$\newcommand{\seq}[1]{\left\{#1_n\right\}}$ 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.

  • $\begingroup$ 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

1 Answer 1


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 $P$.

  • $\begingroup$ How does $E$ uncountable imply that $P$ is nonempty? $\endgroup$
    – Silent
    Jan 9, 2018 at 10:00
  • 1
    $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.