I have the following question regarding metric spaces. The problem is from Fred Croom's book "Principles of Topology".
Show that a finite subset of a metric space has no limit points and is therefore a closed set.
This is what I understand. There has to be infinitely many elements of the subset inside a neighborhood of some point $x$ in the subset, no matter how small the neighborhood is. That being said, this is my attempt.
Proof: We seek to prove a finite subset of a metric space $(X,d)$ has no limit points. Let $A \subset X$ be finite, then for each $x \in X$ there are only finitely many $p\neq x$ in $A$, and thus finitely many distances $d(p,x)$. Thus if $\delta$ is smaller than the least of them then there is no $p \in A$ such that $0 < d(p,x)<\delta$. QED
My only questions is whether or not this is sufficient to claim that the subset is closed.
I want to thank you for taking the time to read this question. I greatly appreciate any assistance you provide.
Best Answer
Your proof is fine. The only bit to complain is that near the end of the proof you write $p\in A$ where it should read $p\in A\setminus\{x\}$.
By the way: A point $x$ is a limit point of a set $A$, when each neighborhood $U$ of $x$ contains at least one point of $A$ distinct from $x$. In metric spaces this is the same as requiring that infinitely many points of $A$ are lying in $U$, but in general spaces there is a difference, though only in "ugly" spaces.