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

Re: sequence with convergent subsequence

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**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 $\displaystyle \{a_{n_k}:n_k\ge N\}\subseteq\{a_n:n\ge N\}$.

Re: sequence with convergent subsequence

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**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 $\displaystyle n_k$

Re: sequence with convergent subsequence

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**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?

Re: sequence with convergent subsequence

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**Plato** Are you sure that you understand what $\displaystyle ~\liminf~$ means?

that is what im not sure about, I dunno what it means

Re: sequence with convergent subsequence

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**wopashui** that is what im not sure about, I dunno what it means

Well then, no wonder you cannot answer questions about it.

Does it make sense to you that if

$\displaystyle A\subseteq B$ then $\displaystyle \inf(B)\le\inf(A)$

and $\displaystyle \sup(A)\le\sup(B)~?$

Suppose that $\displaystyle (a_n)$ is a sequence.

$\displaystyle \limsup(a_n)=\lim _{n \to \infty } \sup \left\{ {a_k :k \geqslant n} \right\}$

Re: sequence with convergent subsequence

we need to first show that $\displaystyle supa_n$ and $\displaystyle infa_n$ exist, do we?