Exercise 185 (Chapter 28, “The category of topological spaces”):

Prove that both the direct product and the direct sum of two Hausdorff topological spaces is Hausdorff.

We will consider Hausdorff topological spaces and and their direct sum (in the first part) or product (in the second part) .

We will proceed without reference to the explicit constructions of direct topological sums and products, using instead the results in “Theorems about the direct product and sum of two topological spaces”.

## Direct sum

Let and be two eventual points of , with . The first result about direct sums in the above page implies that . The second result implies that . Two cases are possible:

- One of and is in and the other in .
- Both and are in , or both are in .

We will consider successively the case where and , and the case where both are in . The other possible cases are similar and will lead to similar results.

First case ( and ):

We have and , with and .

itself is an open neighbourhood of , and hence (last result about direct sums on the above page) is an open neighbourhood of in . Similarly, is an open neighbourhood of in . Since and are disjoint, we have found disjoint neighbourhoods respectively of and .

Second case ():

We have and , with . Since , we have . Since is Hausdorff, there exist and disjoint open neighbourhoods in respectively of and .

//TODO (we need injective)

## Direct product

Let and be two eventual points of , with .

The first result about direct products in the above page implies that or .

Let us suppose that .

Since is Hausdorff and and are distinct points of , there exist open sets and of such that , and .

Then and are disjoint (the image by of a common element would be both in and , which are disjoint). They are open (since and are open and is continuous). Since , we have ; similarly, . Thus there exist two disjoint open neighbourhoods respectively of and in .

The same would result if we had .

This being the case for all with , the topological space is Hausdorff.