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$\{x_n\}$ is a sequence such that every monotonic subsequence of $\{x_n\}$ converges to the limit $x$. Prove that: $x_n\rightarrow x$.

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Do you know that every sequence of reals has a monotone subsequence? –  David Mitra Dec 25 '12 at 16:48
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2 Answers

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Suppose that $x_n$ does not converge to $x$. Then there is a subsequence $x_{n_k}$ such that for every $k$ $$|x_{n_k}-x|>\epsilon$$ for some $\epsilon>0$.

Since every sequence of real numbers has a monotonic subsequence, we can pick a monotonic subsequence $x_{n_{k_l}}$.

And here we encounter a contraduction: $x_{n_{k_l}}$ is a monotonic subsequence of $x_n$ so it converges to $x$, and yet on the other hand it is a subsequence of $x_{n_k}$ and hence cannot converge to $x$.

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Choose a subsequence converging to $\limsup x_n$, by dropping terms you can assume that it is monotonic, so $\limsup x_n = x$. Similarly $\liminf x_n = x$. Therefore the limit of $x_n$ exists and $\lim x_n = x$.

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