Theorem 3. A subset S of a topological space X is compact if and only if every open cover of S by open sets in X has a finite subcover.

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1 Compactess Defiitio 1. A cover or a coverig of a topological space X is a family C of subsets of X whose uio is X. A subcover of a cover C is a subfamily of C which is a cover of X. A ope cover of X is a cover cosistig of ope sets. Defiitio 2. A topological space X is said to be compact if every ope cover of X has a fiite subcover. A subset S of X is said to be compact if S is compact with respect to the subspace topology. Theorem 3. A subset S of a topological space X is compact if ad oly if every ope cover of S by ope sets i X has a fiite subcover. Proof. ( ) Assume that S is a compact subset of a topology space X. Let {G α α Λ} be a collectio of ope subset of X such that S α Λ G α. Let O α = G α S for each α Λ. The O α is ope i S for each α Λ ad α Λ O α = α Λ (G α S) = ( α Λ G α ) S = S. Thus {O α α Λ} is a ope cover of S. It follows that {O α α Λ} cotais a fiite subcover {O α1, O α2,..., O α }. Hece S = i=1o αi = i=1(o αi S) = ( i=1g αi ) S. Thus S i=1g αi. ( ) Let {O α α Λ} be a ope cover of (S, τ s ). The, for each α Λ, there exists a ope set G α i X such that O α = G α S. Thus S = α Λ O α = α Λ (G α S) = ( α Λ G α ) S. Thus S α Λ G α. The there is a fiite subset {G α1, G α2,..., G α } such that S i=1g αi. It follows that S = ( i=1g αi ) S = i=1(g αi S) = i=1o αi. Hece (S, τ s ) is compact. Examples. 1. Ay fiite set is compact. I geeral, (X, τ), where τ is fiite, is compact. I particular, a idiscrete space is compact. 2. Ay ifiite discrete space is ot compact. I fact, if X is a ifiite discrete space, the {{x} x X} is a ope cover of X which has o fiite subcover. 3. R is ot compact. The class {(, ) N} is a ope cover of R which cotais o fiite subcover. 1

2 Theorem 4. A closed subset of a compact space is compact. Proof. Let (X, τ) be a compact space ad F a closed subset of X. Let C = {G α α Λ} be a family of ope subsets of X such that F α Λ G α. The X = F F c = ( α Λ G α ) F c. Thus C {F c } is a ope cover of X ad hece it has a fiite subcover {G α1, G α2,..., G α, F c }, i.e. X = ( i=1g αi ) F c. Sice F X, it follows that F i=1g αi. Thus {G α1, G α2,..., G α } is a fiite subfamily of C such that F i=1g αi. Hece, F is compact. Theorem 5. If A is a compact subset of a Hausdorff space X ad x / A, the x ad A have disjoit eighborhoods. Proof. By the Hausdorffess of X, for each y A, there are ope eighborhoods U y ad V y of x ad y, respectively, such that U y V y =. The {V y y A} is a ope cover for A, which is compact. Hece there are y 1, y 2,..., y A such that A i=1v yi. Let V = i=1v yi ad U = i=1u yi. The U ad V are eighborhoods of x ad A, respectively, ad U V =. Theorem 6. Ay compact subset of a Hausdorff space is closed. Proof. Let A be a compact subset of a Hausdorff space X. To show that A c is ope, let x A c. By the previous theorem, there are eighborhoods U ad V of x ad A, respectively, such that U V =. It follows that x U V c A c. This shows that A c is a eighborhood of x. Hece, A c is a eighborhood of all its elemets. It follows that A c is ope ad that A is closed. Theorem 7. A cotiuous image of a compact space is compact. Proof. Let f : X Y be a cotiuous fuctio from a compact space X ito a space Y. By Theorem 3, it is sufficiet to assume that f maps X oto Y ad we show that Y is compact. To see this, let {G α α Λ} be a ope cover for Y. By cotiuity of f, it follows that f 1 [G α ] is ope i X for each α ad α Λ f 1 [G α ] = f 1 [ α Λ G α ] = f 1 [Y ] = X. Hece, {f 1 [G α ] α Λ} is a ope cover for X. By the compactess of X, there are α 1, α 2,..., α Λ such that i=1f 1 [G αi ] = X. Thus Y = f[x] = f[ i=1f 1 [G αi ]] = f[f 1 [ i=1g αi ]] i=1g αi. This shows that Y is compact. 2

3 Corollary 8. Let f : X Y is a bijective cotiuous fuctio. compact ad Y is Hausdorff, the f is a homeomorphism. If X is Proof. Let f : X Y be a bijective cotiuous fuctio. Assume that X is compact ad Y is Hausdorff. To show that f is a homeomorphism, it suffices to show that f is a closed map. Let F be a closed subset of X. The F is compact by Theorem 4. By Theorem 7, f[f ] is compact i the Hausdorff space Y. Hece, f[f ] is closed i Y by Theorem 6. Theorem 9. A cotiuous fuctio of a compact metric space ito a metric space is uiformly cotiuous. Proof. Let f : (X, d) (Y, ρ) be a cotiuous fuctio betwee metric spaces. Assume that X is compact. To show that f is uiformly cotiuous, let ε > 0. By cotiuity of f, for each x X, there is a δ x > 0 such that d(y, x) < δ x = ρ(f(y), f(x)) < ε 2. The {B d (x, δx ) x X} is a ope cover for a compact space X. Hece, 2 we ca choose x 1, x 2,..., x X such that X = i=1b d (x i, δx i ). Choose 2 δ = 1 mi{d 2 x 1, d x2,..., d x }. Now, let x, y X be such that d(x, y) < δ. The x B d (x i, δx i 2 ) for some i. Hece, ρ(f(x), f(x i)) < ε 2. Moreover, d(y, x i ) d(y, x) + d(x, x i ) δ + δ x i 2 δ x i. It follows that ρ(f(y), f(x i )) < ε. By triagle iequality, 2 ρ(f(x), f(y)) ρ(f(x), f(x i )) + ρ(f(x i ), f(y)) < ε. This shows that f is uiformly cotiuous. Next we itroduce a property that measures boudedess of a subset of a metric space. We say that a o-empty set A is bouded if diam(a) <, i.e., there is a costat M > 0 such that d(x, y) M for ay x, y A. This defiitio of boudedess depeds o the metric rather tha the set itself. For example, R with the discrete metric is bouded eve though i some sese R is large. The followig defiitio of total boudedess captures this spirit. Defiitio 10. A metric space (X, d) is said to be totally bouded or precompact if for ay ε > 0, there is a fiite cover of X by sets of diameter less tha ε. Theorem 11. A metric space (X, d) is totally bouded if ad oly if for each ε > 0, X ca be covered by fiitely may ε-balls. 3

4 Proof. ( ) Let ε > 0. The there exist subsets A 1, A 1,..., A of X such that diam A i < ε for all i {1, 2,..., } ad i=1a i = X. We may assume that each A i is o-empty ad choose x i A i. If x X, the x A i for some i ad hece, d(x, x i ) diam(a i ) < ε. This show that X = i=1b d (x i, ε). ( ) Let ε > 0. The there is a fiite subset {x 1, x 2,..., x } of X such that X = i=1b d (x i, ε). Let A 4 i = B d (x i, ε ) for each i {1, 2,..., }. For each 4 a, b A i, d(a, b) d(a, x i ) + d(x i, b) < ε + ε = ε. Hece, diam(a i) < ε. Theorem 12. A subspace of a totally bouded metric space is totally bouded. Proof. Let (X, d) be a totally bouded metric space ad Y a subspace of X. Let ε > 0. The there exist A 1, A 2,..., A X such that diam A i < ε for all i {1, 2,..., } ad i=1a i = X. The diam(a i Y ) diam A i < ε for each i {1, 2,..., } ad i=1(a i Y ) = ( i=1a i ) Y = X Y = Y. Theorem 13. Every totally bouded subset of a metric space is bouded. Proof. Let (X, d) be a totally bouded metric space. The there exists a fiite subset {x 1, x 2,..., x } of X such that i=1b d (x i, 1) = X. Let K = max{d(x i, x j ) i, j {1, 2,..., }} + 2. Let x, y X. The x B d (x i, 1) ad y B d (x j, 1) for some i, j. Thus d(x, y) d(x, x i ) + d(x i, x j ) + d(x j, y) < 1 + d(x i, x j ) + 1 K, i.e. diam X K, so X is bouded. I geeral, the coverse is ot true. The space R with the discrete metric is bouded because d(x, y) 1 for all x, y R, but it caot be covered by a fiitely may balls of radius 1 2. However, it is true for R with the usual metric. Theorem 14. A bouded subset of R is totally bouded. Proof. We will prove this theorem for the case = 1. The proof for the case > 1 is similar but slightly more complicated. By Theorem 12, it suffices to prove that a closed iterval [a, b] is totally bouded. Let ε > 0. Choose a N such that (b a)/ < ε. For i = 0, 1,...,, let x i = a + i b a. The [a, b] = i=1[x i 1, x i ] ad diam([x i 1, x i ]) = x i x i 1 < ε. 4

5 Theorem 15. A metric space is totally bouded if ad oly if every sequece i it has a Cauchy subsequece. Proof. ( ) Assume that (X, d) is ot tally bouded. The there is a ε > 0 such that X caot be covered by fiitely may balls of radius ε. Let x 1 X. The B d (x 1, ε) X, so we ca choose x 2 X B d (x 1, ε). I geeral, for each N, we ca choose x +1 X i=1b d (x i, ε). If m >, the x m / B d (x, ε), ad thus d(x m, x ) ε. It is easy to see that (x ) has o Cauchy subsequece. ( ) Assume that (X, d) is totally bouded ad let (x ) be a sequece i (X, d). Set B 0 = X. There exist A 11, A 12,..., A 11 X such that 1 diam A 1i < 1 for all i {1, 2,..., 1 } ad A 1i = X. At least oe of these A 1i s, called B 1, must cotai ifiitely may terms of (x ). Let (x 11, x 12,... ) be a subsequece of (x ) which lies etirely i B 1. Sice B 1 X, it is totally bouded. There are A 21,..., A 22 B 1 such that diam A 2i < 1 2 for all i {1, 2,..., 2 2} ad A 2i = B 1. At least oe of these A 2i s, called B 2, must cotai ifiitely may terms of (x 1 ). The diam B 2 < 1 ad we ca choose a subsequece (x 2 21, x 22, x 23,... ) of (x 11, x 12, x 13,... ) i B 2. We ca cotiue this process. To make this argumet more precise, we will give a iductive costructio. Assume that we ca choose B i B i 1 with diam B i < 1 ad a subsequece i (x i1, x i2, x i3,... ) of (x (i 1)1, x (i 1)2, x (i 1)3,... ) i B i for all i {1, 2,..., k 1}. Sice B k 1 B k 2... B 1 B 0 = X, B k 1 is totally bouded. The there exist A k1, A k2,..., A kk B k 1 such that diam A kj < 1 k for all j {1, 2,..., k k} ad A ki = B k 1. At least oe of these A ki s, called B k, must cotai ifiitely may terms of (x k 1, ). Hece B k B k 1, diam B k < 1 ad we ca choose a subsequece k (x k1, x k2, x k3,... ) of (x (k 1)1, x (k 1)2, x (k 1)3,... ) which lies etirely i B k. Now we choose the diagoal elemets (x 11, x 22, x 33,... ) from the above subsequeces. This is to guaratee that the idex of the chose subsequece is strictly icreasig. To see that it is a Cauchy subsequece of (x ), let ε > 0. Choose a N N such that 1 < ε. Let m, N be such that m, N. N The x mm B m B N ad x B B N. Thus d(x mm, x ) diam B N < 1 N < ε. Hece (x ) is a Cauchy subsequece of (x ). 5 i=1 i=1 i=1

6 Defiitio 16. A space X is said to be sequetially compact if every sequece i X has a coverget subsequece. Theorem 17. A metric space X is sequetially compact if ad oly if it is complete ad totally bouded. Proof. ( ) Let (X, d) be a sequetially compact space. By Theorem 15, it is totally bouded because a coverget sequece is a Cauchy sequece. To see that it is complete, let (x ) be a Cauchy sequece i X. Sice X is sequetially compact, it has a coverget subsequece. Hece, (x ) is coverget. It follows that X is complete. ( ) Assume that (X, d) is totally bouded ad complete. To see that X is sequetially compact, let (x ) be a sequece i X. By Theorem 15, it has a Cauchy subsequece (x k ). Sice (X, d) is complete, (x k ) is coverget. Hece, (x ) has a coverget subsequece. This shows that X is sequetially compact. Defiitio 18. Let C be a cover of a metric space X. A Lebesgue umber for C is a positive umber λ such that ay subset of X of diameter less tha or equal to λ is cotaied i some member of C. Remark. If λ is a Lebesgue umber, the so is ay λ > 0 such that λ λ. Example. Let X = (0, 1) R ad C = {( 1, 1) 2}. The C is a ope cover for X, but it has o Lebesgue umber. To see this, let λ > 0. Choose a positive iteger such that 1 < λ. Let A = (0, 1 ) X. The diam(a) = 1 < λ, but A is ot cotaied i ay member of C. Hece C has o Lebesgue umber. Theorem 19. Every ope cover of a sequetially compact metric space has a Lebesgue umber. Proof. Let C be a ope cover of a sequetially compact metric space (X, d). Suppose that C does ot have a Lebesgue umber. The for each N there exists a subset B of X such that diam(b ) 1 ad B G for all G C. For each N, choose x B. Sice X is sequetially compact, the sequece (x ) cotais a coverget subsequece (x k ). Let x X be the limit of this subsequece. The x G 0 for some G 0 C. Sice G 0 is ope, there exists ε > 0 such that B d (x, ε) G 0. Sice (x k ) coverges to x, there exists a iteger N such that d(x k, x) < ε 2 for ay k N. 6

7 Choose M N such that 1 M < ε 2. Let K = max{m, N}. The K K M. Hece, Moreover, For ay y B K, we have d(x K, x) < ε 2 ad x K B K. diam(b K ) 1 K < ε 2. d(x, y) d(x, x K ) + d(x K, y) < ε 2 + ε 2 = ε. Hece y B d (x, ε). The B K that B K G for all G C. B d (x, ε) G 0. This cotradicts the fact Defiitio 20. A space X is said to satisfy the Bolzao-Weierstrass property if every ifiite subset has a accumulatio poit i X. Theorem 21. I a metric space X, the followig statemets are equivalet: (a) X is compact; (b) X has the Bolzao-Weierstrass property; (c) X is sequetially compact; (d) X is complete ad totally bouded. Proof. We have already proved (c) (d) i Theorem 17. (a) (b). Let (X, d) be a compact metric space ad S a ifiite subset of X. Suppose that S has o accumulatio poit. The for each x X there exists a ope eighborhood V x such that V x (S {x}) =. Hece, C = {V x x X} is a ope cover of X. Sice X is compact, there exists a fiite subcover {V x1, V x2,..., V x } of C. Sice each ball cotais at most oe poit of S, X = i=1v xi cotais fiitely may poits of S. Hece, S is fiite, cotrary to the hypothesis. (b) (c). Assume that X has the Bolzao-Weierstrass property. To show that X is sequetially compact, let (x ) be a sequece i X. Case I. The set S = {x N} is fiite. The there is a X such that x = a for ifiitely may s. Choose 1 = mi{ N x = a} ad for ay k 2, let k = mi({ N x = a} { 1, 2,..., k 1 }). The (x k ) is a costat subsequece of (x ) ad hece is coverget. 7

8 Case II. The set S = {x N} is ifiite. By the assumptio, S has a accumulatio poit x (i X). For each N, B d (x, 1 ) (S {x}) ; i fact, B d (x, 1 ) (S {x}) is a ifiite set. Let 1 = mi{ N x B d (x, 1) (S {x})} ad k = mi({ N x B d (x, 1 k ) (S {x})} { 1,..., k 1 }), for ay k 2. The (x k ) is a subsequece of (x ) such that d(x k, x) < 1 k for each k N. Hece (x k ) coverges to x. (c) (a). Assume that (X, d) is a sequetially compact metric space. Let C be a ope cover for X. Hece C has a Lebesgue umber λ > 0. Moreover, X is totally bouded. Thus there exist A 1, A 2,..., A X such that X = i=1a i ad diam(a i ) λ for each i. Hece for each i {1, 2,..., }, there exists G i C such that A i G i. Thus X = i=1g i. This shows that C has a fiite subcover. Hece X is compact. Theorem 22 (Heie-Borel). A subset of R is compact if ad oly if it is closed ad bouded. Proof. By Theorem 21, a metric space is compact if ad oly if it is complete ad totally bouded. Hece, if a subset of R is compact, the it is closed ad bouded. (I fact, we ca prove directly that compactess implies a set beig closed ad bouded without usig Theorem 21.) Coversely, let A be a closed ad bouded subset of R. By Theorem 14, A is totally bouded. Sice A is closed subset of R, which is complete, A is also complete. Hece, A is compact. Corollary 23 (Extreme Value Theorem). A real-valued cotiuous fuctio o a compact space has a maximum ad a miimum. Proof. Assume that f : X R is a cotiuous fuctio ad X is compact. By Theorem 7, f[x] is a compact subset of R. Hece, f[x] is closed ad bouded. Let a = if f[x] ad b = sup f[x]. Sice f[x] is closed, a ad b are i f[x]. Thus a ad b are maximum ad miimum of f[x], respectively. Theorem 24 (Bolzao-Weierstrass). Every bouded ifiite subset of R has at least oe accumulatio poit (i R ). Proof. Let A be a bouded ifiite subset of R. Sice A is bouded, it is cotaied i some closed cube I = [, ] [, ]. Sice I is closed ad bouded, it is compact by Heie-Borel theorem. Sice A is a ifiite subset of a compact set I, it must have a accumulatio poit (i R ) by Theorem 21. 8

9 Theorem 25. Every bouded sequece i R has a coverget subsequece. Proof. Let (x ) be a bouded sequece i R. Hece it is cotaied i some closed cube I = [, ] [, ]. Sice I is closed ad bouded, it is compact by Heie-Borel theorem. The I is sequetially compact. This implies that (x ) has a coverget subsequece. 9