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If X is Hausdorff, the diagonal [tex]\Delta[/tex] is closed in [tex]X\times X[/tex].
Assume X is Hausdorff. Now, we have two distinct points x, y and disjoint open sets U, V containing x, y, respectively. The basis element [tex]U \times V[/tex] containing [tex](x,y) \in X \times X[/tex] should not intersect [tex]\Delta[/tex] by the assumption given to the Hausdorff property.
For every [tex](x,y) \notin \Delta[/tex], we have a basis element in [tex]X \times X[/tex] containing (x,y), which does not intersect [tex]\Delta[/tex].
Thus, [tex]X \times X \setminus \Delta is open[/tex] and we conclude [tex]\Delta[/tex] is a closed set in [tex]X \times X[/tex] .
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If the diagonal [tex]\Delta[/tex] is closed in [tex]X\times X[/tex], X is Haudorff.
Supppose [tex]\Delta[/tex] is closed in [tex]X\times X[/tex]. Then, [tex]X \times X \setminus \Delta[/tex] is open. Let [tex](x,y) \in X \times X[/tex] and [tex]x \neq y[/tex]. For [tex](x,y) \notin \Delta[/tex], we have a basis element [tex]U \times V[/tex] in [tex]X \times X[/tex] containing (x, y).
We remain to show U and V are disjoint. Suppose on the contrary that U and V are not disjoint. Then, there is an element [tex](z,z) \in X times X[/tex] which belongs to both U and V. Contradicting the fact that x and y are distinct.
Thus, X is Hausdorff.