# Verification of the proof of the theorem: “E is infinite subset of a compact set K, prove: E has a limit point in K”

I am self studying real analysis using Rudin "Principles of Mathematical Analysis". I have come up with the proof for one of the theorems, which is slightly different to Rudin one. Can you please help me to confirm or disprove it? Thanks a lot!

Theorem: $E$ is infinite subset of a compact set $K$, then $E$ has a limit point in $K$.

Proof:

• Assume $\forall x \in K$, $x$ is not limit point in $E$. Then $\forall x \in K$, $\exists$ neighborhood $V(x)$ of $x$ which includes up to one member of $E$.

• But $K$ is compact so $\exists$ an open cover of $K$ such that $K \subset \cup_{i=1}^{n}V(x_i)$, where $n$ is natural number. This cover contains at most $n$ members of $E$ and hence can not be an open cover for $E$, which causes contradiction, since $E \subset K$.

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About your notation: Please don't use symbols like that. Especially like in ‘$\exists$ open cover of $K \ni K$’ — $\in$ and $\ni$ are not just abbreviations of ‘in’ and ‘containing’ but have a very specific meaning. – Zhen Lin Aug 23 '11 at 2:57
@Zhen Some people use $\ni$ to mean "such that". See this list. I do think it's a bad idea, though. – Dylan Moreland Aug 23 '11 at 3:24
On the other hand, the proof seems fine. – Dylan Moreland Aug 23 '11 at 3:31
@Leon: The edit I did to the question was only cosmetic; I hope you don't mind. – Arturo Magidin Aug 23 '11 at 3:44

That works (if I were grading, I would probably want to see at least some explanation for the existence of $V(x)$, but I agree that it is straighforward to establish that such a neighborhood exists).
That said, this result can be proven directly, without having to rely on a proof by contradiction (at least, assuming the Axiom of Choice): since $E$ is infinite, it contains a countable subset $\{e_i\mid i\in\mathbb{N}\}$, with $e_i\neq e_j$ if $i\neq j$. This is a sequence of elements of $K$; since $K$ is compact, any sequence contains a converging subsequence, so there is a subsequence of this sequence that converges in $K$. The limit of that sequence is a limit point of $E$.
The proof given by the OP is o.k. If you want to avoid a contradiction you can argue as follows: For any $x\in K$ put $V(x):=K$ if $x$ is a limit point of $E$ and choose an open neighborhood $V(x)$ containing at most one point of $E$ otherwise. Since $K$ is compact there is a finite set $\{x_1,\ldots, x_n\}\subset K$ with $K\subset \bigcup_1^n V(x_k)$, and since $E$ is infinite at least one of the $x_k$ has to be a limit point of $E$.