# limit superior of a sequence proof

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. $\forall n>N_{\epsilon}$, $x_{n}<r+\epsilon$ . 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 $n>N_{\epsilon}$, $x_{n}<r+\epsilon$, there are infinitely such $x_{n}<r+\epsilon$.

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

• No, I think your argument is correct, except you need to be consistent with your notation by changing $r$ to $x_0$. – Paul Jan 23 '12 at 2:45

As Paul said in the comments, your argument is basically just fine, apart from using $r$ where the question has $x_0$. There are a few minor problems with this sentence in addition to the use of $r$ for $x_0$:
Since $\{x_n\}$ is a bounded sequence, there are only $N_\epsilon$ values of $\{x_n\}$ s.t. $x_n>r+\epsilon$.
First, you should omit the opening clause, since you’re not actually using the boundedness of the sequence here: what you’re using is simply the definition of $N_\epsilon$. (The boundedness of the sequence is needed only to ensure that the limit superior is finite, so that the question makes sense.) Secondly, there may not be $N_\epsilon$ terms that exceed $x_0+\epsilon$: some of the first $N_\epsilon$ terms might be $\le x_0+\epsilon$. However, there are definitely at most $N_\epsilon$ such terms. Finally, ‘values of ${x_n}$’ reads a bit oddly: ‘terms of ${x_n}$’ or ‘values of $n$’ would be clearer. After making these changes, the sentence becomes:
By the choice of $N_\epsilon$ there are at most $N_\epsilon$ values of $n$ such that $x_n\ge x_0+\epsilon$ and therefore at most $N_\epsilon$ values of $n$ such that $x_n>x_0+\epsilon$
However, since for all $n>N_\epsilon$, $x_n<x_0+\epsilon$, there are infinitely many $n$ such that $x_n<x_0+\epsilon$.