How may we cahracterize the closed sets of the product topology?

Question

I was wondering if there is a way to characterize the closed sets in a product topology $X:=\prod_i X_i$. These is what I currently know:

1. Given a collection $\{C_j\}$ so that $C_j$ is closed in $X_j$ and $C_j\ne X_j$ for finitely many $j$, we have that $\prod_j C_j$ is closed in $X$.

2. Naturally, arbitrary intersection and finite unions of the sets described in $(1)$ are closed.

Are these all the possible closed sets in $X$? If so, how would one prove it, if not, what other conditions can be included so as to characterize all closed sets in $X$?

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Frunobulax's answer in this post shows that not all closed sets in $X$ are the intersections of sets described in $(1)$, although he did not mention finite unions.

Answer 1

What Frunobulax shows is that the products of closed sets do not form a basis for the product topology on $X$. However, for finite products, it is still true that these sets generate the product topology on $X$ when we consider finite products. So yes, for finite products, the arbitrary intersections of finite unions of products of closed sets give all of the closed sets for $X$.

I will give a proof for the product of two spaces, but it's easily generalised to finitely many spaces. Let $V$ be a closed set in $X := X_1 \times X_2$. Then $X \setminus V$ is open we can write $$X \setminus V = \bigcup_i U_i \times W_i $$ where $U_i$ is open in $X_1$ and $W_i$ is open in $X_2$. We then have

$$V = X\setminus \bigcup_i U_i \times W_i = \bigcap_i X\setminus( U_i \times W_i) $$ which by some basic set theory equals $$\bigcap_i \bigg(\big((X_1\setminus U_i) \times W_i\big) \cup \big(U_i \times (X_2\setminus W_i)\big) \cup \big((X_1\setminus U_i) \times (X_2\setminus W_i)\big) \bigg)$$

This finishes the proof.

For arbitrary products, you just need to restrict the statement so that all but finitely many of the multiplicands equal $X_i$, as you have done in your question.