TITLE: The closed set in the product topology
QUESTION [7 upvotes]: 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 $Y$?

REPLY [10 votes]: 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.