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Let $X$ and $Y$ be topological spaces. Show that if both $X$ and $Y$ satisfy $T_j$ separation axiom with $j = 1, 2, 3$, then $X \times Y$ equipped with the product topology satisfies $T_j$ as well.

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For $T_1$ and $T_2$ this is very easy; have you made any progress on either of those? –  Brian M. Scott Feb 25 '13 at 16:20

1 Answer 1

HINTS: For $T_1$ suppose that $\langle x_0,y_0\rangle,\langle x_1,y_1\rangle\in X\times Y$ and $\langle x_0,y_0\rangle\ne\langle x_1,y_1\rangle$. Then either $x_0\ne x_1$, or $y_0\ne y_1$. If $x_0\ne x_1$, use the fact that $X$ is $T_1$; otherwise, use the fact that $Y$ is $T_1$. The argument for $T_2$ is similar.

For $T_3$ suppose that $\langle x,y\rangle\in X\times Y$, $F\subseteq X\times Y$, $F$ is closed, and $\langle x,y\rangle\notin F$. Then there are open sets $U$ in $X$ and $V$ in $Y$ such that $\langle x,y\rangle\in U\times V$ and $(U\times V)\cap F=\varnothing$ (why?).

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