# Every sequence has at least one limit point

This is, of course, easily proven with the help of the Bolzano-Weierstrass theorem. However, going through the lecture notes in my class, the B-W theorem is only introduced after the claim

Every sequence has at least one limit point

is introduced, which itself is introduced as a corollary of:

Theorem:

1) The element $a\in\mathbb R\cup\{{-\infty,\infty}\}$ is a limit point of a sequence $\{X_n\}$ iff exists its subsequence $\{X_{k_n}\}$ converging to $a$.

2) Limit inferior of $\{X_n\}$ is the lowermost limit point of the sequence.

3) Limit superior of $\{X_n\}$ is the uppermost limit point of the sequence.

(Hopefully the idea is clear, this is only my own translation as the text is not english)

I don't quite see how does that imply that "every sequence has at least one limit point" and therefore would be grateful for any help, thanks.

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Every sequence has at least one limit point because the limit superior and limit inferior are examples of limit points of the sequence. The superior and inferior limits may coincide however (if and only if the sequence converges) so hence "at least one" rather than "at least two".

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Thanks, I see now where my problem with my original question is: I misunderstood the terms limit superior and limit inferior and need to read a bit more on that. Is it then true if I say that every sequence has a limit superior and limit inferior? – Dahn Jahn Dec 10 '12 at 13:27
@DahnJahn Yes, the limit superior and limit inferior of a sequence always exist, because they are defined to be the limits of $a_n=\sup_{k\geq n} \{ x_k \}$ and $b_n=\inf_{k\geq n} \{ x_k \}$ respectively, and both $a_n$ and $b_n$ are monotone (make sure you think about why) thus their limits exist. – Ragib Zaman Dec 10 '12 at 13:32