Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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.

share|improve this question
2  
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?).

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.