Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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?

share|improve this question
2  
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

1 Answer 1

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 $x_n>x_0+\epsilon$

As a matter of taste I’d then make a small change in the wording of the last sentence to match:

However, since for all $n>N_\epsilon$, $x_n<x_0+\epsilon$, there are infinitely many $n$ such that $x_n<x_0+\epsilon$.

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.