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$\newcommand{\seq}[1]{\left\{#1_n\right\}}$ Question 27 says

Suppose $E \subset \mathbb{R}^k$, $E$ is uncountable, and let $P$ be the set of all condensation points of $E$. Prove that $P$ is perfect and that at most countably many points of $E$ are not in $P$.

I seemed to be able to prove the two statements without using the fact that $E$ was uncountable, is this extraneous detail in the question? - it is implicit if we prove $P$ is perfect and is non-empty.


That $P' \subseteq P$ is quite straightforward. My general approach to prove $P \subseteq P'$ was:

  • Assume (to get a contradiction) for a given nbhd $N$ of $p \in P$ that $\not\exists$ any other $q\not=p \in N \cap P$, that is, for every point $e \in N$ $\exists$ a nbhd $N_e$ with at most countably many elements of $E$.
  • Given a countable base $\seq{V}$, $\forall e \in N$ $\exists V_e \in \seq{V} : e \in V_e \subset N_e$
  • $N \subset \bigcup_{e \in N}V_e$, but this is the union of a countable number of $\seq{V}$, each containing an a countable number of elements of $E$, so $E \cap N$ countable contradicting $p \in P$.

Thus $P$ is perfect.

A similar method of counting proves the second statement.


As a mere student of the subject I'm slightly perturbed by the possibility of unnecessary restrictions in the question. I'm used to every concept mattering.

Is this due to a misunderstanding of the definition of perfect sets perhaps? I wonder, since the corollary in question 28 seems to have nothing to with the question, and depends solely on the fact that non-empty perfect subsets of $\mathbb{R}^k$ are uncountable.

Thanks for any clarification.

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  • $\begingroup$ Look closely at the definition of a condensation point. $\endgroup$ Oct 16, 2015 at 11:57
  • $\begingroup$ That every nbhd has uncountably many elements of $E$. But if $E$ has no condensations points then $P$ is empty and perfect. If $P$ not empty then $E$ uncountable just follows from the definition and isn't required in the question? $\endgroup$ Oct 16, 2015 at 12:00

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You are right, the uncountability of $E$ is not necessary to prove any of the conclusions mentioned in the exercise, and your proof of those facts is correct (not excluding the possibility of a glitch in the part you left out, but that's unlikely). As you say, the only thing the uncountability of $E$ is required for is the nonemptiness of $P$.

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  • $\begingroup$ How does $E$ uncountable imply that $P$ is nonempty? $\endgroup$
    – Silent
    Jan 9, 2018 at 10:00
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    $\begingroup$ Via the second countability of $\mathbb{R}^k$. For every $x \notin P$, choose an open neighbourhood $U_x$ of $x$ such that $U_x \cap E$ is countable. Let $U = \bigcup_{x\notin P} U_x$. Since $U$ is second countable (as a subspace of a second countable space), it is Lindelöf, so the open cover $\{ U_x : x \notin P\}$ contains a countable subcover. But then $$E \cap U = \bigcup_{n = 1}^{\infty} \bigl(E \cap U_{x_n}\bigr)$$ is countable, so $E\setminus U$ is nonempty. But $(\mathbb{R}^k\setminus P) \subset U$ [in fact, the two are equal], so $E\setminus U \subset E\cap P$. $\endgroup$ Jan 9, 2018 at 10:11

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