Given $A \subseteq \mathbf{R}$, let $L$ be the set of all limit points of $A$. One can show that the set $L$ is closed. Now, argue that if $x$ is a limit point of $A \cup L$, then $x$ is a limit point of $A$. This can then be used to prove that $\overline{A} = A \cup L$ (the closure of $A$) is closed, i.e., contains its limit points.

The proof uses the fact that $x$ is either a limit point of $A$ or $L$. How do we know this? I understand that $x$ is the limit of a sequence $(x_n)$ with terms in $A \cup L$, but how does this imply that the sequence is either in $A$ or in $L$? Why can't the sequence have some points in $A$ and some in $L$? Besides this, the argument is straightforward.

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    $\begingroup$ How do you define $\overline A$? $\endgroup$ – tomasz Jun 25 '14 at 7:11
  • $\begingroup$ I edited the question: $\overline{A} := A\cup L$. $\endgroup$ – Randy Randerson Jun 25 '14 at 7:16

Hint: ${\bf R}$ is a metric space, so any limit point is the limit of some sequence. If $x_n$ is a sequence of points in $L\cup A$, then either for infinitely many $n$ we have $x_n\in A$ or for infinitely many $n$ we have $x_n\in L$.

  • $\begingroup$ The question is from Understanding Analysis (Abbott), and metric spaces haven't been introduced yet. $\endgroup$ – Randy Randerson Jun 25 '14 at 7:17
  • $\begingroup$ @fctaylor25: So can you also add the definition of limit point you are using in the question? $\endgroup$ – user99914 Jun 25 '14 at 7:24
  • $\begingroup$ A point $x$ is a limit point of a set $A$ if every $\epsilon$-neighborhood of $x$ intersects the set $A$ in some point other than $x$. $\endgroup$ – Randy Randerson Jun 25 '14 at 7:26
  • $\begingroup$ @fctaylor25: That every limit point is the limit of a sequence is elementary from your definition. I believe you can do it yourself. $\endgroup$ – tomasz Jun 25 '14 at 10:53
  • $\begingroup$ I've already done that. What I don't understand is why the points in the sequence (which clearly must be either from $A$ or from $L$) must be either all from $A$ or all from $L$. It isn't true in general that $\forall x[Q(x) \vee P(x)] \equiv [\forall x Q(x) \vee \forall x P(x)]$. $\endgroup$ – Randy Randerson Jun 25 '14 at 17:28

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