We explained in the previous chapter what a topological property (homeomorphic invariant) is but gave few good examples. We now explore some of the most important ones. Recurring themes will be:
If A is a subspace of a space (X, T), U is an open cover for A iff U is a cover for A such that each member of U is open in X.
The classic theorem of Heine-Borel-Lebesgue asserts that, in R, every open cover of a closed bounded subset has a finite subcover. This theorem has extraordinarily profound consequences and like most good theorems, its conclusion has become a definition.
Definition 1 (X, T) is said to be compact iff every open cover of X has a finite subcover.
Theorem 1 [Alexander's Subbase Theorem] Let S be any subbase for (X,T). If every open cover of X by members of S has a finite subcover, then X is compact.
The proof of this deep result is an application of Zorn's lemma, and is not an exercise for the faint-hearted!
(i) (R, Á) is not compact, for consider U = { (-n,n): n Î N }. Similarly, (C, Tusual) is not compact.
(ii) (0,1) is not compact, for consider U = { ([1/n],1) :n ³ 2 }.
(iii) (X, C) is compact, for any X.
(iv) Given x Î X, (X, E(x)) is compact; (X, I(x)) is not compact unless X is finite.
(v) T finite for any X Þ (X, T) compact.
(vi) X finite, T any topology for X Þ (X, T)
compact.
(vii) X infinite Þ (X, D) not compact.
(viii) Given (X, T), if (xn) is a sequence in X convergent to
x, then { xn:n Î N } È{x} is compact.
We call a subset A of (X,T) a compact subset when the subspace (A, TA) is a compact space. It's a nuisance to have to look at TA in order to decide on this. It would be easier to use the original T. Thankfully, we can!
Lemma 1 A is a compact subset of (X,T) iff every T-open cover of A has a finite subcover.
Proof Exercise.
Lemma 2 Compactness is closed-hereditary and preserved by continuous maps.
Proof Exercise.
Example
The unit circle in R2 is compact; indeed, paths in any space are compact.
In any metric space (M, d), every compact subset K is closed and bounded:
(bounded, since given any x0 Î M,
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Neither half is valid in all topological spaces;
Further, in a metric space, a closed bounded subset needn't be compact (e.g. consider M with the discrete metric and let A Í M be infinite; then A is closed, bounded (since A Í B(x,2) = M for any x Î M), yet it is certainly not compact! Alternatively, the subspace (0,1) is closed (in itself), bounded, but not compact.)
However, the Heine-Borel theorem asserts that such is the case for R and Rn; the following is a special case of the theorem:
Theorem 2 Every closed, bounded interval [a,b] in R is compact.
Proof Let U be any open cover of [a,b] and let K = { x Î [a,b]: [a,x] is covered by a finite subfamily of U}. Note that if x Î K and a £ y £ x, then y Î K. Clearly, K ¹ Æ since a Î K. Moreover, given x Î K, there exists dx > 0 such that [x,x+ dx) Í K (since x Î some open U Î chosen finite subcover of U). Since K is bounded, k* = supK exists.
| (i) | k* Î K | : Choose U Î U such that k* Î U; then there exists e > 0 such that (k*-e,k*] Í U. Since there exists x Î K such that k*-e < x < k*, k* Î K. |
| (ii) | k* = b | : If k* < b, choose U Î U with k* Î U and note that [k*,k*+d) Í U for some d > 0 -contradiction! |
Note An alternative proof [Willard, Page 116] is to invoke the connected nature of [a,b] by showing K is clopen in [a,b].
Theorem 3 Any continuous map from a compact space into a metric space is bounded.
Proof Immediate.
Corollary 1 If (X, T) is compact and f:X ® R is continuous, then f is bounded and attains its bounds.
Proof Clearly, f is bounded. Let m = supf(X) and l = inff(X); we must prove that m Î f(X) and l Î f(X). Suppose that m \not Î f(X). Since [`f(X)] = f(X), then there exists e > 0 such that (m-e,m+e) Çf(X) = Æ i.e. for all x Î X, f(x) £ m-e¼contra!
Similarly, if l \not Î f(X), then there exists e > 0 such that [l,l+e) Çf(X) = Æ whence l+e is a lower bound for f(X)!
Definition 2 A topological space (X, T) is said to be sequentially compact if and only if every sequence in X has a convergent subsequence.
Recall from Chapter 1 the definition of convergence of sequences in topological spaces and the cautionary remarks accompanying it. There we noted that, contrary to the metric space situation, sequences in topology can have several different limits! Consider, for example, (X,T0) and (R,L). In the latter space, if xn ® l, then xn = l for all n ³ some n0. Thus the sequence 1,[1/2], [1/4], [1/8], ¼ does not converge in (R,L)!
Lemma 3 Sequential compactness is closed-hereditary and preserved by continuous maps.
Proof Exercise.
We shall prove in the next section that in metric spaces, sequential compactness and compactness are equivalent!
Definition 3 Given a topological space (X,T), a subset A of X and x Î X, x is said to be an accumulation point of A iff every neighbourhood of x contains infinitely many points of A.
Lemma 4 Given a compact space (X,T) with an infinite subset A of X, then A has an accumulation point.
Proof Suppose not; then for each x Î X, there exists a neighbourhood Nx of x such that Nx ÇA is (at most) finite; the family { Nx:x Î X } is an open cover of X and so has a finite subcover { Nxi:i = 1, ¼,n }. But A Í X and A is infinite, whence
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Lemma 5 Given a sequentially compact metric space (M,d) and e > 0, there is a finite number of open balls, radius e, which cover M.
Proof Suppose not and that for some e > 0, there exists no finite family of open balls, radius e, covering M. We derive a contradiction by constructing a sequence (xn) inductively such that d(xm,xn) ³ e for all n, m (n ¹ m), whence no subsequence is even Cauchy!
Let x1 Î M and suppose inductively that x1, ¼,xk have been chosen in M such that d(xi,xj) ³ e for all i, j £ k,i ¹ j.By hypothesis, {B(xi,e):i = 1,¼,k } is not an (open) cover of M and so there exists xk+1 Î M such that d(xk+1,xi) ³ e for 1 £ i £ k. We thus construct the required sequence (xn), which clearly has no convergent subsequence.
Theorem 4 A metric space is compact iff it is sequentially compact.
Proof
$\Rightarrow :$ Suppose (M,d) is compact. Given any sequence
(xn) in M, either A = {x1,x2, ¼} is finite or it is infinite. If
A is finite, there must be at least one point l in A which occurs
infinitely often in the sequence and its occurrences form a subsequence
converging to l. If A is infinite, then by the previous lemma there exists
x Î X such that every neighbourhood of x contains infinitely many points
of A.
For each k Î w, B(x,[1/k]) contains infinitely many xn's:
select one, call it xnk, making sure that nk > nk-1 > nk-2 ¼. We have a subsequence (xn1, xn2, ¼, xnk, ¼)
so that d(x,xnk) < [1/k] ® 0 i.e. xnk ® x. Thus in either case there exists a convergent subsequence and
so (M,d) is sequentially compact.
$\Leftarrow :$ Conversely, suppose (M,d) is sequentially compact
and not compact. Then there exists some open cover {Gi:i Î I}
of M having no finite subcover. By Lemma 2.1.4, with e = [1/n] (n Î w), we can cover M by a finite number of
balls of radius [1/n]. For each n, there has to be one of these, say
B(xn,[1/n]), which cannot be covered by any finite number of the sets
Gi. The sequence (xn) must have a convergent subsequence (xnk)
which converges to a limit l. Yet {Gi:i Î I} covers M, so l Î some Gi0, say.
As k ® ¥, xnk ® l; but also [1/(nk)] ® 0 and 1/nk is the radius of the ball centred on xnk. So eventually B(xnk,[1/(nk)]) is inside Gi0, contradictory to their choice! (More rigorously, there exists m Î w such that B(l,[2/m]) Í Gi0. Now B(l,[1/m]) contains xnk for all k ³ k0 say, so choose k ³ k0 such that nk ³ m. Then B(xnk,[1/(nk)]) Í B(l, [2/m]) Í Gi0.) Hence, M is compact.
Recall that a map f:(X1,d1) ® (X2,d2), where (Xi,di) is a metric space for each i, is uniformly continuous on Xi if given any e > 0,$d > 0 such that d1(x,y) < d for x,y Î X1 Þ d2(f(x),f(y)) < e.
Ordinary continuity of f is a local property, while uniform continuity is a global property since it says something about the behaviour of f over the whole space X1. Since compactness allows us to pass from the local to the global, the next result is not surprising:
Theorem 5 If (X,d) is a compact metric space and f:X ® R is continuous, then f is uniformly continuous on X.
Note Result holds for any metric space codomain.
Proof Let e > 0; since f is continuous,
for each
x Î X, $dx > 0 such that d(x,y) < 2 dx
Þ |f(x) -f(y)| < [(e)/2]. The family {Bdx(x):x Î X } is an open cover of X and so has a finite subcover
{Bdxi(xi):i = 1,¼,n} of X. Let d = min{dxi:i = 1,¼,n}; then, given x,y Î X such that d(x,y) < d, it follows that |f(x) -f(y)| < e
(for x Î Bdxi(xi) for some i, whence d(x,xi) < dxi and so d(y,xi) £ d(y,x) + d(x,xi) < d+ dxi £ 2dxi Þ |f(y) -f(xi)| < [(e)/2].
Thus |f(x)-f(y)| £ |f(x) -f(xi)| + |f(xi) - f(y)| < [(e)/2] +[(e)/2] = e).
Note Compactness is not a necessary condition on the domain for uniform continuity. For example, for any metric space (X,d), let f:X ® X be the identity map. Then f is easily seen to be uniformly continuous on X.
Definition 4 A topological space (X,T) is locally compact iff each point of X has a compact neighbourhood.
Clearly, every compact space is locally compact. However, the converse is not true.
Examples
(i) With X infinite, the discrete space (X,D) is clearly locally
compact (for each x Î X, {x} is a compact neighbourhood of x!) but
not compact.
(ii) With X infinite and x Î X, (X, Á(x)) is locally compact
(but not compact).
(iii) (R,Á) is locally compact (x Î R Þ [x-1,x+1] is a
compact neighbourhood of x).
(iv) The set of rational numbers Q with its usual topology is not a
locally compact
space, for suppose otherwise; then 0 has a compact neighbourhood C in Q so
we can choose e > 0 such that J = QÇ[-e, e] Í C. Now J is closed in (compact) C and is therefore compact in
R. Thus, J must be closed in R-but [`J]R = [-e,e]!
Lemma 6
(i) Local compactness is closed-hereditary.
(ii) Local compactness is preserved by continuous open maps - it is not
preserved by continuous maps in general. Consider idQ:(Q,D) ® (Q,ÁQ) which is continuous and onto; (Q, D) is
locally compact while (Q, ÁQ) isn't!
Definition 5
A topological space (X, T) is said to be
(i) Lindelöf iff every open cover of X has a countable
subcover
(ii) countably compact iff every countable open cover of X has
a finite subcover.
However, for metric spaces, or more generally, metrizable spaces, the conditions
compact, countably compact and sequentially compact are equivalent.
Note Second countable Þ separable; separable + metrizable
Þ second countable ... and so in metrizable spaces, second
countability and separability are equivalent.
A partition of (X, T) means a pair of disjoint, non-empty, T-open subsets whose union is X. Notice that, since these sets are complements of one another, they are both closed as well as both open. Indeed, the definition of 'partition' is not affected by replacing the term 'open' by 'closed'.
Definition 6
A connected space (X, T) is one which has no partition.
(Otherwise, (X, T) is said to be disconnected.)
If Æ ¹ A Í (X, T), we call A a connected set
in X whenever (A, TA) is a connected space.
Lemma 7 (X, T) is connected iff X and Æ are the only subsets which are clopen.
(i) (X, T0) is connected.
(ii) (X, D) cannot be connected unless |X| = 1. (Indeed the only
connected subsets are the singletons!)
(iii) The Sorgenfrey line Rs is disconnected (for [x,¥) is clopen!).
(iv) The subspace Q of (R, Á) is not connected because
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Lemma 8 Æ Ì A Í (X, T) is not connected iff there exist T-open sets G, H such that A Í G ÈH, A ÇG ¹ Æ, A ÇH ¹ Æ and A ÇG ÇH = Æ. (Again, we can replace 'open' by 'closed' here.)
Proof Exercise.
Note By an interval in R, we mean any subset I such that whenever a < b < c and whenever a Î I and c Î I then b Î I. It is routine to check that the only ones are (a,b), [a,b], [a,b), (a,b], [a, ¥), (a, ¥), (-¥,b), (-¥,b], (-¥,¥) = R and {a} for real a, b, a < b where appropriate.
It turns out that these are exactly the connected subsets of (R, Á):-
Lemma 9 In R, if [a,b] = F1 ÈF2 where F1, F2 are both closed and a Î F1, b Î F2 then F1 ÇF2 ¹ Æ.
Proof Exercise.
Theorem 6 Let Æ Ì I Í (R, Á). Then I is connected iff I is an interval.
Proof
$\Rightarrow$: If I is not an interval, then there exist
a < b < c with a Î I, b \not Î I and c Î I. Take A = I Ç(-¥,b) and B = I Ç(b, ¥). Then A ÈB = I, A ÇB = Æ, A ¹ Æ, B ¹ Æ, A Ì I, B Ì I and A, B are both open in I i.e. A and B partition I and so
I is not connected.
$\Leftarrow$: Suppose I is not connected and that I is an
interval. By the `closed' version of Lemma 2.3.2, there exist closed
subsets K1, K2 of R such that I Í K1 ÈK2, I ÇK1 ¹ Æ, I ÇK2 ¹ Æ and I ÇK1 ÇK2 = Æ. Select a Î I ÇK1, b Î I ÇK2; without loss of
generality, a < b. Then [a,b] Í I so that [a,b] = ([a,b] ÇK1) È([a,b] ÇK2), whence by Lemma 2.3.2, Æ ¹ [a,b] ÇK1 ÇK2 Í I ÇK1 ÇK2 = Æ!
Lemma 10 Connectedness is preserved by continuous maps.
Proof Exercise.
Corollary 2 [Intermediate Value Theorem] If f:[a,b] ® R is continuous and f(a) < y < f(b), then y must be a value of f.
Proof Exercise.
Corollary 3 [Fixed point theorem for [0,1]] If f:[0,1] ® [0,1] is continuous, then it has a `fixed point' i.e. there exists some x Î [0,1] such that f(x) = x.
Proof Consider g(x) = f(x)-x. Then g:[0,1] ® R is continuous. Further, g(0) = f(0) ³ 0 and g(1) = f(1)-1 £ 0 so that 0 is intermediate between g(0) and g(1). Thus, by the Intermediate Value Theorem, there exists x Î [0,1] such that 0 = g(x) = f(x)-x i.e. such that f(x) = x.
Note Given continuous h:[a,b] ® [a,b], it follows that h has a fixed point since [a,b] @ [0,1] and `every continuous function has a fixed point' is a homeomorphic invariant.
Lemma 11 Let (X,T) be disconnected with Æ Ì Y Ì X, Y clopen. If A is any connected subset of X, then A Í Y or A Í X \Y.
Proof If A ÇY ¹ Æ ¹ A Ç(X \Y), then Æ Ì A ÇY Ì A and A ÇY is clopen in A. Thus, A is not connected! It follows that either A ÇY = Æ or A ÇX \Y = Æ i.e. either A Í X \Y or A Í Y.
Lemma 12 If the family {Ai:i Î I} of connected subsets of a space (X,T) has a non-empty intersection, then its union Èi Î IAi is connected.
Proof Suppose not and that there exists a non-empty proper clopen subset Y of Èi Î IAi. Then for each i Î I, either Ai Í Y or Ai Í Èi Î IAi \Y. However if for some j, Aj Í Y, then Ai Í Y for each i Î I (since Çi Î IAi ¹ Æ) which implies that Èi Î IAi Í Y!
Similarly, if for some k Î I, Ak Í Èi Î IAi \Y, then Èi Î IAi Í Èi Î IAi \Y!
Corollary 4 Given a family {Ci:i Î I} of connected subsets of a space (X,T), if B Í X is also connected and B ÇCi ¹ Æ for all i Î I, then B È(Èi Î ICi) is connected.
Proof Take Ai = B ÈCi in Lemma 2.3.3.
Lemma 13 If A is a connected subset of a space (X,T) and A Í B Í [`A]T, then B is a connected subset.
Proof If B is not connected, then there exists Æ Ì Y Ì B which is clopen in B . By Lemma 2.3.3, either A Í Y or A Í B \Y. Suppose A Í Y (a similar argument suffices for A Í B \Y); then [`A]T Í [`Y]T and so B \Y = B Ç(B \Y) = [`A]TB Ç(B \Y) Í [`Y]TB Ç(B \Y) = Y Ç(B \Y) = Æ - a contradiction!
Definition 7
Let (X, T) be a topological space with x Î X; we define the
component of x, Cx, in (X, T) to be the union of all connected
subsets of X which contain x i.e.
Cx = È{ A Í X:x Î A and A is connected }.
For each x Î X, it follows from Lemma 2.3.3 that Cx is the maximum connected subset of X which contains x. Also it is clear that if x, y Î X, either Cx = Cy or Cx ÇCy = Æ (for if z Î Cx ÇCy, then Cx ÈCy Í Cz Í Cx ÇCy whence Cx = Cy( = Cz)). Thus we may speak of the components of a space (X, T) (without reference to specific points of X): they partition the space into connected closed subsets (by Lemma 2.3.3) and are precisely the maximal connected subsets of X.
Examples
(i) If (X, T) is connected, (X, T) has only one component,
namely X!
(ii) For any discrete space, the components are the singletons.
(iii) In Q (with its usual topology), the components are the singletons.
(Thus, components need not be open.)
Definition 8 A space (X, T) is totally disconnected iff the only connected subsets of X are the singletons (equivalently, the components of (X, T) are the singletons).
Thus, by the previous examples, we see that the space Q of rationals, the space R \Q of irrationals and any discrete space are all totally disconnected. Further, the Sorgenfrey line Rs is totally disconnected.
Definition 9 A topological space (X, T) is pathwise connected iff for any x, y Î X, there exists a continuous function f:[0,1] ® X such that f(0) = x and f(1) = y. Such a function f is called a path from x to y.
Theorem 7 Every pathwise connected space is connected.
Proof Let (X, T) be pathwise connected and let a Î X; for every x Î X, there exists a path px:[0,1] ® X from a to x. Then, for each x Î X, px([0,1]) is connected; moreover, pa(0) = a Î Çx Î Xpx([0,1]) so that by Lemma 2.3.3, X = Èx Î Xpx([0,1]) is connected.
Note well The converse is false. Consider the following example, the topologist's sine curve :
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(for suppose, w.l.o.g., there exists a path p:[0,1] ® X with p(0) = ([1/(p)],0) and p(1) = (0,0). Then p1°p, being continuous, must take all values between 0 and [1/(p)], in particular [1/((2n+[1/2])p)] for each n i.e. there exists tn Î [0,1] such that p1°p(tn) = [1/((2n+[1/2])p)] for each n. Thus, p(tn) = ([1/((2n+[1/2])p)],1) ® (0,1) as n ® ¥. Now tn Î [0,1] for all n which implies that there exists a subsequence (tnk) in [0,1] with tnk ® l. Then p(tnk) ® p(l) and so p1°p(tnk) ® 0. Thus p(l) = (0,y) for some y, whence y = 0 (since p(l) Î X)!)
Definition 10
A topological space is said to be
(i) separable iff it has a countable dense subset.
(ii) completely separable (equivalently,second countable) iff it has
a countable base.
(i) (R, Á) is separable (since [`Q] = R).
(ii) (X, C) is separable for any X.
(iii) (R, L) is not separable.
Theorem 8
(i) Complete separability implies separability.
(ii) The converse is true in metric spaces.
Theorem 9
(i) Complete separability is hereditary.
(ii) Separability is not hereditary.
(Consider the `included point' topology
Á(0) on R; then (R, Á(0)) is separable, since
[` {0}] = R. However, R \{0} is not separable
because it is discrete.)
Separability does not imply complete separability since, for example,
(R, Á(0)) is separable but not completely separable.(Suppose
there exists a countable base B for its topology. Given x ¹ 0,
{0,x } is an open neighbourhood of x and so there exists Bx Î B such that x Î Bx Í { 0,x }.Thus Bx = {0,x } i.e.
B is uncountable ... contradiction!
Theorem 10 Separability is preserved by continuous maps.
Proof Exercise.
Note Complete separability is not preserved by continuous maps.