4
$\begingroup$

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?

$\endgroup$
  • 1
    $\begingroup$ 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
2
$\begingroup$

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 $V \subset U$.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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