Direct sum of Hausdorff topological spaces
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.