0
$\begingroup$

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$ ?

$\endgroup$
1
$\begingroup$

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.

$\endgroup$
2
$\begingroup$

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.

$\endgroup$
1
$\begingroup$

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.

$\endgroup$
0
$\begingroup$

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

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.