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Let $\{X_n\}_n$ be a sequence bounded sequence. Its convergent proper subsequences converge to the same limit $\ell$. I want to prove that $\{X_n\}_n$ converges to $\ell$.

Notice that proper subsequences are all the sequences except for the sequence itself. Is it enough to say that $\{X_{2n}\}$ and $\{X_{2n+1}\}$ are convergent to $l$ then $\{X_n\}$ is convergent to $\ell$?

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    $\begingroup$ Actually one does not assume that $(x_{2n})$ and $(x_{2n+1})$ converge to $\ell$ since the hypothesis concerns only the convergent subsequences, not every subsequence. $\endgroup$ – Did Oct 1 '17 at 4:29
  • $\begingroup$ math.stackexchange.com/questions/397978/… $\endgroup$ – Guy Fsone Nov 11 '17 at 16:32
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Proof by contradiction

Suppose that $\{X_n\}$ does not converge to $\ell$. Then, there is $\varepsilon_0>0$ such that $$\forall N\in\mathbb N,\exists n=n(N) : n>N~~~and ~~~ |X_n -\ell|>\varepsilon_0 $$

For $N_1=1$ there exists $n_1$ such that $$n_1>N_1 ~~~and ~~~ |X_{n_1} -\ell|>\varepsilon_0 $$ Taking successively $N_{k+1}> \max\{N_k, n_k,k+1\}$ there exists $n_{k+1}>N_{k+1}$ such that,

$$ |X_{ n_{k+1}} -\ell|>\varepsilon_0 $$

It is easy to see that, $\{X_{ n_k}\}_k$ is a subsequence of $\{X_{ n}\}_n$ since $$ n_k< n_{k+1} \quad i.e ~~\text{the map }~~k\mapsto n_k~~~\text{Is one-to-one}$$

However, $$\forall k,~~ |X_{ n_{k}} -\ell|>\varepsilon_0 \qquad \text{and}~~~\{X_{ n_{k}} \}~~~\text{is bounded} $$

Therefore By Bolzano-Weierstrass Theorem's there exists $\{X_{ n_{k_p} }\}_p$ subsequence of $\{X_{ n_{k} }\}_k$ which converges to some limit $\ell_1 $ but $\{X_{ n_{k_p} }\}_p\to \ell_1$ is also a congering subsequence of $\{X_n\}_n$

By assumption, $\ell=\ell_1$ that is together with the fact $\{X_{ n_{k_p} }\}_p$ is a subsequence of $\{X_{ n_{k} }\}_k$ we have

$$0=\lim_{p\to\infty } |X_{ n_{k_p} }-\ell|>\varepsilon_0>0~~~\text{which is a CONTRADICTION}$$

Note that $$\forall p,~~|X_{ n_{k_p} }-\ell|>\varepsilon_0$$ Since $$\forall k,~~|X_{ n_{k}} -\ell|>\varepsilon_0$$

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  • $\begingroup$ Why can one assume that ℓ=ℓ1? Doesn't that lead to loss of generality? $\endgroup$ – Ahmad Lamaa Sep 23 '18 at 20:24
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    $\begingroup$ I did not assume. read the question again. $\ell$ is unique with a specific property which is fulfilled by $\ell_1$ $\endgroup$ – Guy Fsone Sep 26 '18 at 22:58

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