sequence with convergent subsequence

• Sep 24th 2011, 02:31 PM
wopashui
sequence with convergent subsequence
If $\displaystyle (a_{n_k})$ is a convergent subsequence of $\displaystyle (a_n)$, show that lim inf $\displaystyle _{n--> infinity}$ $\displaystyle a_n$ $\displaystyle <= $$\displaystyle lim_{k-->infinity}$$\displaystyle a_{n_k}$ $\displaystyle <=$ lim sup$\displaystyle _{n-->infinity}$$\displaystyle a_n. Im not sure what lim inf \displaystyle _{n--> infinity} \displaystyle a_n [tex] and lim sup\displaystyle _{n-->infinity}$$\displaystyle a_n$ mean, since $\displaystyle inf a_n$ and $\displaystyle sup a_n$ are a number
• Sep 24th 2011, 02:55 PM
Plato
Re: sequence with convergent subsequence
Quote:

Originally Posted by wopashui
If $\displaystyle (a_{n_k})$ is a convergent subsequence of $\displaystyle (a_n)$, show that lim inf $\displaystyle _{n--> infinity}$ $\displaystyle a_n$ $\displaystyle <= $$\displaystyle lim_{k-->infinity}$$\displaystyle a_{n_k}$ $\displaystyle <=$ lim sup$\displaystyle _{n-->infinity}$$\displaystyle a_n$.

This basically an easy question.
By definition of subsequence it must be the case that $\displaystyle k\le n_k$.
Thus $\displaystyle \{a_{n_k}:n_k\ge N\}\subseteq\{a_n:n\ge N\}$.
• Sep 25th 2011, 03:08 PM
wopashui
Re: sequence with convergent subsequence
Quote:

Originally Posted by Plato
This basically an easy question.
By definition of subsequence it must be the case that $\displaystyle k\le n_k$.
Thus $\displaystyle \{a_{n_k}:n_k\ge N\}\subseteq\{a_n:n\ge N\}$.

hmm, but we are talking about the limit here, not just K and $\displaystyle n_k$
• Sep 25th 2011, 03:23 PM
Plato
Re: sequence with convergent subsequence
Quote:

Originally Posted by wopashui
hmm, but we are talking about the limit here, not just K and $\displaystyle n_k$

Are you sure that you understand what $\displaystyle ~\liminf~$ means?
• Sep 25th 2011, 04:14 PM
wopashui
Re: sequence with convergent subsequence
Quote:

Originally Posted by Plato
Are you sure that you understand what $\displaystyle ~\liminf~$ means?

that is what im not sure about, I dunno what it means
• Sep 25th 2011, 05:03 PM
Plato
Re: sequence with convergent subsequence
Quote:

Originally Posted by wopashui
that is what im not sure about, I dunno what it means

$\displaystyle A\subseteq B$ then $\displaystyle \inf(B)\le\inf(A)$
and $\displaystyle \sup(A)\le\sup(B)~?$
Suppose that $\displaystyle (a_n)$ is a sequence.
$\displaystyle \limsup(a_n)=\lim _{n \to \infty } \sup \left\{ {a_k :k \geqslant n} \right\}$
we need to first show that $\displaystyle supa_n$ and $\displaystyle infa_n$ exist, do we?