The closed set in the product topology

Question

We know that for the product topology $X\times Y$, the open sets are generated by $U\times V$,where $U,V$ are open in $X,Y$ respectively. I am considering the closed sets in $X\times Y$, are they generated by the closed sets in $X$ and $Y$?

Answer 1

If $A$ is closed in $X$ and $B$ is closed in $Y$, then $A=X\setminus A'$ and $B=Y\setminus B'$ for open sets $A'$ and $B'$. But $$A \times B = (X\setminus A') \times (Y\setminus B') = (X \times Y) \setminus ((A' \times Y) \cup (X \times B')),$$ so $A \times B$ is the complement of the open set $(A' \times Y) \cup (X \times B')$ in $X \times Y$ and thus closed. This proves that the product of two closed sets is a closed set in the product topology.

But that doesn't mean that the products of closed sets form a basis for the closed sets in the product topology (as the products of the open sets form a basis for the open sets). Consider the cofinite topology on $X=Y=\mathbb N$. Then $(\mathbb{N}\setminus\{1\})\times(\mathbb{N}\setminus\{1\})$ is open in the product topology, so its complement $D=(\mathbb{N}\times\{1\})\cup(\{1\}\times\mathbb{N})$ is closed. But $D$ can't be expressed as the intersection of products of closed sets as the closed sets are either $\mathbb{N}$ or finite.

Answer 2

This is a very old question but I want to make a comment here which I personally found helpful. Frunobulax's answer indeed shows that the products of closed sets do not form a basis for the product topology. However, the question as asked was whether they generate the product topology, and the answer to that question is yes (at least in the case of a finite product).

Let $C$ be a closed set in $Z := X \times Y$. Then $Z \setminus V$ is open we can write $$X \setminus C = \bigcup_i U_i \times V_i $$ where $U_i$ is open in $X$ and $V_i$ is open in $Y$. We then have

$$C = Z\setminus \bigcup_i U_i \times V_i = \bigcap_i Z\setminus( U_i \times V_i) $$ which by some basic set theory equals $$\bigcap_i \bigg(\big((X\setminus U_i) \times V_i\big) \cup \big(U_i \times (Y\setminus V_i)\big) \cup \big((X\setminus U_i) \times (Y\setminus V_i)\big) \bigg)$$

But $X\setminus U_i$, $Y\setminus V_i$ are closed sets, so indeed every closed set in $X \times Y$ is an arbitrary intersection of a finite union of products of closed sets.