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$E'$ is the set of all limit points of $E$.

Proving $E'$ is closed:

$E$ is finite

1) If $E$ is finite, then $E$ has no limit points and hence $E'=\emptyset$ and hence $E'$ is closed.

What if $E$ is infinite?

If $x\in E'$ then $x$ is a limit point of $E$.

Proof

Assume $x\in E'$ and $x$ is not a limit point of $E$.

And above I am lost, how do I prove $E'$ is closed? I know it has to be obviously(since it is literally defined as a set of limit points...).

Source: Rudin's Principles of Mathematical Analysis, Page 43, question 6.

Understand
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  • I believe this is from Rudin, yes? In any case, it is always good practice to disclose the source. You should also declare which definitions you are using. – Alp Uzman Feb 23 '15 at 09:37

3 Answers3

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for metric spaces:

Just show that $\complement E'$ is open. If $y \notin E'$, then there must be an $r>0$ such that $B(y,r)$ does not contain any element of $E$ (since $y$ is not a limit point).

Now show that $B(y,r)$ cannot contain any element of $E'$

Blah
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    Should your $E$ up there be $(E')^c$ [or as you have written $\complement E'$]? Or it is fine since $E\subset (E')^c = X/E'$? Where $X$ is the universe – Understand Feb 24 '15 at 03:15
  • No, it is $X \setminus E'$. If this set is open, then you are done. Just draw a picture: If $B(y,r)$ contains an element of $E'$, then this would be a limit point, so there is some Element $e$ of $E$ near $y$ – Blah Feb 24 '15 at 06:53
  • Thank you, yes I asked a second question to clarify this after writing that comment, and then after that was answered I understood your answer completely and accepted it. Thanks again! – Understand Feb 25 '15 at 03:54
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Suppose $\{L_i\}$ has limit $L$, and the sequence $a_{ij}$ (as $j$ varies) has limit $L_i$ for all $i$.
Take a sequence of open sets $B_i$ whose limit is $L$, and assume $B_{i+1}\subset B_i$. Take a subset of the sequences, so that now $L_i\in B_i$.
Let $C_i$ be an open set with $L_i\in C_i\subset B_i$.
Take one element from each sequence $a_{ik_i}$ so that $a_{ik_i}\in C_i$.
Then $a_{ik_i}\in B_i$, and the sequence $b_i=a_{ik_i}$ tends to $L$.

Empy2
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  • I thought of this at first as well, but Rudin uses punctured neighborhoods for limit points. So we need a slight adjustment for this proof I think. – Alp Uzman Feb 23 '15 at 10:03
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As an alternative to Blah's outline, here is a proof directly from the definition of a limit point:

Let $p\in E''$. Then by definition $\forall\epsilon,\exists q\in B_\epsilon (p)\cap E'\setminus\{p\}$. Let $\epsilon>0$ and take a limit point $q\in B_\epsilon (p)\setminus\{p\}$. Reciting the definition of a limit point (of $E$ this time), we have that $\forall\epsilon',\exists r\in B_{\epsilon'} (q)\cap E\setminus\{q\}$.

I leave it to you to pick $\epsilon'$ so that it is guaranteed that we can find an $r\in E$ for which $r\in B_\epsilon(p)\setminus\{p\}$ holds.

Hint: Draw a picture.

Alp Uzman
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