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Proving projection maps on an infinite product are open

Attempt prove $\pi_\beta:\prod_{\alpha \in \lambda} X_\alpha \rightarrow X_\beta$ is an open map.Where each $X_\alpha$ is a topological space.

Attempt: I was trying to generalize this attempt by someone Projection maps are open but I do not know if this is generalizable as such.

Let $W$ be open in $\prod_{\alpha \in \lambda}X_\alpha$. $W=\bigcup(\prod_{\alpha \in \lambda}U_\alpha)$. Where $U_\alpha$ is open for finitely many $\alpha$ and $X_\alpha$ for the rest of the indices. Then $\pi_\beta(W)=\pi_\beta(\bigcup (\prod_{\alpha \in \lambda}U_\alpha))=\bigcup \pi_\beta (\prod_{\alpha \in \lambda}U_\alpha)=\bigcup U_\beta$ which is open in $X_\beta$. Also is there an index I should use for the union of the elements? Is the link I provided easily generalizable to a situation like this. i.e. Is my solution correct?

Also do I need to assume that the topology defined on the product is necessarily the product topology? I was trying to do this without using excessively burdensome notation involving unions of intersections. Or for the infinite case must we take unions of intersections of subbasic open sets?

Scott Frazier
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1 Answers1

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If $U$ is a basic open set in the product topology (yes that is the topology we must consider on $\prod_{\alpha \in \lambda} X_\alpha$), then $U$ is of the form

$$U= \prod_{\alpha \in \lambda} U_\alpha, \text{ where }U_\alpha = X_\alpha \text{ if } \alpha \notin F$$

where all $U_\alpha$ are open sets and $F \subseteq \lambda$ is finite.

Then $\pi_\beta[U]= U_\beta$ which is either open always.

Fact:

If a function $f: X \to Y$ is open on basic sets for some base $\mathcal{B}$ for $X$, then $f$ is open.

Which holds because $O$ open in $X$ implies that $O = \bigcup_{i \in I} B_i$ for some index set $I$ and where all $B_i \in \mathcal{B}$, so that $$f[O]=f[\bigcup_{i \in I} B_i]=\bigcup_{i \in I} f[B_i]$$

which is by assumption a union of open subsets of $Y$ so open in $Y$ ($f$ is open on the base, and images commute with unions).

So indeed $\pi_\beta$ is open on the standard base and so open on the product. No need to consider subbase elements and the analogous result for subbases instead of bases is false in general anyway (intersections and $f$-images do not commute).

Henno Brandsma
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