Let $A$ be a subset of a metric space. $A'$ be the set of limit points of $A$ and $A^i$ be the set of isolated points of $A$.

Show $A \subset A' \cup A^i$.

The picture on my mind is that I draw a disk plus a point outside of the disk, call it a set $A$, if the point falls on the disk, it is automatically a limit point of $A$. If the point is the point that lies outside of the disk, it is of course an isolated point. But I got trouble formalizing it in a rigorous way. Any hint?

EDIT: Definition added:

Limit Point: $A \subset X$, $X$ a metric space. $b \in X$ is a limit point if every neighborhood (topology sense) of $b$ contains a point of $A$ different from $b$.

Isolated point: $b$ is an isolated point of $A$ if there exists a neighborhood around $b$ that contains no point from $A$.

  • $\begingroup$ How are you defining "limit point" and "isolated point"? $\endgroup$ – Nate Eldredge Oct 5 '12 at 22:00
  • $\begingroup$ If you look closely you can see that the definitions are just the negation of each other $\endgroup$ – Robson Aug 28 '18 at 0:56

This is just a matter of using the definitions of isolated point and limit point, so it’s hard to give a useful hint without doing the problem, but I’ll try.

Let $x\in A$; if $x\in A^i$, we’re done with $x$, so suppose that $x\notin A^i$. This means that there is no open set $U$ in $X$ such that $U\cap A=\{x\}$. In other words, if $U$ is open, and $x\in U$, then $U$ also contains at least one other point of $A$. What does that tell you about $x$?

  • $\begingroup$ It must imply that for every neighborhood around $x$ there must be a point in $A$, since if it is not the case, there will exist a neighborhood that contains no point of $A$. $x$ would be an isolated point and contradicts $x \not \in A^i$ $\endgroup$ – Daniel Oct 5 '12 at 22:18
  • $\begingroup$ @jsk: Exactly. And that’s the definition of a limit point: every nbhd of it contains another point of the set. $\endgroup$ – Brian M. Scott Oct 5 '12 at 22:19
  • $\begingroup$ Merci Beaucoup! $\endgroup$ – Daniel Oct 5 '12 at 22:20
  • $\begingroup$ @jsk: You’re welcome! $\endgroup$ – Brian M. Scott Oct 5 '12 at 22:23

A point in $A$ is either isolated, or it is not. You have to show that if it is not isolated, it is a limitpoint. Let $x\in A$ be a non-isolated point. Then $\forall \varepsilon > 0 \exists y \in A : |x-y| < \varepsilon $. Now try to construct a sequence that converges to x.

  • $\begingroup$ Sequence, not series. $\endgroup$ – Brian M. Scott Oct 5 '12 at 22:03
  • $\begingroup$ Fixed, thanks. German here, sorry. $\endgroup$ – Stefan Oct 5 '12 at 22:04
  • $\begingroup$ Kein Problem! I always want to translate Reihe as sequence. $\endgroup$ – Brian M. Scott Oct 5 '12 at 22:18

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