Clarification on if why one subsequence diverges, it has no further subsequence that converges

Question

I'm having trouble with the last sentence of the proof. Namely, why exactly does $x_{n_{k}}$ have no further subsequences that converge? Perhaps I'm missing knowledge of a theorem.

Thanks.

Answer 1

No, it's not due to any additional theorems. It follows immediately from the construction of the subsequence $\left\{x_{n_k}\right\}_k$. Note that all these numbers $x_{n_k}$ are chosen so that $\left|x_{n_k}-z\right|\ge\varepsilon$, where $\varepsilon>0$ is a certain positive constant value (that was chosen in the beginning of the proof). In other words, all $\left\{x_{n_k}\right\}_k$ lie outside the interval $(z-\varepsilon,z+\varepsilon)$. That's why neither this subsequence $\left\{x_{n_k}\right\}_k$ nor any of its further subsubsequences (is that a word?) could possible converge to $z$ — simply because all these numbers are at least $\varepsilon$ units away from $z$.