Is product of two closed sets closed?

$A$ is closed in $X$, so $A^c$ is open, likewise for $B$ in $Y$.

Moreover, $A^c \times Y$ and $X \times B^c$ are both open in $X \times Y$.

Thus $$(A \times B)^c = (A^c \times Y) \cup (X \times B^c) $$ is open. Hence $A \times B$ is closed.

Product topology: $X×B^c = π_2^{-1}(B^c)$ is open in $X×Y$, and $A^c×Y = π_1^{-1}(A^c)$ is open in $X×Y$. And $(A×B)^c = ?$

Recall that a subset $A$ of a topological space is closed iff limit of every convergent net with terms in $A$ also belongs to $A$.

In metric spaces (and, more generally, in sequential spaces) nets can be replaced by sequences in the above characterization. (So if this context is sufficient for you, the argument below works just the same - and can be followed without any knowledge about nets.)

Let $A\subseteq X$ and $B\subseteq Y$ be closed subsets of topological spaces $X$, $Y$, respectively.

• Let us assume that a net $(a_d,b_d)_{d\in D}$ converges to $(a,b)$ in $X\times Y$ and $(a_d,b_d)\in A\times B$.
• This implies that $a_d\to a$ in $X$, and $b_d\to b$ in $Y$. (For example, using continuity of the projections. If $p_1 \colon X\times Y \to X$ denotes the projection $p_1(x,y)=x$, then we get $p_1(a_d,b_d)\to p_1(a,b)$ which is exactly $a_d\to a$.)
• From this we get $a\in A$ and $b\in B$, i.e., $(a,b)\in A\times B$. (Since both $A$, $B$ are closed.)

Since this shows that $A\times B$ is closed under limits of nets, the above characterization gives us closedness of $A\times B$.