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Is the following argument correct?

If $(x_n)$ is bounded and diverges, then there exist two subsequences of $(x_n)$ that converge to different limits.

Proof. Assume $(x_n)$ is bounded and diverges, then by Bolzano-Weierstrass theorem it follows that $(x_n)$ contains a subsequence $(x_{n_k})$ such that $(x_{n_k})\to\beta$ for some $\beta\in\mathbf{R}$. Now assume that every subsequence converges to the same limit, but then $(x_n)\to\beta$, a contradiction, thus we must have another subsequence $(x_{n_r})$ such that $(x_{n_r})\to\alpha$ and $\alpha\neq\beta$.

$\blacksquare$

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marked as duplicate by rtybase, Adrian Keister, Taroccoesbrocco, Leucippus, Xander Henderson Aug 2 '18 at 1:00

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  • $\begingroup$ I presume this question only really considers sequences in $\mathbb R$ $($or $\Bbb R^n)$. The proposition is not true in general. $\endgroup$ – Fimpellizieri Aug 1 '18 at 21:47
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No, it is not correct. You are trying to prove that some convergent subsequence converges to a number different from $\beta$. Denying this means that every subsequence either diverges or converges to $\beta$ too. But you only considered the second possibility.

You can prove the theorem as follows: since the sequence diverges, $\limsup_nx_n\neq\liminf_nx_n$. But $(x_n)_{n\in\mathbb N}$ has a subsequence whose limit is $\limsup_nx_n$ and a subsequence whose limit is $\liminf_nx_n$.

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