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In Baby Rudin, there is proven that if every infinite subset of a set $E$ has a limit point, then the set is closed:

See Baby Rudin theorem 2.41 for more details and a proof.

I have a quick question. In the part where he proves that E is closed, he mentions that $x_0$ is a limit point of S. I can see that this is true, but I don't understand how it is relevant for the proof. It only seems to matter that $x_0 \notin E$, so it can't be a limit point of $S$ in $E$ regardless whether or not it actually is a limit point.

Am I missing something or am I correct?

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You are right. I guess Rudin wanted to first point out all the limit points of $S$ and after seeing no limit point of $S$ in $E$, he then made the contradict argument.

Tengu
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    Thanks for your answer. However, my question was why Rudin says it is a limit point of S? (You answered for E instead of S) –  Jul 11 '18 at 22:49
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    I understand that ot is a limit point of S, but not how it is relevant for the proof. –  Jul 11 '18 at 22:53
  • @Math_QED I edited my answer. Hope this helps. – Tengu Jul 11 '18 at 23:00
  • Great, it helped. Thanks! –  Jul 11 '18 at 23:01
  • @Matt A Pelto Thanks, I edited. I think the OP wanted to ask about the presentation of the proof. The point is to show that any point in $E$ is not a limit point of $S$. Since we already know that $x_0$ is not in $E$, why did Rudin need to show that "$x_0$ is a limit point of $S$"? That's what I think the question is. – Tengu Jul 11 '18 at 23:34
  • Yes, exactly. Thanks! –  Jul 12 '18 at 07:13
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Rudin is proving by contrapositive here, suppose $E$ is not closed, then there has to be a limit point $x_0$ of $E$ s.t. $x_0 \notin E$, note that $E$ can also be closed when having no limit points at all, i.e. vacuously , and that’s why we need to construct $S \subset E$ s.t. $S$ has a limit point $x_0 \notin S$.

Therefore we can’t say that 'It only seems to matter that $x_0 \notin E$ 'etc..

C_Arietta_C
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