Question about a set without limit points

Question

Let $X$ be a topological space and $A\subset X$ a subset. A limit point $a$ of $A$ is a point such that each neighbourhood of $a$ contains infinitly many points of $A$.

Is the following true: $A$ has no limit points $\Leftrightarrow$ every point $x$ in $X$ has a neighbourhood which contains only finitly many points of $A$.

If we are in a metric space $(X,d)$ and we pick an element $a$ in a subset $A\subset X$ that has no limit point. Can we argue, that $a$ has a neighbourhood that only contains finitely many points of $A$ and therefore there is a minimal distance between $a$ and any other member of the set $A$?

Answer 1

Yes, you can say that there is a neighborhood which contains only $a$ if not every neighborhood of $a$ contains at least two points so, there exists $x_n\in B(a,1/n), n>0, x_n\neq a$ and $lim_mx_n=a$ and $a$ is a limit point.

Let $U$ be a neighborhood of $A$ which contains only $a$, there exists an open ball $B(a,r)\subset U$, this implies that for every $b\in A$, $d(a,b)>r/2$.

Answer 2

Remember a limit point need not be in $A$ itself.

Consider the set $X=\{ 1/n \mid n\in \mathbb{N}\}$ as a subspace of $\mathbb{R}$. This is discrete, so each element has an open neighbourhood that contains only itself. However, $0$ is a limit point of $X$, since every neighbourhood of $0$ intersects $X$.

I suspect the other implication is false in general too, but true for metric spaces.