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!
 A: 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.
