Is product of two closed sets closed? If $A$ is closed in $X$ and $B$ is closed in $Y$, 
does it follow that $A \times B$ is closed in $X \times Y$ ?
 A: 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 $A\times B$.
A: $A$ 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.
A: Let $\pi_i$ denote the projection on $i$-th coordinate
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 $(A×B)^c = ?$
A: If we can find an open neighborhood $U_x \subset {(A \times B)}^c$ given any arbitrary $x \in {(A \times B)}^c$ this would imply that ${(A \times B)}^c$ is open and thus $(A \times B)$ would be closed.
Given $x \in (A \times B), \; where \; x=(x_1,y_1) \; s.t. \; x_1 \in X \; and \; y_1 \in Y.$ Then $x_1 \in X-A$ which is open and $y_1 \in Y-B$ which is also open. Therefore $\exists \; \; {U_x}_1 \in X-A \; and \; {U_y}_1 \in Y-B,$ open neighborhoods about $x_1$ and $y_1$ respectively.
${U_x}_1 \times {U_y}_1$ is an open set in $X \times Y \; and \; {U_x}_1 \times {U_y}_1$ is an open neighborhood of our arbitrary point i $x \in {(A \times B)}^c.$ Therefore, ${(A \times B)}^c$ is open implying that $(A \times B)$ is closed.
A: $A$ 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.
A: Given $A$ and $B$ are closed sets, let $(x,y)$ be a limit point of $A \times B$. Then there exists a sequence $\{(x_n, y_n)\}$ in $A \times B$ such that $\{(x_n, y_n)\}$ converges to $(x, y)$.
Since $\{(x_n, y_n)\}$ is a sequence in $A \times B$, $\{x_n\}$ is a sequence in $A$ and $\{y_n\}$ is a sequence in $B$.
Also, since $\{(x_n, y_n)\}$ converges to $(x, y)$, $\{x_n\}$ converges to $x$ and $\{y_n\}$ converges to $y$.
We obtain a sequence $\{x_n\}$ in $A$ converging to $x$. This implies that $x$ is a limit point of $A$ and hence it is in the closure of $A$. Since $A$ is closed, $A$ equals it closure, i.e. $x$ is in $A$.
Similarly, we can prove $y$ is in $B$.
Therefore, $(x, y)$ is in $A \times B$, i.e. $A \times B$ contains all its limit points.
Hence $A \times B$ is closed.
