# Product of Hausdorff spaces

• Aug 16th 2010, 04:01 PM
Benmath
Product of Hausdorff spaces
I have to prove that if $\displaystyle X_\alpha$ is Hausdorff for each $\displaystyle \alpha \in A$, then $\displaystyle X= \Pi_{\alpha \in A}X_\alpha$ 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?
• Aug 17th 2010, 03:24 AM
HallsofIvy
No. Induction on n shows that "P(n)" is true for all finite n- it does not extend to infinite values.
• Aug 17th 2010, 04:42 AM
yoyoma
Take a pair of distinct points $\displaystyle x, y \in X$. Then for some index $\displaystyle \beta \in A, x_\beta \neq y_\beta$, where $\displaystyle x_\beta, y_\beta \in X_\beta$. Since $\displaystyle X_\beta$ is Hausdorff, we can find disjoint open subsets $\displaystyle U$ and $\displaystyle V$ of $\displaystyle X_\beta$ containing $\displaystyle x_\beta$ and $\displaystyle y_\beta$ respectively.

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