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If $X$ is a metric space and $A\subseteq X$ is a closed subset, are the elements of $A$ only limit and interior points? That is $A = int(A)\cup E(A)$, where $E$ is the set of limit points of A?

$x\in X$ is said to be a limit point of $A$ if for all $\varepsilon>0$, the neighbourhood $N_\varepsilon(x)$ contains infinitely many points of $A$.

My guess it that its true, but I'll appreciate if someone gives a proof or a counter example

Mangostino
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  • @markvs No, a point $x$ is a limit point of $A$ if there is a sequence of points in $A\setminus {x}$ converging to $x$. Equivalently, every neighborhood of $x$ contains a point in $A$ other than $x$. – kccu Apr 09 '22 at 16:35
  • Hi @markvs, thank you for you answer. Can you be more specific on what do you mean by 'some kind of compactness'. – Mangostino Apr 09 '22 at 16:46

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No it is not true. Consider, for instance, $\mathbb{Z} \subseteq \mathbb{R}$ with the usual Euclidean metric. Both the interior and the set of limit points of $\mathbb{Z}$ are empty.

I believe it would be correct to say that $A$ is the union of:

  • The interior of $A$,
  • The set of limit points of $A$, and
  • The set of isolated points of $A$.

Since closed sets contain their limit points, it's clear that the above three sets are all contained in $A$. To show $A$ is a subset of their union, it suffices to show that if $x \in A$ but $x$ is not in the interior of $A$ nor a limit point of $A$, then $x$ is an isolated point of $A$.

Since $x$ is not a limit point of $A$, there exists a neighborhood $N_{\epsilon}$ of $x$ containing only finitely many points of $A$, call them $a_1,\dots,a_n$. Let $\epsilon' = \min_{i} d(x,a_i)>0$. Then $N_{\epsilon'}(x)$ contains no points of $A$ other than $x$ itself, and therefore $x$ is an isolated point of $A$.

kccu
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The statement is false. If $p$ is a point of any metric space $X$, then the closure of $\{p\}$ is $\{p\}$ itself. But $p$ is neither an interior point of $\{p\}$ nor a limit point.

José Carlos Santos
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  • Thank you José. What about if I change the hypothesis for $int(A)$ and $E$ to be non empty? ps. I've seen plenty of your answers in this community and I've found them very helpful. – Mangostino Apr 09 '22 at 16:44
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    @Mangostino That also does not work. Consider $[0,1] \cup{2} \subseteq \mathbb{R}$. Its interior is nonempty but $2$ is neither in the interior nor a limit point. You need to include isolated points for your statement to hold. See my answer for more details. – kccu Apr 09 '22 at 16:45
  • It looks like you've extended too far. If $p$ is isolated in $X$, then $p$ is interior. – Brian Moehring Apr 09 '22 at 16:45
  • @BrianMoehring That is not true. The interior is the largest open set contained in $A$. An isolated point is not contained within any open set contained in $A$. – kccu Apr 09 '22 at 16:46
  • @kccu If $p$ is isolated in $X$, then ${p} \subset X$ is open, hence...? – Brian Moehring Apr 09 '22 at 16:50
  • @kccu Since your response is delayed, I'll have to give the punchline to my comment. You seem to be making the same mistake I was pointing out in the answer of overextending. A point $p \in X$ is isolated in $Y\subseteq X$ if and only if ${p}$ is open in $Y$, so $p$ would be interior to ${p}$ in $Y$. In particular, if $Y=X$, then $p$ is isolated in $X$ if and only if $p$ is an interior point of ${p}$. – Brian Moehring Apr 09 '22 at 17:12
  • @BrianMoehring ${p}$ may be open in $Y$, but it is not necessarily open in the underlying metric space $X$. The interior of $Y$ is the union of all subsets of $Y$ that are open in $X$ (not in $Y$). – kccu Apr 09 '22 at 17:27
  • @kccu Yes, but if $Y=X$? Did you simply stop reading before my last sentence? [I invite you to re-read our comment thread so far, starting with mine. Your comments have been a lot of non sequitur to someone talking about isolated points of $X$] – Brian Moehring Apr 09 '22 at 17:33
  • I see what you are saying now, I read your comments too quickly. But my answer and first comment still hold - the points that are isolated in $Y$ (but not in $X$) are not part of the interior of $Y$ nor the set of limit points of $Y$, even when those sets are non-empty. – kccu Apr 09 '22 at 20:56