Prove that the open ball of radius r about x intersects A is infinite.

Question

Let $x$ be a limit point of $A$. Then prove that for any $r>0, B_r(x)\cap A$ is infinite.

The definition of a limit point: Let A be a subset of a metric space $(X,d)$. A point $x\in X$ is called a limit point of $A$ if for every $r>0, B_r(x)$ the intersection $A\setminus\{x\}$ is not equal to the empty set.

I found a hint saying that the proof proceeds with the idea of induction but I really do not know how to do that. Can someone help me, please? Thanks so much.

Answer 1

Suppose that for some $r>0$, $B_r(x) \cap A$ is finite, say $B_r(x) \cap A = \{a_1, \ldots, a_n\}$. Then we have finitely many distances $\{d(x,a_i): i \in \{1,\ldots, n\} , a_i \neq x\}$ that are all $>0$, so let $r > s>0$ be such that (go below the minimum of these distances):

$$\forall i \in \{a_1, \ldots, a_n\} , (x \neq a_i) \rightarrow s < d(x, a_i)$$

This means $B(x,s) \cap A \subset \{x\}$, or otherwise put: $B(x,s)$ can only intersect $A$ in $\{x\}$ at most. So $x$ is not a limit point of $A$ in your sense, which is a contradiction.

Answer 2

Assume there are only finitely many $x_i\in A\cap B_r(x)$ besides $x$ for some $r>0$. Then take $r'=\frac 12\min_i d(x,x_i)$. Then $A\cap B_{r'}(x)=\{x\}$, in contradiction to your definition of limit points. Why it is $\{x\}$? Because any other point $y\in A\cap B_{r'}(x)$ besides $x$ must also be in $A\cap B_r(x)$, which cannot be as $d(y,x)<\min_i d(x_i,x)$ and the $x_i$ and $x$ are all the points in $A\cap B_r(x)$.