# $V$ sequentially compact iff for every infinite $A \subset V$ there exists a $x \in V$ such that $B(x; \delta) \cap A$ has infinite elements

Please need verification of the first implication, and a little help with the second implication.

Given is a metric space $(V, d)$ Prove the following:

$V$ is sequentially compact iff for every subset $A \subset V$ with infinitely many elements there exists a point $x \in V$ such that for every $\delta > 0$ the set $B(x; \delta) \cap A$ contains infinitely many elements.

$\mathbf{HINTS}$: $\Rightarrow$ Consider a useful sequence $(a_{n})$ in $A$. And $\Leftarrow$ consider a $(a_{n})$ in $V$ and look at the cases that {${a_{n} | n \in \mathbb{N}}$} contains finitely or infinitely many elements.

Proof:

$\Rightarrow$ Let $(a_{n})_{n \in \mathbb{N}}$ be a sequence in $A$ with infinitely many elements. Since $V$ is sequentially compact there is a subsequence $(a_{n_{j}})_{j \in \mathbb{N}}$ of $(a_{n})_{n \in \mathbb{N}}$ which converges to {$x$} $\in V$. So for every $\epsilon > 0$ there exists a $N \in \mathbb{N}$ such that $j > N \Rightarrow (a_{n_{j}}) \in B(x; \epsilon)$. So, $n > j > N$ implies that all $a_{n}$ will eventually be trapped in $B(x; \epsilon)$. But $(a_{n})_{n \in \mathbb{N}} \in A$ for every $n \in \mathbb{N}$ and is infinite by construction. So $B(x; \delta) \cap A$ contains infinitely many elements.

$\Leftarrow$ Let $(a_{n})_{n \in \mathbb{N}}$ a sequence in $V$. First consider that $(a_{n})$ is infinite. Define $A :=$ {$a_{n} | n \in \mathbb{N}$}. Then there exists a $x \in V$ such that for every $\delta > 0$ the set $B(x; \delta) \cap A$ contains infinitely elements. So there must be a subsequence of $(a_{n})_{n \in \mathbb{N}}$ converging to {$x$}? And what about the case that $A$ is finite?

Any help is much appreciated!

The proof for $\Rightarrow$ isn't entirely correct. You should state at the beginning that $a_n = a_m$ if and only if $n=m$ (and why you can construct such a sequence, although that is quite obvious). If this isn't the case, a subsequence $(a_{n_k})$ could still be constant and the proof obviously won't work. Consider for example the sequence $a_{2n} =1$ and $a_{2n+1} = n$ for all $\mathbb{N}$. It contains infinitely many elements and yet it also contains a subsequence that is constant.
To fix the proof for $\Leftarrow$, try to explicitly construct a subsequence. For example, take for any $k \in \mathbb{N}$, an element $a_{n_k} \in B(x, 1/k) \cap A$. As for the finite case, any sequence with only a finite amount of differing elements must have a convergent subsequence. There must be at least one $c \in \mathbb{R}$ such that $a_{n_k} = c$ for $n_1 < n_2 < \ldots < n_k < \ldots$ (If this wasn't true $\{a_n \mid n \in \mathbb{N}\}$ would be infinite).