If a sequence has a convergent sub-sequence, then the sequence converges(fake proof)

Question

Let $(X,d)$ be a metric space and $x_n$ be a sequence such that it has a convergent sub-sequence $x_{n_k} \to x $, then $x_n \to x$

This is false, for example, the sequence $(-1)^n$ has a convergent sub-sequence but it does not converge.

Here's a proof of the above statement, I can't figure out what the flaw is.

Proof:

Suppose that $x_n$ does not converge to $x$, so there exists an $\epsilon$ such that for all $N$ we have $$n \geq N \rightarrow d(x_n, x) \geq \epsilon$$ on the other hand there is an $N_\epsilon$ such that $$n_k\geq N_\epsilon \rightarrow d(x_{n_k},x) < \epsilon$$ since $x_{n_k}$ is a subsequence of $x_n$, its members are also members of the main sequence, therefore there are members of $x_n$ such that $d(x_n,x) < \epsilon$ which is a contradiction.

What is wrong with this proof?

Answer 1

The contrapositive is wrong. To say $x_n \not\rightarrow x$ is to say

\begin{equation} \exists\epsilon>0 \text{ s.t. } \forall N \in \mathbb{N} \: \: \exists n \geq N \text{ s.t. } d(x_n,x) \geq \epsilon \end{equation}

And not

\begin{equation} \exists\epsilon>0 \text{ s.t. } \forall N \in \mathbb{N} \: \: \forall n \geq N \text{ s.t. } d(x_n,x) \geq \epsilon \end{equation}

The first allows the existence of the subsequence because not all $n$ are obligated to satisfy $d(x_{n_k},x) \geq \epsilon $.