Accumulation points, open sets and intersections

Question

Let $(X, \Vert\cdot\Vert)$ be a normed space, with $A, G \subset X$ and $G$ open. If $A \cap G$ contains only a finite amount of points (possibly none), can any point of $G$ be an accumulation point of $A$?

An incomplete attempt

Assume $A \cap G = \emptyset$. Then as $G$ is open, there exists an $\varepsilon > 0$ for every point $x \in G$, for which the open ball $B(x, \varepsilon) \subset G$. But because the intersection $A \cap G$ is empty, no points in this ball can be in $A$, contradicting the definition of the accumulation point.

Now assume that there exists a finite set of points $x_i$ in $A \cap G$, where $i \in \{1, \ldots, N\}$ and $N \in \mathbb N$. How can I proceed from here? The form of the claim suggests a contraposition might work here.

Let a point $x_G \in G$ be an accumulation point of $A$. Then by definition, every punctured open ball $B'(x_G, \varepsilon)$, where $\varepsilon > 0$ should contain a point of $A$. What would be the next step? Should this somehow contradict the finiteness of the intersection $A \cap G$, or the openness of $G$? The first option seems more doable, but I'm not sure how to achieve it.

Answer 1

No, it cannot, because we're in a metric space. I'll use $d(x,y)$ for $\|x-y\|$ as the norm itself is not relevant.

Let $x$ be any point of $G$ and pick $r>0$ such that $B(x,r) \subseteq G$. This can be done as $G$ is open.

Then by assumption $A \cap G$ is finite so say $A \cap G = \{x_1,x_2,\ldots, x_N\}$ for some finite $N$ and pick $r'<r$ such that also $0< r' < \min \{d(x,x_i): x_i \neq x\}$ which can be done as we take a minimum of finitely many strictly positive numbers.

Then $B(x, r') \cap A$ can at most be $\{x\}$ so $x$ is not a limit point of $A$. Any $y \in B(x,r')\cap A$ lies in $G$ (as $r' < r$) and so must be among the $x_i$ and $y$ cannot be $x_i$ as $d(x,y) < d(x,x_i)$ for any $i$, unless $y=x$.

Answer 2

No point of $G$ can be an accumulation point of $A$. Take $p\in G$. Let $r>0$ be such that $B_r(p)\subset G$. Then $B_r(p)\cap A\subset G\cap A$, and therefore $B_r(p)\cap G$ is finite. If$$\bigl(B_r(p)\setminus\{p\}\bigr)\cap G=\{a_1,\ldots,a_N\},$$take $r^\ast\in(0,r]$ such that$$(\forall k\in\{1,2,\ldots,N\}):\lVert x-a_k\rVert<r^\ast.$$Then $B_{r^\ast}(p)\cap A$ is either empty or consists only of $\{p\}$. In any case, $p$ is not an accumulation point of $A$.