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 $X \times Y$ ?

• I think we did have the same question some time ago Aug 30, 2013 at 8:07
• @DominicMichaelis It's the opposite direction. Aug 30, 2013 at 8:12
• Please, try to make the title of your question more informative. E.g., Why does $a<b$ imply $a+c<b+c$? is much more useful for other users than A question about inequality. From How can I ask a good question?: Make your title as descriptive as possible. In many cases one can actually phrase the title as the question, at least in such a way so as to be comprehensible to an expert reader. Aug 30, 2013 at 8:26
• And about the reverse, if the product is closed so each one is closed? Thank you so much Feb 16, 2021 at 1:43
• @Quiet_waters You can see this post among the linked questions: If product of two sets $A\times B$ is closed, are $A$ and $B$ closed? Feb 16, 2021 at 7:20

$A$ 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.

• why did you not include $A^C\times B^C$? Dec 15, 2021 at 6:20
• @Killaspe Because $A^C\times B^C \subseteq A^C\times Y$ or also $A^C\times B^C\subseteq X\times B^C$.
– P-A
Feb 11, 2022 at 16:32

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 $$(A×B)^c = ?$$

• maybe say somewhere that $\pi_1$ and $\pi_2$ are the projections to the first ( to the second) component Aug 30, 2013 at 8:20
• Why do $A$ and $B$ need to be non-empty? Aug 31, 2013 at 0:08
• @StefanSmith I don’t know. Maybe you don’t. I probably was thinking about the converse direction, since it was linked. Aug 31, 2013 at 19:49

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 $A\times B$.

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.

$$A$$ 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.

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.