Cartesian product of closed set is closed

Question

Supposed A and B are closed sets in the Euclidean space. Show that $\mathbf{(A \times B)}$ is closed.

I tried to derive a contradiction but do not know if I am on the right track.

Suppose $(A \times B)$ is not closed. Let $(a_n,b_n)$ be a sequence in $(A \times B)$ where $(a_n,b_n) \to (a,b)$. Because $(A \times B)$ is not closed, it does not contain its limit points. Thus, $(a,b) \not\in (A \times B)$ implying that $a\not\in A$ and $b \not\in B$. Thus $A, B$ are not closed. A contradiction.

Answer 1

The sequential approach is more suited to a "positive" approach, so not by contradiction.

A set $C$ (in any metric space $X$, e.g.) is closed iff for all sequences $(c_n)$ that lie in $C$ and that converge (in $X$) to some $c \in X$, we we have $c \in C$.

So you can reason by starting with an arbitrary convergent sequence for the set you want to show to be closed.

Let $(a_n,b_n) \in A \times B$ (so in particular, all $a_n$ are from $A$ and all $b_n$ are from $B$) and suppose that $(a_n, b_n) \to (a,b) \,(n \to \infty)$ in the total space. Then the fact that we have a Cartesian product (so projections are continuous) implies that $a_n \to a$ and $b_n \to b$. The closedness of $A$ then implies $a \in A$, and the closedness of $B$ implies $b \in B$. So $(a,b) \in A \times B$, as required. That then finished the proof.

It's less convenient to reason from "non-convergence" of sequences, as you see in your own attempt. And in general spaces you could use nets (generalised sequences) to transfer this proof to a more general context, and larger products too.

Answer 2

I presume you mean $A\subseteq X$ closed and $B\subseteq Y$ closed, and that $A\times B$ is supposed to be closed in $X\times Y$.

Choosing a sequential argument prevents the argument from being valid in non-sequential spaces (unless you mean $(a_n,b_n)$ to be a generalised sequence, i.e. indexed by an appropriate directed set). Also:

• you appear, or at least so you declare, to be confusing the negation of "$C$ contains all its limit points", which would be "there is a limit point of $C$ which is not in $C$", with "no limit point of $C$ is in $C$".

• $(a,b)\notin A\times B$ does not imply $a\notin A\land b\notin B$. It implies $a\notin A\lor b\notin B$.

My suggestion is the following:

The maps $\pi_X:X\times Y\to X$ and $\pi_Y:X\times Y\to Y$ are continuous, and $$A\times B=\pi_X^{-1}[A]\cap \pi_Y^{-1}[B].$$

Therefore $A\times B$ is closed.