Product of Hausdorff spaces

**Benmath** · August 16th 2010, 04:01 PM

I have to prove that if $X_\alpha$ is Hausdorff for each $\alpha \in A$ , then $X= \Pi_{\alpha \in A}X_\alpha$ is also Hausdorff.

I have proved this for the two dimensional case, and I know if A is finite then it would be true for all of them by induction, but it doesn't say anything about A being finite or infinite. Is this still true by induction if A is infinite?

**HallsofIvy** · August 17th 2010, 03:24 AM

No. Induction on n shows that "P(n)" is true for all finite n- it does not extend to infinite values.

**yoyoma** · August 17th 2010, 04:42 AM

Take a pair of distinct points $x, y \in X$ . Then for some index $\beta \in A, x_\beta \neq y_\beta$ , where $x_\beta, y_\beta \in X_\beta$ . Since $X_\beta$ is Hausdorff, we can find disjoint open subsets $U$ and $V$ of $X_\beta$ containing $x_\beta$ and $y_\beta$ respectively.

Now let $U_\alpha = X_\alpha$ for $\alpha \neq \beta$ , and $U_\beta = U$ , and similarly take $V_\alpha = X_\alpha$ for $\alpha \neq \beta$ , and $V_\beta = V$ . Then $\Pi_{\alpha \in A}U_\alpha$ and $\Pi_{\alpha \in A}V_\alpha$ are disjoint open subsets of $X$ containing $x$ and $y$ respectively, showing that $X$ is Hausdorff.

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