# i cant find the basic lim inf=lim sup convergence proof

• December 13th 2008, 03:08 AM
transgalactic
i cant find the basic lim inf=lim sup convergence proof
i look every where for that basic proof
that if
lim inf An=lim sup An then the sequence converges

??
• December 13th 2008, 03:32 AM
Moo
Hello,
Quote:

Originally Posted by transgalactic
i look every where for that basic proof
that if
lim inf An=lim sup An then the sequence converges

??

In my lecture notes, it is denoted as a definition.

That is :
$(A_n)$ converges if $\limsup A_n=\liminf A_n$
• December 13th 2008, 03:39 AM
transgalactic
i have to proove this definition
• December 13th 2008, 10:39 AM
Mathstud28
Quote:

Originally Posted by transgalactic
i look every where for that basic proof
that if
lim inf An=lim sup An then the sequence converges

??

Let $S$ be the set of all subsequential limits of $\left\{a_n\right\}$. Also let $\inf S=\sup S=\xi$. So it is clear that if $x\in S$ then $\xi\leqslant x \leqslant \xi\implies x=\xi$. Now let us define a subsequence of $\left\{a_n\right\}$ as $\left\{a_{\varphi}\right\}$ where $\varphi:\mathbb{N}\mapsto\mathbb{N}$ and $\varphi$ is strictly increasing. So since $\left\{a_{\varphi}\right\}\to\xi~\forall\varphi$ it is clear that there exists a $N$ such that $N\leqslant\varphi\implies d(a_{\varphi},\xi)<\varepsilon$.Now based on the above definition we may say that as soon as $\lceil\varphi^{-1}(N)\rceil\leqslant n\implies d\left(a_n,\xi\right)<\varepsilon$ which says exactly that not only does $a_n$ converge, but it converges to $\xi$.

That is how I would do it. Is that acceptable?

• December 14th 2008, 02:42 AM
Jhevon
Quote:

Originally Posted by Mathstud28
Let $S$ be the set of all subsequential limits of $\left\{a_n\right\}$. Also let $\inf S=\sup S=\xi$. So it is clear that if $x\in S$ then $\xi\leqslant x \leqslant \xi\implies x=\xi$. Now let us define a subsequence of $\left\{a_n\right\}$ as $\left\{a_{\varphi}\right\}$ where $\varphi:\mathbb{N}\mapsto\mathbb{N}$ and $\varphi$ is strictly increasing. So since $\left\{a_{\varphi}\right\}\to\xi~\forall\varphi$ it is clear that there exists a $N$ such that $N\leqslant\varphi\implies d(a_{\varphi},\xi)<\varepsilon$.Now based on the above definition we may say that as soon as $\lceil\varphi^{-1}(N)\rceil\leqslant n\implies d\left(a_n,\xi\right)<\varepsilon$ which says exactly that not only does $a_n$ converge, but it converges to $\xi$.

That is how I would do it. Is that acceptable?

i don't like this proof. i don't think it does the job. first, i think you mean to use limsup and liminf where you use sup and inf. secondly, your proof suggests that the sequence is constant, which of course, is not true for convergent sequences in general.
• December 14th 2008, 10:09 AM
Mathstud28
Quote:

Originally Posted by Jhevon
i think you mean to use limsup and liminf where you use sup and inf.

No I don't $\limsup a_n=\sup\left(S\right)$ where $S$ is the set of all subsequential limits. The same goes for $\liminf a_n=\inf\left(S\right)$
Quote:

secondly, your proof suggests that the sequence is constant, which of course, is not true for convergent sequences in general.
I'm not quite sure what you mean by this? Do you mean the f maps to a single value kind of constant or some other kind...either way I don't think I assumed anything
• December 14th 2008, 02:39 PM
Jhevon
Quote:

Originally Posted by Mathstud28
No I don't $\limsup a_n=\sup\left(S\right)$ where $S$ is the set of all subsequential limits. The same goes for $\liminf a_n=\inf\left(S\right)$

I'm not quite sure what you mean by this? Do you mean the f maps to a single value kind of constant or some other kind...either way I don't think I assumed anything

oh, ok, i missed your definition of S... sorry

i will look over the proof again later. i don't have time now