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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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To get the best possible answers, you should explain what your thoughts on the problem are so far. That way, people won't tell you things you already know, and they can write answers at an appropriate level; also, people are much more willing to help you if you show that you've tried the problem yourself. Also, many would consider your post rude because it is a command ("Prove..."), not a request for help, so please consider rewriting it. – Zev Chonoles Dec 25 '12 at 16:44
Do you know that every sequence of reals has a monotone subsequence? – David Mitra Dec 25 '12 at 16:48
up vote 3 down vote accepted

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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