Limit Points and Infinite Subsets of a Compact Set (Proof Verification)

Question

I'm practicing my proof-writing and was hoping you could let me know if this proof looks good. I would like to know if the proof is incorrect, if there are parts that are overly wordy/complicated, or if I'm missing some element of a proof that is helpful to see, if not strictly necessary.

The Prompt:

Prove that if $E$ is an infinite subset of a compact set $K$, then $E$ has a limit point in $K$.

My Proof:

Suppose that $E$ has no limit point in $K$. Now define $\{V_\alpha\}$ as a collection where every $e \in E$ has one element in the collection, which is an open neighborhood around $e$ that contains no other elements of $E$. Since $E$ is infinite, we can see that $\{V_\alpha\} \cup E^c$ is an infinite open cover of $K$, and since $K$ is compact there is a finite sub-collection of $\{V_\alpha\} \cup E^c$ that still covers $K$, which we will name $G$ and define as:

$G = V_1 \cup \dots \cup V_n \cup E^c$

for $n$ elements of $\{V_\alpha\}$. Briefly note that $E^c$ does not necessarily have to be part of the finite subcover (say if $K = E$), but that this does not change our analysis. Note then, that $V_{n+1} \subset V_1 \cup \dots \cup V_n \cup E^c$. $V_{n+1}$ is an open neighborhood of some $e \in E$, and since $e \notin E^c$, it must be an element of some $V_k$ for $k$ between $1$ and $n$, which contradicts our assumption that each element of $\{V_\alpha\}$ contained no other elements of $E$. Thus $E$ has a limit point in $K$.

Answer 1

You have the right idea, but you get off the track a bit at the end. I’d do it something like this:

Suppose that $E$ has no limit point in $K$. Then $E$ is closed, and each $e\in E$ has an open nbhd $V_e$ such that $V_e\cap E=\{e\}$, so $\mathscr{V}=\{V_e:e\in E\}\cup\{K\setminus E\}$ is an open cover of $K$. Note that if $e\in E$, $V_e$ is the only member of $\mathscr{V}$ that contains $e$, so any subcover of $\mathscr{V}$ must include $V_e$. This is true for every $e\in E$, so any subcover of $\mathscr{V}$ must include all of the sets $V_e$ and therefore cannot be finite, contradicting the compactness of $K$.

Alternatively, you could conclude by saying that $\mathscr{V}$ must have a finite subcover, since $K$ is compact, but it must include all of the sets $V_e$ with $e\in E$, contradicting the assumption that $E$ is infinite. What the argument really shows is that $E$ can be finite, or $K$ can be compact, but not both at once.

There are a couple of very specific problems with what you wrote. You’ve used $G$ for two different things: first you say that it’s the finite subcover, which means that it’s the collection $\{V_1,V_2,\ldots,V_n,K\setminus E\}$, but then you say that it’s the union of the sets in this collection. A little later you refer to a set $V_{n+1}$ that is never defined. What you mean, I think, is that if we take any one of the nbhds $V_\alpha$ other than $V_1,\ldots,V_n$ and call it $V_{n+1}$, then we can argue that $V_{n+1}\subseteq V_1\cup\ldots\cup V_n$ and hence that if $e$ is the point of $E$ of which $V_{n+1}$ is an open nbhd, then $e\in V_k$ for some $k\in\{1,\ldots,n\}$, contradicting the choice of the nbhds $V_\alpha$. All of this is easier if you index the nbhds by points of $E$ as I did.

Answer 2

You can conclude quickly once you have your sets $\{V_\alpha\} \cup E^c$, because $\{V_\alpha\}_{\alpha\in I}$ is by hypothesis infinite, and by construction, $\{V_\alpha\} \cup E^c$ has no finite subcover: if any $V_{\alpha}$ is removed from the collection, then the corresponding point $e_{\alpha}\in V_{\alpha}$ is not covered by any element in the remaining collection of sets.