# Why is this set closed?[metric-spaces]

I am reading a note, where part of it is this:

Why is S' closed?

I have tried to argument like this, but I am not able to finish the argument:

Let $\{x_n\}$ be a convergent sequence in S', then it is also a convergent sequence in S, since S is closed the point it converges to must be in S. But how do we know that this point is in S'?

Another way I thought is that the dfiniton of S' might look like a closed ball, however it might not be that sincewe only choose values of S it might be more complicated.

Your idea works. Let $(x_n)$ be a sequence in $S'$, such that $x_n \to x \in \mathbb R^n$. Then $x \in S$, as $x$ is closed, moreover, $\|\cdot - z\| \colon S \to \mathbb R$ is continuous, even Lipschitz, as $$\bigl| \|a-s\| - \|b - s\|\bigr| \le \|a-b\|$$ by the triangle inequality. So $\|x_n - z\| \to \|x-z\|$ and hence $\|x-z\| \le d_S + 1$. So $x \in S'$.