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From Rudin's Principles of Mathematical Analysis (Chapter 2, Exercise 6)

Let $E'$ be the set of all limit points of a set $E$. Prove that $E'$ is closed.

I think I got it but my argument is a bit hand wavy:

If $x$ is a limit point of $E'$, then every neighborhood of $x$ contains some $y\in E'$, and every neighborhood of $y$ contains some $z\in E$. Therefore every neighborhood of $x$ contains some $z\in E$, and so $x$ is a limit point of $E$. Then $x\in E'$, so $E'$ is closed.

The thing that's bugging me is the leap from one neighborhood to another. Is this formally correct?

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  • $\begingroup$ You can see an answer here. $\endgroup$ – hrkrshnn Jun 8 '14 at 8:40
  • $\begingroup$ If your space is Hausdorff the leap is easy $\endgroup$ – user135229 Aug 1 '15 at 17:16

10 Answers 10

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Your argument is correct, but incomplete: All you need to finish it is to ensure that you can find a neighborhood of $y$ contained in the neighborhood of $x$ that you began with (any will do, since all contain elements of $E$). Use the triangle inequality to find an appropriate radius for the neighborhood of $y$.

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  • $\begingroup$ Let's see if I got this straight: Let $r'=d(x,y)<r$, for some $y$ in $x$'s neighborhood of radius $r$. Then consider the neighborhood of $y$ with radius $r-r'$. This one is contained in $x$'s neighborhood and contains some $z\in E$. Is this correct? Thanks a lot! $\endgroup$ – F M Dec 29 '10 at 5:58
  • $\begingroup$ @FernandoMartin the neighborhood of $y$ with radius $r - r'$ contains $x$'s neighborhood, not the other way around. $\endgroup$ – Don Larynx Nov 5 '13 at 17:51
  • $\begingroup$ @DonLarynx: I'm pretty sure it is correct. $\endgroup$ – F M Nov 6 '13 at 2:44
  • $\begingroup$ Just a question: I managed to find a proof that the set a limitpoints of a set is closed, but this under the extra condition that singletons are closed. Is the statement true in general? The answers on this page are all dealing with metric spaces. $\endgroup$ – drhab Aug 8 '14 at 13:06
  • $\begingroup$ Never mind. I have found math.stackexchange.com/q/490968/75923 $\endgroup$ – drhab Aug 8 '14 at 13:22
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Let $\hat S$ be the set of all limit points of $S$. Prove that $\hat S$ is a closed set.

Proof: Suppose $x_0$ is a limit point of $\hat S$. Then given $\varepsilon > 0$ there exists $x \in \hat S$ with $\vert x - x_0\vert < \frac\varepsilon2$. Now $x \in \hat S$ is a limit point of $S$ so there exists $x' \in S$ such that $\vert x' - x \vert < \frac\varepsilon2$. Now $$\vert x' - x_0 \vert = \vert x' - x + x - x_0 \vert \leq \vert x' - x \vert + \vert x - x_0 \vert < \frac\varepsilon2 + \frac\varepsilon2 = \varepsilon$$ Thus $x_0$ is a limit point of $S$ and by definition is contained in $\hat S$. We have shown that $\hat S$ contains all of its limit points. By theorem that states that a set is closed if and only if it contains all its limit points, we have just shown that $\hat S$ is a closed set.

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    $\begingroup$ You should use the metric form $\mathrm{d}$ and not $|\cdot|$. $\endgroup$ – Steven Gamer Jan 22 '13 at 14:27
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    $\begingroup$ It is more general; you may have metric spaces that do not have addition or subtraction defined. However, if take merely as a notational convenience, then that is fair enough. $\endgroup$ – Hayden Nov 20 '13 at 23:32
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    $\begingroup$ I think that there is something missing in your proof. In the definition of a limit point you need $x' \neq x_0$ and your proof doesn't ensure that. $\endgroup$ – Ludolila Mar 7 '14 at 19:25
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    $\begingroup$ Is there any way to ensure that they are not equal? $\endgroup$ – CuriousKid7 Nov 2 '16 at 0:29
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    $\begingroup$ We have that |x′−x| is strictly less than min(ε/2, |x−$x_0$|). This will ensure that x′ ≠ $x_0$. $\endgroup$ – CuriousKid7 Nov 2 '16 at 0:52
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Let $x$ be a limit point of $E'$, and let $\varepsilon >0$. Then (by definition) there exists $y\in E'$ such that $0<d(x,y)<\frac{\varepsilon}{2}$. Since $y\in E'$ there exists $z\in E$ such that $0<d(y,z)<d(x,y)$ (here one uses $d(x,y)$ as the epsilon from the definition of a limit point).

By triangle inequality we have $d(x,z)\leq d(x,y)+d(y,z)<\varepsilon$, and note that indeed $x\neq y$. So (by definition) $x$ is a limit point of $E$. That is $E'' \subseteq E'$, which proves that $E'$ is closed.

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If x is a limit point of E', then $$\forall x \forall r > 0 ( d(x, y) < r \to \exists y \in E' )$$ There exists a positive real number h such that $d(x, y) = r - h$.

y is a limit point of E, then $$\forall y ( d(y, z) < h \to \exists z \in E )$$

So, $$\forall x \forall r > 0 ( d(x, z) < d(x, y) + d(y, z) = r \to \exists z \in E )$$ Thus x is a limit point of E, $x \in E'$.

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An equivalent definition of a neighborhood of $x$ is that it is an open set containing $x$. If you adopt this definition, then your proof is perfect.

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    $\begingroup$ did you mean, "a set that contains an open set containing $x$"? If not, what is the definition of nbd that you take to begin with? $\endgroup$ – Arturo Magidin Dec 28 '10 at 22:07
  • $\begingroup$ @Arturo. Edited my answer. I meant some authors define "neighborhood" this way. $\endgroup$ – TCL Dec 28 '10 at 22:11
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    $\begingroup$ What do you mean by “equivalent” here? $\endgroup$ – Carsten S Nov 24 '13 at 14:58
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I thought the following, could someone validate if I'm not mistaken or mixed up? ;)

Proving that $A'$ is closed is equivalent to prove that $\mathbb{R}^{n}\setminus A'$ is open

Let $x\in\mathbb{R}^{n}\setminus A'\iff\exists B_{r}\left(x\right):B\cap A\setminus\left\{ x\right\} =\emptyset$ for some $r>0$

Let $B*:=B\setminus\left\{ x\right\} $ and take any $y\in B*\Longrightarrow B*\cap A\setminus\left\{ y\right\} =\emptyset\Longrightarrow y\in\mathbb{R}^{n}\setminus A'\Longrightarrow B\subset\mathbb{R}^{n}\setminus A'\Longrightarrow\mathbb{R}^{n}\setminus A'=\left(\mathbb{R}^{n}\setminus A'\right)^{o}\Longrightarrow A'=\overline{A'}$ So $A'$ is closed

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You want to use that every neighbourhood of $x$ contains an open neighbourhood of $x$. Now an open set is, by definition, a neighbourhood of each of its elements.

In the case of a metric space, $\varepsilon$-neighbourhoods can be used, and depending on your definitions and what has already been proved, you may need to use the triangle inequality to show that they are open.

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I found the proofs in previous answers confusing, so here is an alternative proof by contradiction:

Let $x$ be any limit point of $E'$.

Assume $x \notin E'$. By the definition of limit points, there exists $\epsilon > 0$ such that $(x - \epsilon, x + \epsilon) \cap E$ contains at most one point, $x$, in $E$. But, since $x$ is an limit point of $E'$, then $\exists y \in E'$ such that $|x - y| < \epsilon_1$. Similarly, $y$ is an limit point of $E$, so $\forall \epsilon_2$, $\exists z \in E$ such that $|y - z| < \epsilon_2$. If we take the interval $|x - z|$, then: \begin{align*} |x - z| &= |x + (y - y) - z| \\ &= |(x - y) + (y - z)| \\ &\leq |x - y| + |y - z| \tag{by triangle inequality} \\ &< \epsilon_1 + \epsilon_2 \end{align*} We can choose $\epsilon_1, \epsilon_2 < \frac{\epsilon}{2}$, which means $|x - z| < \epsilon$, and every interval around $x$ contains at least two points in $E$, so $x$ is a limit point of $E$ and $x \in E'$, which contradicts our initial assumption. Therefore, every limit point of $E'$ is in $E'$.

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Let x is limit point of E'.So $\forall\epsilon>0\exists y\in E'$ such that |x-y|<$\epsilon$/2
Now as E' contatain all limit points of E therefore $\forall y\in E'\forall \epsilon>0\exists z\in E$ such that |y-z|<$\epsilon$/2
Now consider |x-z|$\le$|x-y|+|y-z|$\le$ $\epsilon$By Triangular Inequality.
x $\in E'$.This is true for every limit point of E'.
This implies Set of limit point is closed set

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It is actually enough to assume that your space is $T_1$. If this is the case, then $x$ is a limit point of a set $A$ iff every open set $V$ around $x$ contains infinitely many points of $A$. Let $x\in\overline{A'}$, and take $V\in\mathscr{O}_x$. Then there is $y\in V\cap A'$, which means that $V\cap A$ is infinite, and so $x\in A'$.

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