We have observed instances of topological statements which, although true for all metric (and metrizable) spaces, fail for some other topological spaces. Frequently, the cause of failure can be traced to there being `not enough open sets' (in senses to be made precise). For instance, in any metric space, compact subsets are always closed; but not in every topological space, for the proof ultimately depends on the observation
`given x ¹ y, it is possible to find disjoint open sets G and H with x Î G and y Î H'
which is true in a metric space (e.g. put G = B(x,e), H = B(y,e) where e = [1/2] d(x,y)) but fails in, for example, a trivial space (X, T0).
What we do now is to see how `demanding certain minimum levels-of-supply of open sets' gradually eliminates the more pathological topologies, leaving us with those which behave like metric spaces to a greater or lesser extent.
Definition 1 A topological space (X,T) is T1 if, for each x in X, {x} is closed.
Comment 1
(i) Every metrizable space is T1
(ii) (X,T0) isn't T1 unless |X| = 1
(i) T1 is hereditary
(ii) T1 is productive
(iii) T1 Þ every finite set is closed. More precisely,
(X, T) is T1 iff T Ê C, i.e. C
is the weakest of all the T1 topologies that can be defined on X.
The respects in which T1-spaces are `nicer' than others are mostly concerned with `cluster point of a set' (an idea we have avoided!). We show the equivalence, in T1 spaces, of the two forms of its definition used in analysis.
Theorem 2
Given a T1 space (X,T), p Î X and A Í X, the
following are equivalent:
(i) Every neighbourhood of p contains infinitely many points of
A
(ii) Every neighbourhood of p contains at least one point of A
different from p.
Hence, (i) Û (ii).
Definition 2 A topological space (X,T) is T2 (or Hausdorff) iff given x ¹ y in X, $ disjoint neighbourhoods of x and y.
Comment 2
(i) Every metrizable space is T2
(ii) T2 Þ T1 (i.e. any T2 space is T1,
for if x, y Î T2 X and y Î [`{x}], then every
neighbourhood of y contains x, whence x = y.)
(iii) (X,C), with X infinite, cannot be T2
(i) T2 is hereditary
(ii) T2 is productive.
{i} The proof is left to the reader.
{ii} Let (X, T) = Õi Î I(Xi, Ti) be any product
of T2 spaces. Let x = (xi)i Î I and y = (yi)i Î I be
distinct elements of X. Then there exists i0 Î I such that
xi0 ¹ yi0 in Xi0. Choose disjoint open sets G,
H in (Xi0, Ti0) so that xi0 Î G, yi0 Î H. Then x Î pi0-1(G) Î T, y Î pi0-1(H) Î T and since G ÇH = Æ, pi0-1(G)Çpi0-1(H) = Æ. Hence result.
Theorem 4 In a T2-space (X, T), if C is a compact set and x \not Î C, then there exist T-open sets G and H so that x Î G, C Í H and G ÇH = Æ.
Proof A valuable exercise: separate each point of C from x using disjoint open sets, note that the open neighbourhoods of the various elements of C, thus obtained, make up an open covering of C, reduce it to a finite subcover by appealing to compactness ...
Corollary 1 In a T2-space, any compact set is closed.
Corollary 2 In a T2-space, if C and K are non-empty compact and disjoint, then there exist open G, H such that C Í G, K Í H and G ÇH = Æ.
A basic formal distinction between algebra and topology is that although the inverse of a one-one, onto group homomorphism [etc!] is automatically a homomorphism again, the inverse of a one-one, onto continuous map can fail to be continuous. It is a consequence of Corollary 5.2 that, amongst compact T2 spaces, this cannot happen.
Theorem 5 Let f:(X1,T1) ® (X2,T2) be one-one, onto and continuous, where X1 is compact and X2 is T2. Then f is a homeomorphism.
Proof It suffices to prove that f is closed. Given closed K Í X1, then K is compact whence f(K) is compact and so f(K) is closed. Thus f is a closed map.
Theorem 6 (X,T) is T2 iff no net in X has more than one limit.
Proof
(i) $\Rightarrow$ (ii): Let x ¹ y in X; by hypothesis, there exist
disjoint neighbourhoods U of x, V of y. Since a net cannot eventually
belong to each of two disjoint sets, it is clear that no net in X can
converge to both x and y.
(ii) $\Rightarrow$ (i): Suppose that (X, T) is not Hausdorff
and that x ¹ y are points in X for which every neighbourhood of x
intersects every neighbourhood of y. Let Nx (Ny) be the
neighbourhood systems at x (y) respectively. Then both Nx and
Ny are directed by reverse inclusion. We order the Cartesian
product Nx ×Ny by agreeing that
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Corollary 3 Let f:(X1, T1) ® (X2, T2), g:(X1, T1) ® (X2, T2) be continuous where X2 is T2. Then their `agreement set' is closed i.e. A = {x:f(x) = g(x)} is closed.
Definition 3
A space (X, T) is called T3 or regular provided :-
(i) it is T1, and
(ii) given x \not Î closed F, there exist disjoint open
sets G and H so that x Î G, F Í H.
(i) Every metrizable space is T3; for it is certainly T1 and
given x \not Î closed F, we have x Î open X \F so there exists e > 0 so that x Î B(x, e) Í X \F. Put G = B(x, [(e)/2]) and H = { y :d(x,y) > [(e)/2] }; the result now follows.
(ii) Obviously T3 Þ T2.
(iii) One can devise examples of T2 spaces which are not T3.
(iv) It's fairly routine to check that T3 is productive and hereditary.
(v) Warning: Some books take T3 to mean Definition 5.3(ii)
alone, and regular to mean Definition 5.3(i) and (ii);
others do exactly the opposite!
Definition 4
A space (X, T) is T3[1/2] or completely
regular or Tychonoff iff
(i) it is T1, and
(ii) given x Î X, closed non-empty F Í X such that
x \not Î F, there exists continuous f:X ® [0,1] such that
f(F) = {0} and f(x) = 1.
(i) Every metrizable space is T3[1/2]
(ii) Every T3[1/2] space is T3 ( such a space is certainly
T1 and given x \not Î closed F, choose f as in the definition;
define G = f-1([0, [1/3])), H = f-1(([2/3],1]) and observe
that T3 follows.)
(iii) Examples are known of T3 spaces which fail to be Tychonoff
(iv) T3[1/2] is productive and hereditary.
Definition 5
A space (X, T) is T4 or normal if
(i) it is T1, and
(ii) given disjoint non-empty closed subsets A, B of X, there
exist disjoint open sets G, H such that A Í G, B Í H.
Proof Certainly, X is T1; choose a metric d on X such that T is Td. The distance of a point p from a non-empty set A can be defined thus:
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G = {x:d(x, A) < d(x,B) }
H = {x:d(x,B) < d(x,A)}.
Clearly, G ÇH = Æ. Also, each is open (if x Î G and e = [1/2]{d(x,B) - d(x,A)}, then B(x,e) Í G, by the triangle inequality.) Now, if d(p, A) = 0, then for all n Î N, there exists xn Î A such that d(p, xn) < [1/n]. So d(p,xn)® 0 i.e. xn ® p, whence p Î [`A]. Thus for each x Î A, x \not Î B = [`B] so that d(x, B) > 0 = d(x,A) i.e. x Î G. Hence A Í G. Similarly B Í H.
It's true that T4 Þ T3[1/2] but not very obvious. First note that if G0, G1 are open in a T4 space with [`(G0)] Í G1, then there exists open G[1/2] with [`(G0)] Í G[1/2] and [`(G[1/2])] Í G1 (because the given [`(G0)] and X \G1 are disjoint closed sets so that there exist disjoint open sets G[1/2], H such that [`(G0)] Í G[1/2], X\G1 Í H i.e. G1 Ê (closed) X\H Ê G[1/2]).
Lemma 1 [Urysohn's Lemma] Let F1, F2 be disjoint non-empty closed subsets of a T4 space; then there exists a continuous function f:X ® [0,1] such that f(F1) = {0}, f(F2) = {1}.
Proof Given disjoint closed F1 and F2, choose disjoint open G0 and H0 so that F1 Í G0, F2 Í H0. Define G1 = X \F2 (open). Since G0 Í (closed) X \H0 Í X \F2 = G1, we have [`(G0)] Í G1.
By the previous remark, we can now construct:
(i) G[1/2] Î T: [`(G0)] Í G[1/2], [`(G[1/2])] Í G1.
(ii) G[1/4], G[3/4] Î T:
[`(G0)] Í G[1/4], [`(G[1/4])] Í G[1/2], [`(G[1/2])] Í G[3/4], [`(G[3/4])] Í G1.
(iii) ... and so on!
Thus we get an indexed family of open sets
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Observe that the index set is dense in [0,1]: if s < t in [0,1], there exists some [m/(2n)] such that s < [m/(2n)] < t. Define
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Well, f(x) < a iff there exists some r = [m/(2n)] such that f(x) < r < a. It follows that f-1([0, a)) = Èr < aGr, a union of open sets.
Again, f(x) > a iff there exist r1, r2 such that a < r1 < r2 < f(x), implying that x \not Î Gr2 whence x \not Î [`(Gr1)]. It follows that f-1((a,1]) = Èr1 > a(X \[`(Gr1)]), which is again open.
Corollary 4 Every T4 space is T3[1/2].
Proof Immediate from Lemma 5.5. (Note that there exist spaces which are T3[1/2] but not T4.)
Theorem 8 Any compact T2 space is T4.
Proof Use Corollary 5.2 to Theorem 5.2.
Note Unlike the previous axioms, T4 is neither hereditary
nor productive. The global view of the hierarchy can now be filled in as an
exercise from data supplied above:-
| Metrizable | Hereditary? | Productive? |
| T4 | ||
| T3[1/2] | ||
| T3 | ||
| T2 | ||
| T1 |
The following is presented as an indication of how close we are to having `come full circle'.
Theorem 9 Any completely separable T4 space is metrizable!
Sketch Proof
Choose a countable base; list as { (Gn,Hn): n ³ 1} those pairs of
elements of the base for which [`(Gn)] Í Hn. For each
n, use Lemma 5.5 to get continuous fn:X ® [0,1] such
that fn([`(Gn)]) = {0}, fn(X \Hn) = {1}. Define
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