Question: Let X be an infinite Hausdorff space. Suppose x is a limit point of A which is a subset of X (or x is the limit point of $A$ $\subset$ $X$), s.t. x is a limit point of A. Then for every open set U $\in$ x, U contains infinitely many points of A.
How to do this?
If A is the limit point of $A$ $\subset$ $X$, then we know that there $\forall$ U open in A, U $\in$ x, s.t. U $\cap$ A = $\emptyset$, and from definition of Hausdorff spaces, we can say that $\forall$ different x,y $\in$ X, $\exists$ U, V $\in$ T, s.t. $x \in U$ ^ $y \in V$ ^ $U \cap V$ = $\emptyset$. But I am not sure how to bring the concept of infinite Hausdorff space to show that U contains infinitely many points of A.
Need help here, appreciate your help and support.
Best Answer
Suppose $A \subseteq X$ and $x$ is a limit point of $A$, and $X$ is Hausdorff.
This means that for every open neighbourhood $U$ of $x$, $U$ intersects $A \setminus \{x\}$.
Suppose now that $U \cap A$ were finite for some open $U$. Fact: in a Hausdorff space all finite sets are closed. (In fact the latter is equivalent to $X$ being $T_1$ and Hausdorff implies $T_1$) so $V:= X\setminus ((U \cap A) \cup \{x\})$ is open and contains $x$. Now $(V \cap U) \cap A \subseteq \{x\}$ : $y \in (V \cap U) \cap A$ implies $y \in U \cap A$ but as $y \in V$ too, $y=x$. This shows that the open neighbourhood $U \cap V$ of $x$ witnesses that $x$ is not a limit point of $A$, contrary to assumption.
So $U \cap A$ can never be finite, for $U$ an open neighbourhood of $x$.
We only use Hausdorff mildly; $T_1$ is all we need.