4
$\begingroup$

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.

$\endgroup$
  • $\begingroup$ 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. $\endgroup$ – Wortel Jun 9 '13 at 11:43
2
$\begingroup$

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.

$\endgroup$
1
$\begingroup$

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 $T_1$.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.