If the subsequence converge, then will the sequence has at least upper bound or lower bound [closed]

By the Bolzano-Weierstrass theorem, we know that any bounded sequence $\{x_n\}$ has a convergent subsequence. Also, if there exists a converge subsequence, it's not always true that the sequence $\{x_n\}$ is bounded($\{x_n\}$ doesn't always have both upper bound and lower bound), but is it possible to conclude that $\{x_n\}$ at least has one of them(have at least upper bound or lower bound) if there exists a converge subsequence $\{x_{n_k}\}$?

Solution 1:

No, consider the sequence $$x_n= \begin{cases} 1 & n\equiv 1\pmod{2}\\ n & n\equiv 0\pmod{4}\\ -n & n\equiv 2\pmod{4} \end{cases}$$ So the sequence would look like $1,-2,1,4,1,-6,1,8,1,-10,...$, so all the odd terms are $1$, and all the even terms are either there positive or negative versions (alternating at each even number). It is clear that the subsequence of all the odd terms converges, so $x_n$ has a convergent subsequence, but the sequence $x_n$ is not bounded above nor below.