The use of compactness in "compact plus closed equals closed" for metric spaces

Question

I have seen some proofs/explanations for why "a compact set plus a closed set is a closed set". Some examples are: here, here, here, and here. However, I am confused about how we "leverage" compactness in these proofs. I will lay out the definition and proof (attempt) below and then ask my main questions.

Let $X$ be a compact subset of $\mathbb{R}^n$ and $Y$ be a closed subset of $\mathbb{R}^n$ and let $\mathbb{R}^n$ be a metric space with metric $d( \cdot, \cdot )$. Then, we can define: $X + Y := \{ x + y : x \in X, y \in Y \}$ and show that $X + Y$ is closed.

I believe I can show that $X + Y$ is closed with the following "proof" (please feel free to critique the proof and give pointers -- I am still a novice!):

Suppose there is a sequence $\{z_n\}$ that converges to a point $z \in \mathbb{R}^n$. By definition, $z_n = x_n + y_n \in X + Y$ for $x_n \in X$ and $y_n \in Y$. Since $X$ is compact, there is a convergent subsequence of $x_n$ in $X$, meaning that $\{x_{n_k} \} \rightarrow x \in X$. Thus, we can also write a subsequence of $y_n$, as $y_{n_k} = x_{n_k} + y_{n_k} − x_{n_k} = z_{n_k} − x_{n_k}$, which converges to $z − x$, i.e., $\{y_{n_k} \} \rightarrow z − x$. This is due to the fact that subsequences of convergent sequences also converge to the same limit point, meaning $\{ z_{n_k} \} \rightarrow z$. Moreover, for this argument, we require the fact (not proved here) that $\lim_{n_k \rightarrow \infty} \left( a_{n_k} + b_{n_k} \right) = \lim_{n_k \rightarrow \infty} \left( a_{n_k}\right) + \lim_{n_k \rightarrow \infty} \left(b_{n_k} \right)$ if $a_{n_k} \in A \subset \mathbb{R}^n$ and $b_{n_k} \in B \subset \mathbb{R}^n$. Since $X$ is closed and we know that for closed sets, all convergent sequences converge to a point contained in the set, then $z−y \in X$. We therefore have that $z=(z−x)+x \in X+Y$,which means that the set $X + Y$ is closed.

In the proof, I "use" compactness of $X$, but I'm not sure how the "use" of compactness can't be substituted for a "use" of closedness. It seems (link) "closed plus closed" isn't necessarily closed, but I am failing to understand what mechanistically in the proof is different if $X$ is only closed, because we also know that every subsequence of a convergent sequence is also convergent.

So, in essence, I'm saying that we know we can pick any subsequence $\{ x_{n_k} \}$ of $\{ x_n \}$, and we know it will converge to $x$. If we also take a subsequence $\{ y_{n_k} \}$ of $\{ y_n \}$ for the same indices $n_k$, shouldn't they both converge to points in their respective subspaces?

Thank you very much for your help!

P.S. My background is in metric spaces (I have not yet taken a topology course), so if we could please avoid topological arguments and use metric space-based arguments, I'd greatly appreciate it -- it will help my intuition.

Answer 1

Recall that a set $A$ is closed if and only if every convergent sequence $\{x_n\}_{n \in \mathbb{N}} \subseteq A$ converges to some point $x \in A$. In your proof, you correctly assume that the sequence $\{z_n = x_n + y_n\}_{n \in \mathbb{N}}$ is convergent. But notice that the sequence $\{x_n\}_{n \in \mathbb{N}}$ need not necessarily converge. As you likely know, a set $A$ in a metric space $(X,d)$ is compact if and only if it is sequentially compact, i.e. every sequence $\{x_n\}_{n \in \mathbb{N}}$ has a subsequence converging to a limit $x \in A$. If we lose the compactness assumption, then the sequence $x_n$ may not have a convergent subsequence, and so the closedness does not help us.

Consider, for instance $X = Y = \mathbb{N}$. Note that $\mathbb{N}$ is closed vacuously since it has no cluster points. Then the sequence $\{n + n\}_{n \in \mathbb{N}} \subseteq X + Y$ has no convergent subsequence. Please see this post for some nice counter-examples regarding the Minkowski sum of two closed sets. This answer suggests taking $X = \mathbb{Z}$ and $Y =\sqrt2\mathbb{Z}$, which is a simple counter-example.