# Let $(x_n)$ be a divergent sequence in a compact subset of $\mathbb{R}^n$. Prove there are two subsequences that converge to different limits.

Let $(x_n)$ be a divergent sequence in a compact subset of $\mathbb R^n$. Prove that there are two subsequences of $(x_n)$ that are convergent to different limit points.

Some ideas that might be helpful:

Heine-Borel theorem states that a subset of $\mathbb R^n$ is compact if and only if it is closed and bounded.

Bolzano-Weierstrass Theorem, every bounded sequence contains a convergent subsequence

A number $c$ is a limit point of $(x_n)$ if there exists a subsequence of $(x_n)$ convergening to $c$

• You can find some inspiration here: math.stackexchange.com/questions/298817/… You just have to replace the absolute value by the norm in $\mathbb{R}^n$. – Julien Mar 2 '13 at 22:13
• @Ludolila: I don't understand why you have removed the part about the Bolzano-Weierstrass theorem from the post. – Asaf Karagila Mar 2 '13 at 22:18
• – Asaf Karagila Mar 2 '13 at 22:20
• @AsafKaragila : you're right... Sorry, OP... have no idea how it happened... I'll put it back... =) – Ludolila Mar 2 '13 at 22:21

By Bolzano Weierstrass you can pull out a convergent subsequence whose limit is some point $c$. Since your original sequence cannot converge, there's going to be a subsequence that doesn't converge to $c$. Now carefully pluck this subsequence so that it stays away at some fixed positive distance away from $c$. But this subsequence also satisfies Bolzano Weierstrass...
By Bolzano-Weierstrass $(x_{n})$ has a subsequence converging to some $x$. Now we know that $(x_n)$ does not converge to $x$ because it is divergent. Hence for some $\varepsilon>0$ we have $x_{n}\notin B(x,\varepsilon)$ for infinitely many $n$. Take these indices and denote the obtained subsequence by $(x_{n_{k}})$. This sequence is bounded and has a limit point (by Bolzano-Weierstrass) and it is different from $x$.
Hence $(x_n)$ has two different limit points.
• Note that we are in $\mathbb{R}^{n}$ and not in $\mathbb{R}$. How do you define limsup and liminf of a vector valued sequence? – T. Eskin Mar 3 '13 at 10:46