If all proper subsequences converge to same limit then the sequence converges.

Let $$\{X_n\}_n$$ be a 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 $$\ell$$?

• 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.
– Did
Oct 1 '17 at 4:29
• math.stackexchange.com/questions/397978/… Nov 11 '17 at 16:32

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 converging 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 $$\forall k,~~|X_{ n_{k}} -\ell|>\varepsilon_0$$

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