Proof of limit point ( accumulation point )

Question

Let $(X, d)$ be a metric space. Let $A ⊂ X$ and $x ∈ X$. Prove the following:

$x$ is a limit point of $A$ if and only if $∀ v ∈ V $, $V$ neighborhood of $x$, then $V ∩ A$ is infinite.

I know this:

Definition of a limit point: $x$ is a limit point of $S$ provided that every neighborhood of $x$ contains elements from $S$ distinct from $x$.

But I don't know how to prove the claim.

Answer 1

Imagine graphically in a neighborhood (open ball) $V$ of $x$ a finite number of points of A. This contradict the notion of limit point. In fact, you can reduce the initial neighborhood $V$ so that it no contains the finite number of common points $A$ and $V$.

For contradiction, suppose that there exists $V$ neighborhood of $x$ that contains a finite number points of $A$. Enumerate the points $$ (A\setminus \{x\})\cap V= \{a_1,a_2,\dots,a_n\}$$ then $$ r=\min \{ d(a_i,x)\colon i=1,\dots,n\}$$ is strictly positive. Let $\tilde{V}=\left\{ y\in X\colon d(y,x)<\frac{r}{2} \right\}$ then we have that $$ (A\setminus \{x\})\cap \tilde{V} = \emptyset.$$ I.e., $\tilde{V}$ is neighborhood of $x$ that no contains some point of $A$ distinct of $x$, this a contradiction. Therefore, every neighborhood $V$ of $x$ contains a infinite number points of $A$.