Let ${(a_n)}$ be a bounded sequence that does not converge.

Question

Let ${(a_n)}$ be a bounded sequence that does not converge. Show that ${(a_n)}$ has two subsequences that converge to different limits.

I think I am supposed to prove $(a_n)$ does not converge and for some reason all I can think of doing is to show it is not a monotone sequence but I do not know how, or maybe using the Bolzano Weierstrass Theorem. I am completely confused any help would be much appreciated.

Are you familiar with the fact that the limit exists iff $\liminf_n a_n=\limsup_n a_n$? — Alex R., Sep 29 '20 at 20:29
What space do you have at hand ? metric, vector (finite or infinite dimensional) ,topological ,.. — Physor, Sep 29 '20 at 20:31

score 2 · Answer 1 · answered Sep 29 '20 at 20:30

Hint :

Apply Bolzano-Weierstrass to get a subsequence that converges to a limit point $a$.
Because $(a_n)$ does not converge to $a$, show that there is a subsequence of $(a_n)$ which stays at a distance $> \varepsilon >0$ of $a$, for a certain $\varepsilon$.
Apply Bolzano-Weierstrass again to this new subsequence, and conclude.

score 1 · Answer 2 · answered Sep 29 '20 at 20:28

1

Hint. If it's bounded and does not converge, show that the $\limsup$ must be strictly greater than the $\liminf$.

answered Sep 29 '20 at 20:28

Ethan Bolker

2 Answers2