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.

Is my proof correct? Prove: $a_n \leq b_n \implies \limsup a_n \leq \limsup b_n$


Let $a_n$ and $b_n$ be sequences such that $a_n \leq b_n \forall_n$. Suppose $\limsup a_n \nleq \limsup b_n$.

That is: $\limsup a_n > \limsup b_n$. From this we know:

$\forall_{\epsilon > 0} \exists_N \forall_{n>N} \implies |b_n - b| <\epsilon$. Where $b$ is the $\limsup b_n$.

$\forall_{\epsilon_1 > 0} \exists_{N_1} \forall_{n > N_1} \implies |a_n - a| < \epsilon_1$. Where $a$ is the $\limsup a_n$

So, let $a^* = a + \dfrac{\epsilon_1}{2}$ and let $b^* = b - \dfrac{\epsilon}{2}$.

Hence, $a^* \in |a_n - a| <\epsilon_1$ and $b^* \in |b_n - b| < \epsilon$. And clearly we see that $b^* < a^*$.Thus, we have found an element of $b_n$ namely $b^* < a^*$ an element of $a$. This is contradiction since we are given $a_n \leq b_n \forall_n$.

Therefore, the supposition is false, and $\limsup a_n \leq \limsup b_n$.

share|improve this question
This proof is for limits, not fot limsups –  Norbert Oct 14 '12 at 17:28
How do you know that $a^*$ and $b^*$ are elements of the sequences? –  Pedro Tamaroff Oct 14 '12 at 17:29
Good point... How can I reformulate that piece then? –  CodeKingPlusPlus Oct 14 '12 at 17:50
Could I choose the minimum of the interval for b and the maximum of the interval for a. And then conclude that I have found b < a? –  CodeKingPlusPlus Oct 14 '12 at 17:52

1 Answer 1

up vote 3 down vote accepted

Recall that given a bounded sequence, we define $\limsup a_n$ as $$\lim a_n^+$$ where $$a_n^+=\sup\{a_n,a_{n+1},\dots,\}$$

Now, if $a_n\leq b_n$ for each $n$, what can you say about the relaton among each of the values:

$$a_n^+=\sup\{a_n,a_{n+1},\dots,\}$$ $$b_n^+=\sup\{b_n,b_{n+1},\dots,\}$$

What can you then say about their limits?

share|improve this answer
Well I can say that each $b_n \geq a_n$. And then it follows that: $\lim a_n^+ \leq \lim b_n^+$ –  CodeKingPlusPlus Oct 14 '12 at 18:08
I mean each $a_n^+ \leq b_n^+$ And then it folows by taking the limit on both sides of the inequality: $\lim a_n^+ \leq \lim b_n^+$ –  CodeKingPlusPlus Oct 14 '12 at 18:25
@CodeKingPlusPlus Informally, yes. –  Pedro Tamaroff Oct 14 '12 at 18:38

Your Answer


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.