Proof correctness of: $\limsup a_n = \infty \implies \exists{a_{n_k}}$ such that $a_{n_k} \to \infty$

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.

-
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
$a^{*}_{n}$ 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.) –  Sangchul Lee 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

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 $S=\lim_{k\to\infty} a_{n_k}$.

-
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

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 $a_{2n+1}=n$).

-