This is exercise 13 on page 101 in Chapter 17 of James R. Munkers’ Topology 2nd Edition textbook.

Question

Show that \(X\) is Hausdorff if and only if the diagonal \(\bigtriangleup = \{x \times x : x \in X\}\) is closed in \(X \times X\).

(=>)

By definition of a Hausdorff space, for each pair \(x, y\) of distinct points of \(X\), there exists neighborhoods \(U\) and \(V\) of \(x\) and \(y\) respectively. \(U \cap V = \emptyset\).

To form a basis for \(X \times X\), the sets \(U \times V\) may be used. Because \(U \cap V = \emptyset\), the ordered pairs generated by such cartesian product may not contain any elements which are the same. In other words, for any \(a \in U\), \((a, a)\) may not exist in the open sets. Same for \(V\). Thus, the infinite union of these sets may never contain any ordered pairs with the same elements. Finally, because the infinite union of open sets are open, the closed set looks like \(\bigtriangleup\), completing the conditional.

(<=)

The open set \(X - \bigtriangleup\) contains every every point except those in which the elements are equal. We may denote subsets of \(X - \bigtriangleup\) in the form of \(U \times V\) in which \(U\) and \(V\) are open sets of X. Because any ordered pair of the same element can not be in any \(U\) or \(V\), the open sets \(U\) and \(V\) must not have any elements in common. Thus \(U \cap V = \emptyset\).

For any distinct pair of points \(x_1, x_2 \in X\), it is possible to find a \(U\) and \(V\) that contain \(x_1\) and \(x_2\) respectively and hold the condition of \(U \cap V = \emptyset\). Thus, \(X\) is Hausdorff.

\[\tag*{$\blacksquare$}\]

Extras

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