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.

Prove: $\limsup a_n = \infty \implies \exists{a_{n_k}} (a_{n_k} \to \infty)$

Is this correct?

Proof:

Consider $a_n^* =\inf\{a_k : k \geq n\}$. Thus, $a_n^*$ is monotone non-decreasing. $\limsup a_n = \infty \implies a_n^*$ is unbounded.

Thus, $a_n^*$ is unbounded and montone non-decreasing.

Therefore $a_n^* \to \infty$

Let me know if I need stronger justifications or how I can improve the clarity of my proof.

share|improve this question
2  
For me it is a pure matter of definition and, thus, there's nothing to prove. You must have different definitions. –  DonAntonio Oct 14 '12 at 4:18
    
Does what I am saying make sense though? –  CodeKingPlusPlus Oct 14 '12 at 4:41
1  
$a^{*}_{n}$ need not be a subsequence of $a_n$, putting aside the fact that it also need not diverge to $\infty$. (Consider an enumeration of the rational numbers as a counterexample.) –  sos440 Oct 14 '12 at 6:52
    
What if instead, I just took the maximum instead of the supremum? –  CodeKingPlusPlus Oct 14 '12 at 18:00
    
The [limsup] tag already includes questions about $\liminf$. –  Asaf Karagila Oct 14 '12 at 18:58
add comment

2 Answers 2

up vote 1 down vote accepted

First of all, you should use the fact that the sequence $$a_k^*=\sup\{a_n; n\ge k\}$$ converges to $+\infty$. (Since $\limsup a_n =\lim_{k\to\infty} \sup\{a_n; n\ge k\}$.) The same is not true about $\inf\{a_n; n\ge k\}$.

What we know about the sequence $a_k^*$.

  • It is non-increasing: $a_k^*\ge a_{k+1}^*$.
  • Since it is non-increasing, we know that $\lim_{k\to\infty} a_k^* = \inf_k a_k^*$. * The equation $\inf_k a_k^*=\infty$ implies that each $a_k^*$ is equal to $+\infty$.

If supremum of some set of real numbers is $+\infty$, then it contains arbitrarily large numbers. So we can construct $a_{n_k}$ inductively as follows: If we know $a_{n_1},\dots,a_{n_k}$, then we choose $a_{n_{k+1}}$ in a such way that:

  • $n_{k+1} > n_k$;
  • $a_{n_{k+1}} \ge k+1$.

We know existence of such $n_{k+1}$ from the fact that $\sup\{ a_n; n>n_k\}=+\infty$.


Maybe it is worth mentioning that this result is true not only for $+\infty$. If $S=\limsup a_n$, then there is a subsequence $(a_{n_k})$ such that $S=\lim_{k\to\infty} a_{n_k}$.

share|improve this answer
    
What if I just took the maximum instead of the supremum? This would tend to infinity and the elements selected by max would be elements of the sequence. Therefore I would have my subsequence with the desired properties. –  CodeKingPlusPlus Oct 14 '12 at 18:12
    
You cannot say $\max\{a_n; n\ge k\}$, since maximum need not exist. I've corrected the proof I've posted before - it was incorrect. The mistake was that all $a^*_k$'s are equal to infinity and I was working with them as if they were finite numbers. –  Martin Sleziak Oct 14 '12 at 18:55
    
Why is it that $\lim_{k \to \infty} a_k^* = \inf_k a_k^*$. Also, what is $\inf_k$? In the inductive construction, What is the reason behind $a_{n_{k+1}} \geq k+1$ –  CodeKingPlusPlus Oct 14 '12 at 19:44
    
If $(b_k)$ is a sequence, then $\inf_k b_k$ is just a different notation for $\inf\{b_k; k\in\mathbb N\}$. If a sequence is decreasing (or non-increasing) then the limit is equal to infimum. (See Wikipedia, where a proof is given for an increasing sequence and supremum.) The reason why I chose that $(k+1)$-th term of the subsequence is $\ge k+1$, is that this implies that the subsequence tends to $+\infty$. –  Martin Sleziak Oct 14 '12 at 19:57
add comment

I would suggest to define $n_k=\inf\{t\in\mathbf{N},a_t>k\}$. For each $k$, the set $\{a_t>k\}$ is non-empty because of your assumption on limsup. The sequence $(n_k)_k$ is increasing.

The sequence $(a_{n_k})_k$ is a subsequence of $(a_n)$ and its limit is $+\infty$.

In the original post, your $(a_n^*)_n$ might be constant (define $a_{2n}=1$ and $a_{2n+1}=n$).

share|improve this answer
add comment

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.