(Hausdorff) Spaces

 

i

The proof is left to the reader.

ii

Let be any product of spaces. Let and be distinct elements of X. Then there exists such that in . Choose disjoint open sets G, H in so that , . Then , and since , . Hence result.

The axiom is particularly valuable when exploring compactness. Part of the reason is that implies that points and compact sets can be `separated off' by open sets and even implies that compact sets can be `separated off' from other compact sets in the same way.

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...

 

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 2 that, amongst compact spaces, this cannot happen.

It suffices to prove that f is closed. Given closed , then K is compact whence f(K) is compact and so f(K) is closed. Thus f is a closed map.

(i) (ii):

Let 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) (i):

Suppose that is not Hausdorff and that are points in X for which every neighbourhood of x intersects every neighbourhood of y. Let ( ) be the neighbourhood systems at x (y) respectively. Then both and are directed by reverse inclusion. We order the Cartesian product by agreeing that

Evidently, this order is directed. For each , and hence we may select a point . If is any neighbourhood of x, any neighbourhood of y and , then

That is, the net eventually belongs to both and and consequently converges to both x and y!