$p\notin A$ a limit point of $A \iff$ for every $\delta > 0$ the intersection $B(p;\delta )\cap A$ is infinite

Question

I need to prove that $p\notin A$ a limit point of $A \iff$ for every $\delta > 0$ the intersection $B(p;\delta )\cap A$ is infinite. With $A\subset \mathbb{R}^n$ and $p\subset \mathbb{R}^n$.

To prove that the first implies the second, I wrote the following down, but I got stuck:

"$p$ is a limit point of $A$, so for every $\delta > 0$ it is true that $B(p;\delta) \cap A \neq \emptyset$. This means that there is at least one point in $B(p,\delta)\cap A$."

I also think $p\notin A$ and $p$ a limit point of $A$ means $A$ is an open set, is this correct?

My common sense tells me that if there is one point in the intersection of $A$ and $B(p,\delta)$, and $p$ a limit point of $A$ (an open set), then $p$ must be on the boundary of the open set $A$, is this correct? Then because "for every $\delta>0$ there is a point in $B(p,\delta)\cap A$", there must also be a point in $B(p,\frac{\delta}{2})\cap A$, so then there are definitely multiple points? Could I continue this argument to end up with infinite points in the intersection?

And for the implication the otherway around, is this line of thinking correct and complete (I'm not too sure):

For every $\delta>0$, the intersection $B(p;\delta )\cap A$ is infinite $\implies$ For every $\delta>0$ it is true that $B(p;\delta) \cap A \neq \emptyset$ $\implies$ $p$ a limit point of $A$.

Answer 1

If there is only one point $a$ in $B(p,\delta)\cap A$ then for $\delta'\in(0,|p-a|)$ there are no points in $B(p,\delta')\cap A$, contradicting that $p$ is a limit point.

This contradiction also arises if $B(p,\delta)\cap A$ is finite by taking $\delta'\in(0,\min\{|p-a|\mid a\in B(p,\delta)\cap A\})$.

Conversely if for every $\delta>0$ the set $B(p,\delta)\cap A$ is infinite then it will be non-empty and elements in it will not equal $p$. This tells us that $p$ is a limit point of $A$.

Answer 2

Your idea is alright. You can use contradiction to get a rigorous proof.

If there are finitely many points in the set $B(p, \delta) \cap A$, then you can choose point $q \in B(p, \delta) \cap A$ s.t. the disance between the points from $B(p, \delta) \cap A$ and $p$ is minimum. Then we have that $B(p,\frac{d(p,q)}{2}) \cap A = \emptyset$, which contradicts the fact that $p$ is a limit point of $A$>

Also as $p \not \in A$ we have that $\frac{d(p,q)}{2} > 0$, so everything is well-defined.