Let $(X, \tau)$ be $T_2-Space$ and $A \subseteq X$ and $x \in \overline A$ (closure of $A$ = $\overline A$)
Then prove that $x$ is an accumulation point of $A$ iff $\forall V \in V(x) , V \cap A\setminus \{x\}$ is infinite ($V(x)$ is all neighborhoods of x)
For first part of theorem can we prove by contradiction? Let $\forall V \in V(x)$ $V \cap A\setminus \{x\}$ be finite. How can I pursue this proof?
For converse direction can we say
$x$ is an accumulation point iff $\forall V \in V(x) , V \cap A\setminus \{x\} \neq \emptyset $
Since $V \cap A\setminus \{x\}$ is an infinite set it can't be $\emptyset $ because $\emptyset $ is finite. Is it wrong?
If someone can help me I'll be really glad.
Actually $T_{1}$ suffices. For a $V\in V(x)$, pick an $x_{2}\in V\cap A$ and $x_{2}\ne x$. Now find another $V_{2}\in V(x)$ such that $x_{2}\notin V_{2}$. Pick a $x_{3}\in V_{2}\cap A$, $x_{3}\ne x$, proceed infinitely.
Or, following the idea of the OP proof, assume that $F:=V\cap A-\{x\}$ is finite, as finite sets are closed in $T_{1}$ space, then so is $F$, so $F^{c}$ is open which containing $x$, hence $(F^{c}\cap V)\cap A-\{x\}\ne\emptyset$, say, $z\in (F^{c}\cap V)\cap A$, $z\ne x$, in particular, $z\in V$ and $z\in F^{c}$, then $z\notin V\cap A-\{x\}$, as $z\ne x$ and $z\in A$, we have $z\notin V$, a contradiction.