Showing that if $x$ is a limit point of $A$ in a $T_1$ space, then every neighborhoood of $x$ contains infinitely many points of $A$? Let $X$ be a space satisfying the $T_1$ axiom, and let $A$ be a subset of $X$. Then the point $x$ is a limit point of $A$ if and only if every neighborhood of $x$ contains infinitely many points of $A$.
Here is my attempt at one part of this proof. Suppose $x$ is a limit point of $A$. This means that for if we take some neighborhood $U$ of $x$, then $U$ intersects $A$ at at least one point other than $x$ itself. Let $y_1 \in U\cap A$, with $y_1\neq x$. Is $\{y_1\} \cup \{u\} = U?$ No, because it is union of closed sets and $U$ is open. So take another point $y_2$ and check if $\{y_1\} \cup \{y_2\} \cup \{u\} = U?$. Again, no, because it is a union of closed sets. Ultimately, this means we need to take an infinite number of closed point sets in order that their union equals $U$.
Is this argument correct?
 A: Your argument is not correct, because:


*

*you did not assume that $U$ is open, but then you assert that it is (not a serious problem, of course);

*you say that $U\neq\{y_1\}\cup\{u\}$ because $U$ is open and $\{y_1\}\cup\{u\}$ is closed. Can't a set be closed and open simultaneously?


You can prove it as follows. Suppose that there is a neighborhood $U$ of $x$ with only a finite number of elements of $A$. Let $x_1,x_2,\ldots,x_n$ be these elements. Since the space is $T_1$, each $\{x_k\}$ is closed and therefore the set $C=\bigcup_{k=1}^n\{x_k\}$ is closed. But then $U\setminus C$ is a neighborhood of $x$ which does not intersect $A$, which is impossible, since $x$ is a limit point of $A$.
A: Note first that "$X$ is a $T_1$ space" is equivalent "to all finite subsets of $X$ are closed".
Then suppose that $x$ is a limit point of $A$ and suppose, for a contradiction, that there is some open neighbourhood $O$ of $x$ such that $F = O \cap A$ is finite.
Then define $F'  =F\setminus \{x\}$, which is still finite, and so $F'$ is closed, hence 
$O' = O\setminus F' = O \cap (X\setminus F')$ is also an open neighbourhood of $x$ ($x \in O$ but by definition $x \notin F'$, so $x \in O'$).
Then
$$(O'\setminus \{x\} \cap A )= O  \cap (X\setminus \{x\}) \cap (X\setminus F') \cap A = (O \cap A ) \cap (X\setminus F) = F\setminus F = \emptyset $$
because we cut out the finitely many non-$x$ members of $O \cap A$.
So then $O'$ shows that $x$ is not a limit point of $x$, which is the required contradiction.
Hence every open neighbourhood $O$ of $x$ has $O \cap A$ infinite.
