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I am following a proof of the statement

The derived set(the set of accumulation points) $A'$ of an arbitrary subset $A$ of $\mathbb{R}^2$ is closed.

in a book.

It starts with

Let $q$ be a limit point of $A'$. If it is proved that q $\in A'$, then the proof is done.

Let $G_q$ be the open set containing $q$. Since $q$ is a limit point of $A'$,$G_q$ contains at least one point $r\in A'$ different from $q$. But $G_q$ is an open set containing $r\in A'$; (Up to this I understood) hence $G_q$ contains infinitely many points of $A$ (How? I did not get this.)

So there exist $a \in A$ such that $a \neq q,a \neq r$ and $a \in G_q$. That is,each open set containing $q$ contains infinitely many points of $A$. Hence $q \in A'$.

Can you help me out.

Stefan Hamcke
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Vinod
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  • If there are not infinitely many points, then you can take the minimum distance between these finitely many points as $m = \min{x_1,x_2,\ldots,x_n}$. You will then have an open ball $B(x,m)$ about $x$ which does not contain any other points. So $G_q$ must have infinitely many points. – Wortel Jun 09 '13 at 11:43

3 Answers3

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In $\Bbb R^2$, the points are closed. So, $G_q\setminus\{q\}$ is an open neighborhood of $r\in A'$, so it contains some $a_1\in A$ such that $a_1\ne r$. But, going on, $G_q\setminus\{q,a_1\}$ is also an open neighborhood of $r$, so it contains an $a_2\in A$. And so on.

Berci
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Your question is: Why does $G_q$ contain infinite many points of $A$?

We will use the following theorem. It can be seen in the books on general topology.

Theorem 1: Let $X$ be a $T_1$ space and $A$ is an infinite subset of $X$. Then $x$ is an accumulation if and only if for any nbhd $U$ of $x$, $U$ contains infinite points of $A$.

Proof: Since $G_q$ contains $r$, where $r$ is an accumulation of $A$, it is not difficult to see $U_q$ is a nbhd of $r$. By the theorem 1, we can conclude that $G_q$ contains infinite many points of $A$.

Note that the condition that $\Bbb R^2$ is not necessary. The space only need to be $T_1$.

Community
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Paul
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If $A'=\emptyset$ then we are done. If not let $x\in (A')^c$ then $x$ is not limit point of $A$ this means $\exists~\delta>0$ such that $(x-\delta,x+\delta)\setminus\{x\})\cap A=\emptyset.$ Let $y\in (x-\delta,x+\delta),$ if $y=x$ then $y\in (A')^c$. Now, $(x-\delta,x+\delta)$ is a neighborhood of $y\neq x \Rightarrow (x-\delta,x+\delta)\setminus\{y\}\cap A=\emptyset \Rightarrow (x-\delta,x+\delta)\subseteq (A')^c \Rightarrow (A')^c$ is open set thus $A'$ is closed.

I proved for $\mathbb{R}^1$. Same proof for $\mathbb{R}^2$.

RIPAN DAS
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