Let $(X, \Vert \cdot\Vert)$ be a normed space and $A \subset X$. If $x \in X$ is an accumulation point of $A$, does every open ball $B(x, r)$, where $r > 0$ contain infinitely many points of $A$?
A Contraposition
Suppose there exists an open ball $B(x, r)$, that only contains a few or no points of $A$. In the case of no points, the point $x$ is by definition not an accumulation point of $A$, contradicting the initial assumption.
Now the question is, how do I handle the case of some points in $A$? I know that $X$ is a normed space, meaning I might be able to use the norm axioms to somehow contradict either the initial assumption that $A \subset X$, or the limit point status of $x$, but I'm not sure how to achieve that.
Any suggestions?
Yes, you can construct a sequence $x_n \to x$ from which the result follows.
Choose $x_1 \in B(x,1)\setminus \{x\}$.
Now suppose we have constructed $x_1,...,x_n$, then choose $x_{n+1} \in B\big(x, \min( \|x-x_n\|, {1 \over n})\big) \setminus \{x\}$.
Since $\|x_{n+1}-x\| < \|x_n-x\|$ we see that all the points are distinct and since $\|x_n-x\| < {1 \over n}$ we see that $x_n \to x$.