Let $\{X_{\alpha}\}_{\alpha\in A}$ be a family of topological spaces. The product topology on $X=\prod_{\alpha\in A}X_{\alpha}$ is the weak topology generated by the coordinate maps $\pi^{}_{\alpha}:X\to X_\alpha$. The following is an exercise about open sets in $X$ endowed with the product topology:

If $A$ is infinite, a product of nonempty open sets $\prod_{\alpha\in A}U_{\alpha}$ (where $U_\alpha$ is open in $X_\alpha$) is open in $X$ iff $U_\alpha=X_\alpha$ for all but finitely many $\alpha$.

Observing that the sets of the form $\bigcap_1^n \pi^{-1}_{\alpha_j}(U_{\alpha_j})$ form a base for the product topology, I can show the following direction:

If the open sets $U_\alpha=X_\alpha$ for all but finitely many $\alpha$, then the product of nonempty open sets $\prod_{\alpha\in A}U_{\alpha}$is open in $X$.

Could anyone suggest an idea for the other direction?

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    $\begingroup$ Consider sets of the form $\Pi_{\alpha} U_{\alpha}$, where $U_{\alpha} = X_{\alpha}$ for all but finitely many $\alpha$. Use the fact that these sets form a basis for the product topology, and so if if $V_{\alpha} \neq X_{\alpha}$ for infinitely many $\alpha$, then $\Pi_{\alpha} V_{\alpha}$ contains no basis set, which means that it cannot be open. $\endgroup$ – Shalop Jun 30 '15 at 1:53

You need to prove the converse now, i.e, if $\prod_{\alpha} U_{\alpha}$ is open, then $U_{\alpha} = X_{\alpha}$ for all but finitely many $\alpha \in A$.

We will argue by contradiction, so suppose $U_{\alpha} \neq X_{\alpha}$ for infinitely many $\alpha$, and also that $\prod_{\alpha} U_{\alpha}=:U$ is open.

Since any open set is a union of basis sets, it follows that any open set contains a basis set. Therefore, we can find some set of the form $V:=\bigcap_1^n \pi_{\alpha_j}^{-1}(V_{\alpha_j})$ such that $ V \subset U$, where $\alpha_1,...,\alpha_n \in A$ and $V_{\alpha_j} \subset X_{\alpha_j}$ are open.

Now, since $\{\alpha \in A: X_{\alpha} \neq U_{\alpha} \}$ is an infinite set, there exists some $\alpha_0 \notin \{\alpha_1,...,\alpha_n\}$ such that $U_{\alpha_0} \neq X_{\alpha_0}$. Choose a point $x_{\alpha_0} \in X_{\alpha_0} \backslash U_{\alpha_0}$.

Now let $y = (y_{\alpha})_{\alpha \in A} \in \prod_{\alpha} X_{\alpha}$ be any point such that $y_{\alpha_0} = x_{\alpha_0}$, and such that $y_{\alpha_j} \in V_{\alpha_j}$ for all $1 \leq j \leq n$.

Then $y \notin U$ since $y_{\alpha_0} \notin U_{\alpha_0}$. However, $y \in V$ since $y_{\alpha_j} \in V_{\alpha_j}$ for all $1 \leq j \leq n$. Thus $y \in V \backslash U$, which contradicts the fact that $V \subset U$.


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