercise 29.8. Let - = {1/= | = ∈ Z+}. As subspaces of R, the map 5 : Z+ → - given by5 (=) = 1/= is a homeomorphism. Since-∪{0} is closed and bounded inR, it is compact (Theorem27.3). It is also Hausdor�, so (up to homeomorphism) it is the one-point compacti�cation of - .By Exercise 29.5, 5 extends to a homeomorphism between the one-point compacti�cation of Z+and - ∪ {0}.

Exercise 29.9. Let ? : � → �/� be the quotient map. Then ? is continuous, and it is open byExercise 5(c) of Supplementary Exercises of Chapter 2. Hence ? (�) = �/� is locally compact byExercise 29.3.

Exercise 29.10. Let * be a neighbourhood of G . Since - is locally compact at G, there exists acompact subspace � of - containing a neighbourhood, of G . Then * ∩, is open in -, hencein�. Thus,� \ (* ∩, ) is closed in�, hence compact by Theorem 26.2. Since - is Hausdor�, byLemma 26.4 there exist disjoint open sets+1 and+2 of- containing G and� \ (* ∩, ) respectively.Let + = +1 ∩ * ∩, . Since + is closed in �, it is compact. Furthermore, + is disjoint from� \ (* ∩, ) ⊂� \* , so + ⊂ * .

Exercise 29.11. (a) Since ? and 8/ are continuous and surjective, it follows that c is continuous(Exercise 18.10) and surjective. It remains to prove that � is open in . × / if c−1(�) is open in- × / . Let ~ × I ∈ � with ~ × I = c (G × I). Then G × I ∈ c−1(�), so we can take *1 ×, a basisneighbourhood of G × I contained in c−1(�) . Since, is a neighbourhood of ~, by Exercise 29.10there is a neighbourhood + of ~ such that + is compact and + ⊂ , . Hence G × I ∈ *1 × + ⊂c−1(�) . Note that c (*1 × + ) = ? (*1) × + ⊂ c (c−1(�)) ⊂ �, so ?−1(? (*1)) × + ⊂ c−1(�) .Now, for each D ∈ ?−1(? (*1)) use the tube lemma (Lemma 28.6) to �nd a neighbourhood,D ofD such that ?−1(? (*1)) × + contains the tube,D × + . Then *2 ≔

⋃D,D contains ?−1(? (*1))

and *2 × + ⊂ c−1(�) . For each 8 ≥ 3 we repeat this process to �nd an open set *8+1 containing?−1(? (*8)) such that *8+1 × + ⊂ c−1(�). Let * =

⋃8 *8 . We show that * is saturated. We have

* ⊂ ?−1(? (* )) by elementary set theory. Now, if G ∈ ?−1(? (* )) then ? (G) = ? (F) for somesome F ∈ * . Then F ∈ *8 for some 8, so G ∈ ?−1(? (*8)) ⊂ *8+1 ⊂ * . Thus ?−1(? (* )) ⊂ * .



Therefore * = c−1(? (* )), so * is saturated. Since ? is a quotient map, ? (* ) is open in . . NowG × I ∈ * ×+ ⊂ c−1(�), so c (* ×+ ) = ? (* ) ×+ is an open set containing ~ × I and containedin c (c−1(�)) ⊂ �, so � is open in . × / . We conclude that c is a quotient map.

(b) Since ? : � → � is a quotient map and � is locally compact Hausdor�, it follows from (a)that ? × 8� : �×� → � ×� is a quotient map. Similarly, 8� ×@ : � ×� → � ×� is a quotient map.Therefore, ? × @ = (8� × @) ◦ (? × 8� ) is a quotient map being the composite of quotient maps.

Supplementary Exercises: Nets

Supplementary Exercise 1. (a) If 0 and 1 are elements of a simply ordered set, then either 0 ≤ 1or 1 ≤ 0, so that 0 ≤ max{0, 1} and 1 ≤ max{0, 1}.

(b) If � ⊂ ( and � ⊂ (, then � � � ∪ � and � � � ∪ �.(c) If �, � ∈ �, then � � � ∩ � and � � � ∩ �.(d) Similarly as in (b), as �nite unions of closed sets is closed.

Supplementary Exercise 2. We have that inherits the partial order of � . Let V,W ∈ . Thenthere exists U ∈ � such that V � U and W � U, and in turn X ∈ such that U � X. Thus isdirected.

Supplementary Exercise 3. The set Z+ is simply ordered, hence directed by SupplementaryExercise 1(a). The given de�nitions are precisely the usual ones in the case of sequences.

Supplementary Exercise 4. Let* be a neighbourhood of G and+ a neighbourhood of~, so that* ×+ is a (standard) basis neighbourhood of G × ~ in - × . . By assumption, there exist U, V ∈ �such that GW ∈ * for all W � U and ~W ∈ * for all W � V. As � is directed, there exists X ∈ � suchthat X � U and X � V. Then GW ×~W ∈ * ×+ for all W � X. It follows that (GU ×~U )U ∈� → G ×~ in- × . .

Supplementary Exercise 5. Let (GU )U ∈� be a net in - and suppose that GU → G and GU →~, where G ≠ ~. Since - is Hausdor�, there exist disjoint neighbourhoods * of G and + of ~respectively. Then there exist U, V ∈ � such that GW ∈ * for all W � U and GW ∈ + for all W � V.Since � is directed, there is X ∈ � such that X � U and X � V. But then GW ∈ * ∩ + = ∅ for allW � X, a contradiction.

Supplementary Exercise 6. First suppose (GU )U ∈� is a net in - of points of � converging to G .If * is a neighbourhood of G, then it contains some GU ∈ �. Thus G ∈ �. Conversely, assumethat G ∈ �. We follow the hint. Let � denote the collection of all neighbourhoods of G, partiallyordered by reverse inclusion. Given* ∈ �, there exists G* ∈ * ∩� by assumption. Then (G* )* ∈�is a net in - of points of �. Moreover, if + is a neighbourhood of G, then G* ∈ + for all * ⊂ + ,that is, for all* � + . Thus G* → G .



Supplementary Exercise 7. First assume 5 is continuous and let (GU )U ∈� be a net in - conver-ging to G . Let* be a neighbourhood of 5 (G). Then 5 −1(* ) is a neighbourhood of G, so there existsU ∈ � such that GW ∈ 5 −1(* ) for all W � U, so that 5 (GW ) ∈ * for all such W . Thus 5 (GU ) → 5 (G).

Conversely, suppose that 5 is not continuous. Then there exists* open in. such that 5 −1(* )is not open in - . Then there exists G ∈ 5 −1(* ) such that no neighbourhood of G is containedin 5 −1(* ), so that G ∈ - \ 5 −1(* ) . It follows from Supplementary Exercise 6 that there is a net(GU )U ∈� in - of points of - \ 5 −1(* ) converging to G . As no point 5 (GU ), U ∈ � , belongs to * , itfollows that the net (5 (GU ))U ∈� in . does not converge to 5 (G) ∈ * .

Supplementary Exercise 8. Let 5 : � → - be a be a net in -, 5 (U) = GU for U ∈ � , convergingto G and let 6 : → � be a function form a directed set such that 5 ◦6 : → - is a subnet of 5 .Let* be a neighbourhood of G in - . Then there exists U ∈ � such that GW ∈ * for all W � U. Since6( ) is co�nal in � , there exists V ∈ such that 6(V) � U. Thus, for all W ∈ such that W � V, wehave U � 6(V) � 6(W), so that G6 (W ) ∈ * for all such W . It follows that 5 ◦ 6 converges to G .

Supplementary Exercise 9. Let (GU )U ∈� be a net in- . �rst assume that the net has the point G asan accumulation point. For each neighbourhood* of G, let �* denote the set of those U for whichGU ∈ * ,which is co�nal in � by assumption. We follow the hint. Let be the set of all pairs (U,* )where U ∈ � and * is a neighbourhood of G containing GU . De�ne (U,* ) � (V,+ ) if U � V and+ ⊂ * . Then is partially ordered. Moreover, let (U,* ) and (V,+ ) belong to . As � is directed,there exists W ∈ � such that W � U and W � V. Since �*∩+ is co�nal in � , there exists X ∈ �*∩+ suchthat X � W . Then (U,* ) � (X,* ∩ + ) and (V,+ ) � (X,* ∩ + ) and therefore is directed. Nowde�ne 6 : → � by 6(U,* ) = U. Then clearly 6(U,* ) � 6(V,+ ) for (U,* ) � (V,+ ). Moreover,given U ∈ � and * any neighbourhood of G, then we may �nd V ∈ �* such that U � V = 6(V,* ) .Thus 6( ) = � is co�nal in � . It remains to prove that the subnet (G6 (U,* ) )(U,* ) ∈ of (GU )U ∈�converges to G . So let* be a neighbourhood of G . Then there exists U ∈ �* , so that (U,* ) ∈ . If(V,+ ) � (U,* ), then G6 (V,+ ) = GV ∈ + ⊂ * . It follows that G6 (U,* ) → G .

Conversely, assume that some subnet (G6 (V) )V∈ (for some directed and 6 : → � ) of(GU )U ∈� converges to G . Let * be a neighbourhood of G and let �* denote the set of those U ∈ �for which GU ∈ * . By assumption, there exists V ∈ such that G6 (W ) ∈ * for all W � V. Let U ∈ � .By co�nality of 6( ) in � , there exists X ∈ such that 6(X) � U. Then, taking W ∈ such thatW � V and W � X, we have 6(W) ∈ �* and 6(W) � 6(X) � U. It follows that �* is co�nal in � . Wededuce that G is an accumulation point of the net (GU )U ∈� .

Supplementary Exercise 10. We follow the hint. First assume that- is compact and let (GU )U ∈�be a net in - . Given U ∈ � , let �U = {GV | U � V}. Given indices U1, . . . , U= ∈ � , there exists W suchthat W � U8 for all 8 = 1, . . . , =, so that GW ∈ ∩=8=1�U8 . Thus, the collection {�U }U ∈� has the �niteintersection property. Since - is compact, it follows from Theorem 26.9 that there exists a pointG ∈ ∩U�U . Let * be a neighbourhood of G . Then, given U ∈ � we have G ∈ �U , so there existsV � U such that GV ∈ * . Thus G is an accumulation point of the net (GU )U ∈� by de�nition, and itfollows from Supplementary Exercise 9 that the net (GU )U ∈� has a convergent subnet.



Conversely, suppose every net in - has a convergent subnet. Let � be a collection of closedsets having the �nite intersection property, and let ℬ denote the collection of all �nite intersec-tions of elements of �, partially ordered by reverse inclusion. Then, for each � ∈ ℬ there exists apoint G� ∈ �. Then (G�)�∈ℬ is a net in - . By assumption and Supplementary Exercise 9, (G�)�∈ℬhas an accumulation point G . Let � ∈ � ⊂ ℬ and let * be a neighbourhood of G . Then thereexists � ∈ ℬ such that G� ∈ * and � � �, i.e. � ⊂ �. Thus G� ∈ * ∩ �. Since � and * werearbitrary, it follows that G ∈ � = � for all � ∈ �, so that G ∈ ∩�∈��. We conclude that - iscompact by Theorem 26.9.

Supplementary Exercise 11. (Note that this was proved in Exercise 26.13(a).) Let G ∈ � · �. BySupplementary Exercise 6, there is a net (GU )U ∈� of points of� ·� converging to G . For each U ∈ � ,we have GU = ~U · IU for some ~U ∈ � and IU ∈ �. Now (IU )U ∈� is a net in �, so by SupplementaryExercise 10 it has a convergent subnet (I6 (V) )V∈ (for some directed and6 : → � ), say I6 (V) →1 ∈ �. Recall from Supplementary Exercise 1 of Chapter 2 that � × � → �, (6 × ℎ) ↦→ 6 · ℎ−1,is continuous. It follows from Supplementary Exercise 7 that (~6 (V) )V∈ = (G6 (V) · (I6 (V) )−1)V∈ converges to G · 1−1. Since � is closed, we have 0 B G · 1−1 ∈ � by Supplementary Exercise 6, sothat G = 0 · 1 ∈ � · �. We conclude that � · � is closed in �.

Supplementary Exercise 12. None of the solutions given above for these exercises made useof condition (2).

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