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?
Best Answer
Your argument is not correct, because:
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 $A$.