Proving that the set of limit points of a set is closed

Question

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?

Answer 1

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 $y$.

Answer 2

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.

Answer 3

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.

Answer 4

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, $x \in E'$.

Answer 5

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.

Answer 6

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

Answer 7

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

Answer 8

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 $x\in A'$.

Answer 9

Let $X$ be a $T_1$ space and $A \subseteq X$. Notice that $A'$ is closed $\iff \overline{A'} = A' \iff A'' \subseteq A' $ since $\overline{A'} = A' \cup A'' $. Hence, we show that $A''\subseteq A'$.

Let $x \in A''$. Thus, for every open nbhd $U_x$ of $x$, one has $U_x \cap (A' \setminus \{x\}) \neq \emptyset $.

Pick $y \in U_x \cap (A' \setminus \{x\})$. As $y \in U_x$ and $U_x$ is an open set, there exists an open nbhd $V_y$ of $y$ entirely contained in $U_x$. Moreover, as $X$ is $T_1$, singletons are closed, so $\{x\}$ is a closed subset of $X$. Therefore, $V_y \setminus \{x\} = V_y \cap \{x\}^C$ is the intersection of open subsets of $X$ and so, is also open in $X$, contains $y$, and is entirely contained in $U_x$. (This is the subtlety missed by OP and accepted answer)

Now, as $y \in A'$, then for every open nbhd $U_y$ of $y$, one has $U_y \cap (A \setminus \{y\}) \neq \emptyset $. In particular, this property also applies to the nbhd $V_y \setminus \{x\} $ of $y$.

Picking now $z \in (V_y \setminus \{x\} ) \cap (A \setminus \{y\}) $, we have that $z \in V_y \setminus \{x\} \subset U_x$, where $z \neq x$. Therefore, for every open nbhd $U_x$ of $x$, one has $U_x \cap (A \setminus \{x\}) \neq \emptyset$.

Hence, $x \in A'$.

Answer 10

The answers above I found good but I also found there was a step that needed further justification (I elaborate after proof for context reasons), I also write this in the spirit of "Understanding analysis" by Stephen Abbott for there is a similar exercise in the book, the notation is kept similar for those who like me tried searching for answers. This proof is similar to the one found in the instructor manual for the aforementioned book.

Notation used below

• Epsilon neighbourhood

$V_\epsilon(a):= \{x\in\mathbb{R}: \lvert x-a\rvert<\epsilon\} = (a-\epsilon, a+\epsilon)$

• Punctured epsilon neighbourhood

$V_\epsilon(a)\setminus\{x\}:= \{x\in\mathbb{R}: 0<\lvert x-a\rvert<\epsilon\} = (a-\epsilon,a)\cup(a, a+\epsilon)$

Let $A\subseteq\mathbb{R}$ and let $L$ be a set containing all of the limit points of $A$. Let $x$ be an arbitrary limit point of $L$, we want to show that $x\in L$ or in other-words that $x$ is a limit point of $A$.

Given an arbitrary $\epsilon>0$, we know by definition of a limit point that $(V_\epsilon(x)\setminus\{x\})\cap L\neq \emptyset$ so there is at least a $l\in V_\epsilon(x)\cap L$ where $l\neq x$. Now $l\in L$ so $l$ is a limit point of $A$. We can now pick $\epsilon'>0$ small enough that $V_{\epsilon'}(l)\subseteq V_\epsilon(x)$ and $x\notin V_{\epsilon'}(l)$, then again since $l$ is a limit point of $A$, there is a $a\in V_{\epsilon'}(l)\cap A$ where $a\neq l$.

Now $a\in V_{\epsilon'}(l)\subseteq V_\epsilon(x)$ so $a\in V_\epsilon(x)$ but $x\notin V_{\epsilon'}(l)$ so $a\neq x$. Thus in conclusion for all $\epsilon>0$, there is a $a\in A$ such that $a\in V_\epsilon(x)\cap A$ and $a\neq x$ so $x$ is a limit point of $A$. $\blacksquare$.

What confused me about the previous answers was the lack of motivation for $a\neq x$, so I wanted to add this one for clarity if others found that not trivial like me.

Answer 11

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.

Answer 12

Here is an alternative proof, using a 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 there exists $y \in E'$ such that $|x - y| < \epsilon_1$. Similarly, $y$ is an limit point of $E$, so for all $\epsilon_2$, there 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 $\epsilon_1, \epsilon_2 < \epsilon / 2,$ 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 $E'$.