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 X and Y and their direct sum (in the first part) or product (in the second part) (Z, \alpha, \beta).

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 c_1 and c_2 be two eventual points of Z, with c_1 \neq c_2. The first result about direct sums in the above page implies that \alpha^\rightarrow(X) \cap \beta^\rightarrow(Y) = \emptyset. The second result implies that \alpha^\rightarrow(X) \cup \beta^\rightarrow(Y) = Z. Two cases are possible:

  • One of c_1 and c_2 is in \alpha^\rightarrow(X) and the other in \beta^\rightarrow(Y) .
  • Both c_1 and c_2 are in \alpha^\rightarrow(X), or both are in \beta^\rightarrow(Y).

We will consider successively the case where c_1 \in \alpha^\rightarrow(X) and c_2 \in\beta^\rightarrow(Y), and the case where both are in \alpha^\rightarrow(X). The other possible cases are similar and will lead to similar results.

First case (c_1 \in \alpha^\rightarrow(X) and c_2 \in\beta^\rightarrow(Y)):

We have c_1 = \alpha(a) and c_2 = \beta(b), with a \in X and b \in Y.

X itself is an open neighbourhood of a, and hence (last result about direct sums on the above page) \alpha^\rightarrow(X) is an open neighbourhood of c_1 = \alpha(a) in Z. Similarly, \beta^\rightarrow(Y) is an open neighbourhood of c_2 in Z. Since \alpha^\rightarrow(X) and \beta^\rightarrow(Y) are disjoint, we have found disjoint neighbourhoods respectively of c_1 and c_2.

Second case (c_1, c_2 \in \alpha^\rightarrow(X)):

We have c_1 = \alpha(a_1) and c_2 = \alpha(a_2), with a_1, a_2 \in \alpha^\rightarrow(X). Since c_1 \neq c_2, we have a_1 \neq a_2. Since X is Hausdorff, there exist U_1 and U_2 disjoint open neighbourhoods in X respectively of a_1 and a_2.

//TODO (we need \alpha injective)

Direct product

Let c_1 and c_2 be two eventual points of Z, with c_1 \neq c_2.

The first result about direct products in the above page implies that  \alpha(c_1) \neq \alpha(c_2) or \beta(c_1) \neq \beta(c_2).

Let us suppose that \alpha(c_1) \neq \alpha(c_2).

Since X is Hausdorff and \alpha(c_1) and \alpha(c_2) are distinct points of X, there exist open sets U_1 and U_2 of X such that \alpha(c_1) \in U_1, \alpha(c_2) \in U_2 and U_1 \cap U_2 = \emptyset.

Then \alpha^\leftarrow(U_1) and \alpha^\leftarrow(U_2) are disjoint (the image by \alpha of a common element would be both in U_1 and U_2, which are disjoint). They are open (since U_1 and U_2 are open and \alpha is continuous). Since \alpha(c_1) \in U_1, we have c_1 \in \alpha^\leftarrow(U_1); similarly, c_2 \in \alpha^\leftarrow(U_2). Thus there exist two disjoint open neighbourhoods respectively of c_1 and c_2 in Z.

The same would result if we had \beta(c_1) \neq \beta(c_2).

This being the case for all c_1, c_2 \in Z with c_1 \neq c_2, the topological space Z is Hausdorff.

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