# Help with limsup proof?

• Oct 31st 2009, 10:25 AM
paupsers
Help with limsup proof?
Let $\displaystyle {a_n}$ be a sequence and let $\displaystyle S_n={a_k:k\geq n}$. Prove:

limsup a_n = inf{sup Sn | n in N}

Now, since sup Sn is a single point, isn't inf{sup Sn} the same point?
So then limsup a_n = sup Sn which implies Sn = T, where T is the set of all subsequential limits?

Is that the correct way to interpret this problem?
• Oct 31st 2009, 10:32 AM
Laurent
Quote:

Originally Posted by paupsers
Let $\displaystyle {a_n}$ be a sequence and let $\displaystyle S_n=\{a_k:k\geq n\}$. Prove:

limsup a_n = inf{sup Sn | n in N}

Now, since sup Sn is a single point, isn't inf{sup Sn} the same point?

What do you mean? $\displaystyle \sup S_n=\sup_{k\geq n} a_k$depends of course on $\displaystyle n$.

If you want us to help you, you should give your definition of $\displaystyle \limsup_n a_n$ (because usually, it is $\displaystyle \limsup_n a_n = \inf_n \sup_{k\geq n} a_k$, which is your question...)
• Oct 31st 2009, 10:43 AM
tonio
Quote:

Originally Posted by paupsers
Let $\displaystyle {a_n}$ be a sequence and let $\displaystyle S_n={a_k:k\geq n}$. Prove:

limsup a_n = inf{sup Sn | n in N}

Now, since sup Sn is a single point, isn't inf{sup Sn} the same point?
So then limsup a_n = sup Sn which implies Sn = T, where T is the set of all subsequential limits?

Is that the correct way to interpret this problem?

I think there's a mistake in the notation: it should be $\displaystyle S_n=\{a_k\}_{k\geq n}=\{a_k, a_{k+1},...\}$,

Now you loot at $\displaystyle \inf (\sup_{\substack{n\in \mathbb{N}}}\,S_n)$ , and of course that, in case it exists and is finite, the supremum is a single point, but you're taking the infimum over a possibly infinite set of these numbers.

Tonio
• Oct 31st 2009, 02:56 PM
Laurent
Quote:

Originally Posted by tonio
I think there's a mistake in the notation: it should be $\displaystyle S_n=\{a_k\}_{k\geq n}=\{a_k, a_{k+1},...\}$,

Now you loot at $\displaystyle \inf (\sup_{\substack{n\in \mathbb{N}}}\,S_n)$ , and of course that, in case it exists and is finite, the supremum is a single point, but you're taking the infimum over a possibly infinite set of these numbers.

Tonio

Sorry to correct you: it is $\displaystyle \inf_{\substack{n\in \mathbb{N}}}\,(\sup S_n)$, i.e. we take the supremum of the set $\displaystyle S_n$, and then the infimum of these values when $\displaystyle n$ varies. Maybe it was good to clarify this again, since this was the first cause of problem in the original post, I think.

• Oct 31st 2009, 03:29 PM
tonio
Quote:

Originally Posted by Laurent
Sorry to correct you: it is $\displaystyle \inf_{\substack{n\in \mathbb{N}}}\,(\sup S_n)$, i.e. we take the supremum of the set $\displaystyle S_n$, and then the infimum of these values when $\displaystyle n$ varies. Maybe it was good to clarify this again, since this was the first cause of problem in the original post, I think.

Of course: that \substack thing did whatever it wanted (in fact, what I told it to but that was not what I meant)
It also should be $\displaystyle S_n=\{a_k\}_{k\geq n}=\{a_n, a_{n+1},...\}$.

Tonio
• Nov 1st 2009, 03:52 PM
paupsers
Quote:

Originally Posted by Laurent
What do you mean? $\displaystyle \sup S_n=\sup_{k\geq n} a_k$depends of course on $\displaystyle n$.

If you want us to help you, you should give your definition of $\displaystyle \limsup_n a_n$ (because usually, it is $\displaystyle \limsup_n a_n = \inf_n \sup_{k\geq n} a_k$, which is your question...)

Sorry for the confusion. I can't figure out how to get a bracket ({ }) to show in latex.

I'm just trying to prove that limsup $\displaystyle a_n=inf${$\displaystyle sup S_n|n \in N$} where $\displaystyle S_n=${$\displaystyle a_k|k \geq n$} and $\displaystyle a_n$ is a bounded sequence using only the fact that limsup $\displaystyle a_n=sup(T)$, where T is the set of all subsequential limits of $\displaystyle a_n$.

I realize that what I'm asking is a "definition," but I'm just trying to find a way to derive it, basically. Any help on this?
• Nov 2nd 2009, 11:40 AM
Laurent
Quote:

Originally Posted by paupsers
I'm just trying to prove that limsup $\displaystyle a_n=inf${$\displaystyle sup S_n|n \in N$} where $\displaystyle S_n=${$\displaystyle a_k|k \geq n$} and $\displaystyle a_n$ is a bounded sequence using only the fact that limsup $\displaystyle a_n=sup(T)$, where T is the set of all subsequential limits of $\displaystyle a_n$.

Let $\displaystyle L=\inf_n \sup S_n$. You should note first that the sequence $\displaystyle (\sup S_n)_n$ is decreasing hence $\displaystyle L=\lim_n \sup S_n$.

Since $\displaystyle \limsup_n a_n = \sup T$, you must prove:

a) that $\displaystyle L\geq\lim_n a_{\varphi(n)}$ for any convergent subsequence $\displaystyle (a_{\varphi(n)})_n$
b) that, for any $\displaystyle \varepsilon>0$, there is a convergent subsequence $\displaystyle (a_{\varphi(n)})_n$ such that $\displaystyle \lim_n a_{\varphi(n)}>L-\varepsilon$.

I let you try a) (compare $\displaystyle a_{\varphi(n)}$ with terms from the sequence $\displaystyle (\sup S_n)_n$)

As for b), you have to "construct" a convergent subsequence. In fact, instead of using $\displaystyle \varepsilon$ like above, you can directly construct a subsequence that converges toward $\displaystyle \sup T$ (which means that in fact $\displaystyle \sup T=\max T$). To do so, intuitively, choose $\displaystyle a_{\varphi(1)}$ close to $\displaystyle \sup S_1$ (it is a sup, hence it is a limit of values $\displaystyle a_k$ in $\displaystyle S_1$), then $\displaystyle a_{\varphi(2)}$ close(r) to $\displaystyle \sup S_{\varphi(n)+1}$, and so on: the subsequence is constituted by values that almost match the supremums. I let you formalize this idea. (If you don't succeed, ask for additional help)

NB: you'll probably have to distinguish cases depending whether $\displaystyle \sup T$ is finite or not.

--
About brackets in LaTeX: they're obtained through the commands \{ and \}. There's the same pattern for other restricted characters, like \# for #.
• Nov 2nd 2009, 12:23 PM
paupsers
Quote:

Originally Posted by Laurent
Let $\displaystyle L=\inf_n \sup S_n$. You should note first that the sequence $\displaystyle (\sup S_n)_n$ is decreasing hence $\displaystyle L=\lim_n \sup S_n$.

Since $\displaystyle \limsup_n a_n = \sup T$, you must prove:

a) that $\displaystyle L\geq\lim_n a_{\varphi(n)}$ for any convergent subsequence $\displaystyle (a_{\varphi(n)})_n$
b) that, for any $\displaystyle \varepsilon>0$, there is a convergent subsequence $\displaystyle (a_{\varphi(n)})_n$ such that $\displaystyle \lim_n a_{\varphi(n)}>L-\varepsilon$.

I let you try a) (compare $\displaystyle a_{\varphi(n)}$ with terms from the sequence $\displaystyle (\sup S_n)_n$)

As for b), you have to "construct" a convergent subsequence. In fact, instead of using $\displaystyle \varepsilon$ like above, you can directly construct a subsequence that converges toward $\displaystyle \sup T$ (which means that in fact $\displaystyle \sup T=\max T$). To do so, intuitively, choose $\displaystyle a_{\varphi(1)}$ close to $\displaystyle \sup S_1$ (it is a sup, hence it is a limit of values $\displaystyle a_k$ in $\displaystyle S_1$), then $\displaystyle a_{\varphi(2)}$ close(r) to $\displaystyle \sup S_{\varphi(n)+1}$, and so on: the subsequence is constituted by values that almost match the supremums. I let you formalize this idea. (If you don't succeed, ask for additional help)

NB: you'll probably have to distinguish cases depending whether $\displaystyle \sup T$ is finite or not.

--
About brackets in LaTeX: they're obtained through the commands \{ and \}. There's the same pattern for other restricted characters, like \# for #.

Sigh... all of that seems way above my head. Any way you could elaborate a bit more?