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Let $ \{ t_n \} $ be a bounded sequence on the real line. If all the convergent subsequence $\{ t_{n_k} \} $ of $\{ t_n \} $ converge to same value $t_0$, then can I say that $ \lim_{n \to \infty} t_n = t_0$ ?

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I would argue as follows... Since $\{t_n\}$ is bounded, there exists $R>0$ such that $t_n\in [-R,R]$ for every $n\in\mathbb N$. Then pick a subsequence $\{t_{n_j}\}_j$. As $[-R,R]$ is sequentially compact, there exists a sub-subsequence $\{t_{n_{j_k}}\}_k$ which converges. As this is a particular convergent subsequence of ${t_n}$, then it must converges to some $\bar t$, which is unique for all these subsequences by hypothesis. As $\mathbb R$ is Hausdorff, it follows that $\{t_n\}$ converges to $\bar t$.

We have used the well known fact that if we have a sequence, and we know that from every subsequence we can extract a convergent sub-subsequence, and all of these sub-subsequences have the same limit $t_0$, then also the sequence converges to $t_0$.

This is called Urysohn property and it is valid if the notion of convergence comes directly from the topology. Remember this is not valid e.g. for the almost everywhere convergence.

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The question can be restated as: If there is a single limit point, then does the sequence converge to that point? The answer is yes: the limit inferior and limit superior are limit points of any sequence, so if there is only a single limit point the inferior and superior limits coincide, which implies that the limit of the sequence exists and equals that common point.

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Since the sequence $(t_n)_{n \in \mathbb{N}}$ is bounded,has a convergence subsequence to $t_0$.
If $t_n \not \to t_0$, then $\exists \epsilon >0$ s.t. $\forall k \in \mathbb{N}$ exist $n_k>k$ with $|t_{n_k}-t_0|>\epsilon.$ You can take $n_k$ to be strictly increasing. So the subsequence $(t_{n_k})_{k \in \mathbb{N}}$ is bounded and does not have subsequence converging to $t_0$. This is a contradiction.

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Yes. Depending on the things you know, you can use Bolzano-Weierstraß or limit superior/inferior or ... to prove this.

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