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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? Thanks.

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You can see an answer here. –  Moron Jun 8 at 8:40

7 Answers 7

up vote 13 down vote accepted

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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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! –  Fernando Martin Dec 29 '10 at 5:58
@FernandoMartin the neighborhood of $y$ with radius $r - r'$ contains $x$'s neighborhood, not the other way around. –  Don Larynx Nov 5 '13 at 17:51
@DonLarynx: I'm pretty sure it is correct. –  Fernando Martin Nov 6 '13 at 2:44
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. –  drhab Aug 8 at 13:06
Never mind. I have found math.stackexchange.com/q/490968/75923 –  drhab Aug 8 at 13:22

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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You should use the metric form $\mathrm{d}$ and not $|\cdot|$. –  Steven Gamer Jan 22 '13 at 14:27
Any specific reason @StevenGamer? I like this form. –  Don Larynx Nov 5 '13 at 17:53
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. –  Hayden Nov 20 '13 at 23:32
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. –  Ludolila Mar 7 at 19:25

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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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? –  Arturo Magidin Dec 28 '10 at 22:07
@Arturo. Edited my answer. I meant some authors define "neighborhood" this way. –  TCL Dec 28 '10 at 22:11
What do you mean by “equivalent” here? –  Carsten Schultz Nov 24 '13 at 14:58

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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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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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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