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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.2011-08-23
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    @Zhen Some people use $\ni$ to mean "such that". See [this list](http://www.math.ucdavis.edu/~anne/WQ2007/mat67-Common_Math_Symbols.pdf). I do think it's a bad idea, though.2011-08-23
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    On the other hand, the proof seems fine.2011-08-23
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    @Leon: The edit I did to the question was only cosmetic; I hope you don't mind.2011-08-23

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