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For a metric space $(M,d)$ and a subset $A \subset M$, is it true that the set of limit points of the closure of $A$ is a subset of the limit points of $A$? (I have managed to prove the reverse direction without much difficulty).

I can't think of a counter-example, and am working towards a proof. It seems provable in metric spaces, but I have yet to work out the details, and am still unsure. Could someone verify whether this statement is true or not? If it is true, is it true for topological spaces as well as for metric spaces? I will work the details out myself. If it is not true, what's a counter-example, as I can't think of one?

Note: The definition of closure I am using is: $\bar{A} = A \cup \partial A $ where $ \partial A $ denotes the set of boundary points of $A$.

mechanodroid
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user308485
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3 Answers3

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Let $x$ be a limit point of $\overline A$ that isn't a limit point of $A$. Then there exists an $\epsilon > 0$ such $B_{\epsilon}(x)$ will contains no point of A. But as $x$ is a limit point $\overline A$ it must contain a point $y \in \overline A/A $ so $y$ is a limit point of $A$. Let $0< \epsilon_2 = \epsilon - d(x,y)$. $B_{\epsilon_2}(y)$ must contain a point $z \in A$ as $y$ is a limit point of $A$.

$d(x,z) \le d(x,y) + d(y,z) < d(x,y) + \epsilon_2 = d(x,y) + \epsilon - d(x,y) = \epsilon$.

So $z \in B_{\epsilon}(x)$ which is a contradiction that $B_{\epsilon}(x)$ has no points of $A$.

So there is no limit point of $\overline A$ that isn't a limit point of $A$

fleablood
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  • Ah, I assume then, that this holds for metric spaces, but not necessarily for topological spaces? – user308485 Jan 29 '16 at 00:49
  • No, I'm pretty sure it is true for Topological space by the same argument. Every neigborhood of a limit point of the closure has a point of the closure and that point is either of A or a limit point of A. If a limit point than every of its neighborhood has a point of A. The intersections of the neighborhoods is a neighbor hood of the limit point of A so it has a point of A and is in the arbitrary first neighborhood of the limit point of closure A. So every limit point of A closure is a limit point of A. But didn't ASK about topological spaces. – fleablood Jan 29 '16 at 02:14
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I know it's late, but, just for future reference, I think an alternative intuitive argument (for the metric space) can be as follows.

Suppose $x\in \overline A'$. Then by definition every neighborhood $\mathcal N$ (of radius $r$) of $x$ includes a point $p\in \overline A$ and $p\ne x$. Since $\overline A=A\cup A'$, we have $p\in A$ or $p \in A'$. If $p\in A'$, then every neighborhood $\hat {\mathcal N}$ (of radius $\hat r$) of $p$ must includes $q\in A$. Let $d\triangleq d(x,p)$. With $\hat r=\min\{d, r-d\}$, we have $q\in \mathcal N$, $q\ne x$. So $\mathcal N$ must include a point of $A$ (distinct from $x$). Thus $x\in A'$, proving $\overline A'\subset A'.$

syeh_106
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Supose $(a_n)_{n\in N}$ is a sequence of points of $\bar A$ converging to $a,$ where $a\not \in \{a_n:n\in N\}.$ For each $n\in N$ let $b_n\in A\cap B_d(a_n,d(a,a_n)).$ Then $(b_n)_{n\in N}$ is a sequence of points of $A,$ and $a\not \in \{b_n :n\in N\}.$ We have $\lim_{n\to \infty}d(a,a_n)=0,$ and for each $n\in N$ we have $d(a,b_n)\leq d(a,a_n)+d(a_n,b_n)<2 d(a,a_n).$ So $\lim_{n\to \infty}d(a,b_n)=0.$ So $a$ is a limit point of $A$.

DanielWainfleet
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