Proof that if $x$ is an accumulation point, then $B(x,r)\cap A$ has infinite points Let $(M,d)$ be a metric space, where $A\subset M$. If $x$ is an accumulation point of $A$, by definition $\forall r > 0,\; B(x,r)\setminus \{x\}\cap A \neq \emptyset$, so $\forall r > 0,\; B(x,r)\cap A \neq \emptyset$. Any hint?
 A: Let $A \subseteq M$, and let $x \in A$ be a point of accumulation of $A$, i.e.
for all $r>0$ $B_r(x) \cap A \neq \emptyset$. Let $r_1 = 1$. Then chose $x_1 \in B_r(x) \cap A$. We construct an infinite sequence of elements of $A$ inductively. Suppose $x_1, x_2, \ldots, x_k$ have all been chosen in $A$ where each $x_i$ for $1 \leq i \leq k$ are distinct. Now let $r = \min\{d(x_1,x),d(x_2,x),\ldots,d(x_k,x)\}/2$. Now let $x_{k+1} \in B_r(x)$. Now $x_{k+1}$ is distinct from $x_1,x_2, \ldots,x_k$. Hence, we get a mapping
$\mathbb{N} \to A$, $k \mapsto x_k$. This is injective, and we're done.
Also note that if we take $r=\min\{\frac{1}{k+1}, d(x_1,x), \ldots,d(x_k,x)\}/2$ then we get that $x_k \to x$ for free.
Hope this helps.
A: Suppose $r>0$ and suppose $B(x,r)\setminus\{x\}\cap A = \{x_1,\dots,x_n\}$ (It is nonempty, because $x$ is an accumulation point). Choose $r'=\min\left\{\frac{d(x,x_i)}{2}\mid i=1\dots n\right\}$. Then there exists $x' \in B(x,r')\setminus\{x\}\cap A$. But $B(x,r')\subset B(x,r)$, so $x'=x_i$ for one $i$. Thats a contradiction.
A: Here is a hint:
Choose $r_1=\frac{1}{2^1}$, and $x_1 \in B(x,r_1)$.
Now let $r_2 = \min(\frac{1}{2}d(x,x_1), \frac{1}{2^2})$, and pick $x_2 \in B(x,r_2)$. Continue the process. What can you say about $r_k$ and hence about $x_k$?
