Question about bounded subsequences Let $(x_n)$ be a sequence of real numbers satisfying the following:
Every bounded subsequence $(x_{n_j})$ of $(x_n)$ is convergent.
Prove:

*

*Is possible to $(x_n)$ have convergent subsequences but $(x_n)$ to be not convergent.

*If $(x_n)$ is bounded, then it is convergent.

Number (2) follows from the property because every sequence is a subsequence of itself. For number (1) I do not know if it is enough to find an example or if I have to prove it in general.
thank you
 A: For first argument yes it is possible.
Suppose $((−1)^n)$ is convergent. The  every subsequence of
$((−1)^n)$ converges to the same limit. The subsequence of even terms converges to $1$, whereas the subsequence of odd terms converges to $−1$. Hence we conclude that $((−1)^n)$ is not convergent. Because the convergence of a sequence is unique.
This example also works for second argument.
A: As "unknown" has pointed out, $\big((-1)^n\big)_{n=1}^\infty$ has convergent subsequences and does not converge.
The following is a sequence that does not converge but in which every bounded subsequence (and it has many bounded subsequences) converges to the same limit:
$$
1,2,1,3,1,4,1,5,1,6,1,7,1,8,\ldots
$$
So this has convergence subsequences although it does not converge. For example:
$$
\underbrace{1,2,1,3,1,4,1,5,1,6,\ldots,1,500\,000}_\text{the first one million terms},\,\,\underbrace{1,1,1,1,\ldots\ldots\ldots}_{\begin{smallmatrix} \text{all later terms} \\ \text{whose indices are} \\ \text{multiples of 7}\end{smallmatrix}}
$$
This is a convergent subsequence.
A: @player3236 gave a nice answer in the comments, in which I will expand upon. Some of the other suggestions such as $x_n=(-1)^n$ do not satisfy the assumption that every bounded subsequence converges.
Consider the sequence $(0,1,0,2,0,3,0,4,...)$.
Claim: Let $(y_{k})=(x_{n_k})$ be a bounded subsequence. Then there exists $k_0$ such that $y_{k}=0$ for all $k> k_0.$ Therefore, $y_{k}\to 0$.
Proof: Suppose not. Starting with $k=1$, there exists $k_1> 1$ such that $y_{k_1}>0$. Then there exists $k_2>k_1$ such that $y_{k_2}>y_{k_1}>0$. Given $k_n,$ there exists $k_{n+1}>k_n$ such that $y_{k_{n+1}}>y_{k_n}>\dots>y_{k_1}>0$. But this sequence is clearly unbounded, which is a contradiction. Hence, eventually $y_k=0$ for all $k\geq k_0$.
This claim shows that the sequence satisfies the assumption. Moreover, very clearly, this sequence has convergent subsequences. However, the sequence does not converge.
