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Let $(x_{n})\in\mathbb{R}^{+}$ be bounded and let $x_{0}=\lim\sup_{n\rightarrow\infty}x_{n}$. $\forall\epsilon>0$, prove that there are infinitely many elements less than $x_{0}+\epsilon$ and finitely many terms greater than $x_{0}+\epsilon$.

My attempt: By definition of limit superior, $\forall\epsilon>0$, $\exists N_{\epsilon}\in\mathbb{N}$ s.t. $\forall n>N_{\epsilon}$, $x_{n} . Since $\{x_{n}\}$ is a bounded sequence, there are only $N_{\epsilon}$ values of $\{x_{n}\}$ s.t. $x_{n}>r+\epsilon$. However, since for all $n>N_{\epsilon}$, $x_{n}, there are infinitely such $x_{n}.

I think my proof is probably incomplete/too informal. What do you think?

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    No, I think your argument is correct, except you need to be consistent with your notation by changing $r$ to $x_0$.2012-01-23

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