Re: prove limsup-liminf<2

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**wopashui** Suppose that there exists $\displaystyle N \in N$, such that $\displaystyle |a_n-a_m|<1 $for all $\displaystyle n,m>=N$ (this will be true, in particular, if $\displaystyle (a_n)$ is cauchy). Prove that $\displaystyle (limsup_{n-->\infty}a_n)-(liminf_{n-->\infty}a_n)$$\displaystyle <=2$.

it seems like an easy proof, but i can't find the connection between limsup and cauchy sequence, need some help.

I will leave the epsilons to you but here is an idea. First there exists a subsequence

$\displaystyle a_{n}_k$

that converges to the limit supreemum and another subsequence

$\displaystyle a_{n}_j$

that converges to the lim inf

Now

$\displaystyle |L_{\text{sup}}-L_{\text{inf}}| \le |L_{\text{sup}}-a_n+a_n-L_{\text{inf}}| \le |L_{\text{sup}}-a_n|+|a_n-L_{\text{inf}}| = |L_{\text{sup}}-a_{n}_k+a_{n}_k-a_n|+|a_n-a_{n}_j+a_{n}_j-L_{\text{inf}}| $

Now just use the triangle inequality again and find the n,j, and k's to show what you want

Re: prove limsup-liminf<2

Quote:

Originally Posted by

**wopashui** Suppose that there exists $\displaystyle N \in N$, such that $\displaystyle |a_n-a_m|<1 $for all $\displaystyle n,m>=N$ (this will be true, in particular, if $\displaystyle (a_n)$ is cauchy). Prove that $\displaystyle (limsup_{n-->\infty}a_n)-(liminf_{n-->\infty}a_n)$$\displaystyle <=2$.

it seems like an easy proof, but i can't find the connection between limsup and cauchy sequence, need some help.

I think that you have the wrong idea here. The given does not prove that the sequence is Cauchy. Rather it asks you to note that the conclusion would apply to Cauchy sequences.

First, you must prove the the $\displaystyle \liminf~\&~\limsup$ both exist.

To do that notice that from the given $\displaystyle \left( {\forall n \geqslant N} \right)\left[ {\left| {a_n - a_N } \right| \leqslant 1\, \Rightarrow \, - 1 + a_N \leqslant a_n \leqslant 1 + a_N } \right]$

Thus they exist and if $\displaystyle \lambda=\liminf{a_n}~\&~\sigma= \limsup{a_n} $ then $\displaystyle -1+a_N\le\lambda~\&~\sigma\le 1+a_N$.

Now you finish

Re: prove limsup-liminf<2

thanks, so is TheEmptySet's approach correct?

Re: prove limsup-liminf<2

Quote:

Originally Posted by

**Plato** I think that you have the wrong idea here. The given does not prove that the sequence is Cauchy. Rather it asks you to note that the conclusion would apply to Cauchy sequences.

First, you must prove the the $\displaystyle \liminf~\&~\limsup$ both exist.

To do that notice that from the given $\displaystyle \left( {\forall n \geqslant N} \right)\left[ {\left| {a_n - a_N } \right| \leqslant 1\, \Rightarrow \, - 1 + a_N \leqslant a_n \leqslant 1 + a_N } \right]$

Thus they exist and if $\displaystyle \lambda=\liminf{a_n}~\&~\sigma= \limsup{a_n} $ then $\displaystyle -1+a_N\le\lambda~\&~\sigma\le 1+a_N$.

Now you finish

why are you using N not m here? is it because if $\displaystyle a_m $is not bounded, then $\displaystyle a_n$ will not be bounded, but $\displaystyle a_N $is fixed, so $\displaystyle a_n$ is bounded. but why do we have <= not strictly < here?

Re: prove limsup-liminf<2

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Originally Posted by

**wopashui** why are you using N not m here? is it because if $\displaystyle a_m $is not bounded, then $\displaystyle a_n$ will not be bounded, but $\displaystyle a_N $is fixed, so $\displaystyle a_n$ is bounded. but why do we have <= not strictly < here?

Do you understand any of this?

It was convenient to use a fixed $\displaystyle a_N~,~N=m$.

Surely you understand that if $\displaystyle |a-b|<1$ then it is also true that $\displaystyle |a-b|\le 1$.