At odd times when I am not writing my Ph.d thesis I am solving some problems from Polya-Szego book which seems to me really interesting. Here is the one of the problems (Problem 106 from Volume 1) which I've solved and would be grateful if you can take a look.

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> A convergent sequence has either a maximum or a minimum or both.
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Suppose $\{a\_n\}$ be a convergent sequence but there is no $\min \limits\_{n\in \mathbb{N}} a\_n$ and there is no $\max \limits\_{n\in \mathbb{N}}a\_n$.

It is easy to show the following fact: If the sequence $\{a\_n\}$ does not have a minimum then $\forall n\in \mathbb{N}$ $\exists k>n$ such that $a\_k<a\_n$.

In the same way you can show that if the sequence $\{a\_n\}$ does not have a maximum then $\forall n\in \mathbb{N}$ $\exists m>n$ such that $a\_m>a\_n$.

Then using it we can construct two increasing subsequences $n\_1<n\_2<n\_3<\dots$ and $m\_1<m\_2<m\_3<\dots$ such that $n\_1=m\_1=1$ and $a\_{n\_1}>a\_{n\_2}>a\_{n\_3}>\dots $ and $a\_{m\_1}<a\_{m\_2}<a\_{m\_3}<\dots$.

Since $\{a\_n\}$ is convergent then $a\_{n\_k}\to a$ and $a\_{m\_k}\to a$ as $k\to \infty$.
Then $a=\inf \limits\_{k\in \mathbb{N}}a\_{n\_k}$ and $a=\sup \limits\_{k\in \mathbb{N}}a\_{m\_k}$.

Hence $a\_{n\_1}>a$ and $a\_{m\_1}<a$ but this is a contradiction because $n\_1=m\_1=1$.

Please let me know is everything valid in this short proof?