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Let $A$ be a set, $X$ a metric space, $x$ an accumulation point of $A$ that is, every nbhood of $x$ contains a point $a \in A$, $a \neq x$.

I wrote a proof of the fact "every nbhood of $x$ contains infinitely many points of $A$". But for one I don't know if it is correct and also if it is correct whether it can be improved. Here is my proof:

The open ball $B(x,\varepsilon)$ is a neighbd of $x$ and by assumption contains $a \in A, a \neq x$. Since $B(x, \varepsilon)$ is open there is $\delta$ such that $B(x, \delta) \subseteq B(x, \varepsilon)$ and $a \notin B(x, \delta)$. Since $B(x, \delta)$ is a nbhood of $x_0$ by assumption there is $a' \in A$ with $a' \in B(x, \delta)$.

Then I wrote: "Repeat process to obtain infinitely many points $a \in A$ in $B(x, \varepsilon)$".

Is this valid? And if it is not: how to finish the proof? And if it is correct: is there a neater way to write this proof? Thank you.

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3 Answers 3

up vote 3 down vote accepted

The idea is sort of OK, but can be made more formally correct.

Assume there is a neighbourhood $U = B(x_0, \epsilon)$ such that $U \cap A$ is a finite set, (and we leave out $x_0$ if $x_0 \in A$) say it equals $\{ a_1,\ldots, a_k\}$ for some finite $k$. We have to show this leads to a contradiction. We now apply the idea you also use: take a ball around $x_0$ with radius $r < \min(d(x_0,a_1), d(x_0,a_1),\ldots,d(x_0, a_k))$. Then $B(x_0,r) \subset U$ and by construction the $a_i$ are not in this ball, but there must be some $a \neq x_0$ with $a \in A$ in this ball, as $x_0$ is an accumulation point. This shows that $A \cap U$ could not have been this finite set (we have found a new point). Contradiction.

We can avoid radii and metric, by noting that a finite set is closed (so we need $T_1$ in a general topology setting) and note that $U \setminus (A \cap U)$ is also open and should contain some new point of $A$.

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In $T_1$ space singleton sets are closed? I thought this was only true for Hausdorff space (=$T_2$). –  blue Feb 28 '13 at 17:02
    
Yes, this is exactly equivalent to $T_1$. Say we have $\{p\}$, then for each $q \neq p$, $p,q \in X$, we have some $U_q$ that contains $q$ but misses $p$, using $T_1$. Then $X\setminus\{p\} = \cup_{q \neq p} U_q$, showing the complement of a singleton is open. And if singletons are closed, so are finite sets. –  Henno Brandsma Feb 28 '13 at 17:57

It's fairly good, though there are a few issues to fix (subscripts, and the like).

Alternatively, and more simply, you could show that if there is a neighborhood of $x_0$ with only finitely-many points of $A$, then $x_0$ is not an accumulation point of $A$--in particular, there is an open ball about $x_0$ with no points of $A$ in it (except $x_0$, in the case that $x_0\in A$)--proving the desired result by contrapositive.

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I use the method...

A $\subset$ (X,d) and $x_0$ $\epsilon$ X then by def U $\subset$ X is called neighborhood of $x_o$ if there exist G $\epsilon$ $\tau$ $x_o$ $\epsilon$ G $\subset$ U

as $x_o$ is an accumulation point of A then

$G | ${x$_o$}$ $ $\bigcap$ A $\neq$ $\phi$

A $\subset$ (X,d)

and U $\subset$ X

then

A $\subset$ U

Tell me if my method is correct....

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