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 $c$
 A: 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.
A: 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...
A: There is a subsequence which converges to limsup of the original sequence and another sequence which converge to liminf of original sequence. As the original sequence diverges, its limsup and liminf must be different.
