Prove that the product of a family of topological vector space is Hausdorff space

Question

I want to prove that: let $\Lambda \neq \varnothing$ and $E_{\alpha}$ topological vector space, for all $\alpha \in \Lambda$. Then $$E:=\prod_{\alpha \in \Lambda} E_{\alpha}$$ is a Hausdorff space if, and only if, $E_{\alpha}$ is Hausdorff space, for all $\alpha \in \Lambda$.

I know, by definition

$$E=\left\{f:\Lambda \rightarrow \bigsqcup_{\alpha \in \Lambda }E_{\alpha}\ ; \ f(\alpha)\in E_{\alpha}\right\}.$$

But I don't know how to use this $E$ characterization to prove what I want.

Answer 1

Assume that $E$ is hausdorff. Observe that if $E$ is hausdorff then each of its subspaces are hausdorff. If $x,y\in E_\alpha$ and $x\ne y$ then $\exists f,g \in E$ such that $f(\alpha)=x$ , $g(\alpha)=y$ and $f(\beta)=g(\beta)$ for $\alpha\ne \beta\in \Lambda$ then observe that the space $$E_\alpha\times\left(\prod_{\alpha \ne\beta\in\Lambda}f(\beta)\right)\cong E_\alpha$$ So $E_\alpha$ is hausdorff.

Now the converse, if $f,g\in E$ and $f\ne g$, then for some $\alpha\in\Lambda$ such that $f(\alpha)\ne g(\alpha)$. $E_\alpha$ is Hausdorff, so there are disjoint open sets $V$ and $W$ in $E_\alpha$ such that $f(\alpha)\in V$ and $g(\alpha)\in W$. Now we shall take $V$, $W$, and use the definition of product topology on $E$ to find disjoint open sets in $E$, one containing $f$ and the other containing $g$. Now $\pi_\alpha:E\to E_\alpha$ be the canonical projection map. Then we have both $\pi_\alpha^{-1}(V)$ and $\pi_\alpha^{-1}(W)$ open in $E$ . Also $$f\in \pi_\alpha^{-1}(V) \text{ and } g\in \pi_\alpha^{-1}(W)$$ and also $V\cap W=\varnothing\implies \pi_\alpha^{-1}(V)\cap \pi_\alpha^{-1}(W)=\varnothing$. Hence we are done.

Answer 2

HINT: If $f,g\in E$ and $f\ne g$, there is some $\alpha\in\Lambda$ such that $f(\alpha)\ne g(\alpha)$. $E_\alpha$ is Hausdorff, so there are disjoint open sets $U$ and $V$ in $E_\alpha$ such that $f(\alpha)\in U$ and $g(\alpha)\in V$. Now use $U$, $V$, and the definition of the product topology on $E$ to find disjoint open sets in $E$, one containing $f$ and the other containing $g$. I’ve left a further hint in the spoiler protected block below; mouse over to see it.

Use the projection map $\pi_\alpha:E\to E_\alpha$.