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

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?

• The problem is in the very first line. $d(x_n,x) < \epsilon$ for all $n\ge N$ being false does not in any way mean $d(x_n,x) \ge \epsilon$ for all $n\ge N$. It just means there is at least one $n\ge N$ where $d(x_n,x)\ge \epsilon$. – fleablood Oct 24 '19 at 22:36
• This equivalent to the following prove. If some horse are brown than all horses are brown. Suppose all horses are not brown; then there is some other color that all horses are. Call it "grue". But some horse are brown and therefore those are not grue. That's a contradiction. So all horse are brown. – fleablood Oct 24 '19 at 22:42
• I like that analogy, so I couldn't figure out what the flaw was because I failed to notice that the negation of convergence was wrong! How silly – clementine Oct 24 '19 at 22:49
• When, as in this question, you have a fake proof and a specific counterexample, an effective way to find the flaw in the proof is to apply it to your counterexample. In this case, that would mean writing the proof with every $x_n$ replaced with $(-1)^n$. Then, if necessary, check line by line to see where the proof for this specific example goes wrong. – Andreas Blass Oct 24 '19 at 22:52

The contrapositive is wrong. To say $$x_n \not\rightarrow x$$ is to say
$$$$\exists\epsilon>0 \text{ s.t. } \forall N \in \mathbb{N} \: \: \exists n \geq N \text{ s.t. } d(x_n,x) \geq \epsilon$$$$
$$$$\exists\epsilon>0 \text{ s.t. } \forall N \in \mathbb{N} \: \: \forall n \geq N \text{ s.t. } d(x_n,x) \geq \epsilon$$$$
The first allows the existence of the subsequence because not all $$n$$ are obligated to satisfy $$d(x_{n_k},x) \geq \epsilon$$.