# If x is a limit point of a set S, then every open ball centered at x contains infinitely many points of S?

I'm told that the following statement is true:

"If a limit point of the set S is defined as x, then every open ball that is centered at x contains infinitely many points of S."

Yet I can't begin to imagine how the proof is completed. Any assistance would be very welcomed. Thanks!

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What kind of space are you talking about? This is false in general, but true in, for example, metric spaces. – Arturo Magidin Mar 31 '11 at 18:42
Yeah, I meant metric space. Sorry about that! – user8951 Mar 31 '11 at 18:48

I assume that "$x$ is a limit point of $S$" means that:

• Every open ball that is centered at $x$ contains some point of $S$ other than $x$.

For metric spaces.

Suppose there is an $\epsilon\gt 0$ such that $B(x,\epsilon)\cap S-\{x\} = \{s_1,\ldots,s_n\}$ is finite. Let $\delta_i = d(x,s_i)$ be the distance from $x$ so $s_i$ for each $i$. Now let $\delta = \frac{1}{2}\min(\delta_1,\ldots,\delta_n)$. Then $B(x,\delta)\cap S - \{x\} = \emptyset$, which means that $x$ cannot be a limit point of $S$.

By contrapositive, if $x$ is a limit point of $S$, then every open ball centered at $x$ contains infinitely many points of $S$.

For topological spaces.

Modifying the proposition by replacing "open ball containing $x$" with "open set containing $x$" (i.e., "open neighborhood of $x$"), the result is false in general but true in $T_1$ (or better) spaces. A space is $T_1$ if and only if for every $x$ and $y$, $x\neq y$, there are open subsets $U$ and $V$ such that $x\in U-V$ and $y\in V-U$.

One argues as above: suppose there is an open set $U$ containing $x$ such that $U\cap S-\{x\} = \{s_1,\ldots,s_n\}$ is finite. For each $i$, let $U_i$ be an open set containing $x$ but not containing $s_i$ (the existence of $U_i$ is guaranteed by the separation property). Then $V=U\cap U_1\cap\cdots\cap U_n$ is a finite intersection of open sets, hence open, it contains $x$, and $V\cap S-\{x\}=\emptyset$, so $x$ is not a limit point of $S$.

However, the proposition fails if $X$ is not $T_1$. Let $x$ and $y$ be elements that witness the fact that $X$ is not $T_1$. So either all open sets that contains $x$ also contain $y$, or all open sets that contain $y$ also contain $x$. Assume without loss of generality that all open sets that contain $x$ also contain $y$. Then $S=\{y\}$ has $x$ as a limit point, but every open set that contains $x$ intersects $S$ at a single point, not infinitely many.

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Sorry, I'm trying to follow the logic here (and having a wee bit of trouble!). Why do we assign δ as one-half the smallest δi value? – user8951 Mar 31 '11 at 19:33
@Jill: Overkill; this ensure that the distance from $x$ to any $s_i$ is definitely larger than $\delta$; you could have just taken the minimum of the $\delta_i$ instead. – Arturo Magidin Mar 31 '11 at 19:34
That's what I thought. Thanks! I am humbled in your presence :) – user8951 Mar 31 '11 at 19:38