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I am stuck on proofs with subsequences. I do not really have a strategy or starting point with subsequences.

NOTE: subsequential limits are limits of subsequences

Prove: $a_n$ is bounded $\implies \liminf a_n \leq \limsup a_n$

Proof:

Let $a_n$ be a bounded sequence. That is, $\forall_n(a_n \leq A)$.

If $a_n$ converges then $\liminf a_n = \lim a_n = \limsup a_n$ and we are done.

Otherwise $a_n$ has a set of subsequential limits we need to show $\liminf a_n \leq \limsup a_n$:

This is where I am stuck...

  • 6
    The boundedness hypothesis is irrelevant.2012-10-14

3 Answers 3

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Let $\{a_n\}$ be a bounded sequence. Then we define the sequences $\{a_n^+\}$ and $\{a_n^-\}$ by

$a_n^+=\sup\{a_n,a_{n+1}\dots\}$

$a_n^-=\inf\{a_n,a_{n+1}\dots\}$

We (may) then define

$\lim a_n^+=\limsup a_n$ $\lim a_n^-=\liminf a_n$

Now, you need two things to work this out:

$(1)$ Let $A$ be any bounded nonempty subset of $\Bbb R$. Then

$\inf A\leq \sup A$

$(2)$ Let $\{\alpha_n\}$ be a sequence such that $a_n\geq0 $ for each $n\in \Bbb N$. Then $\lim a_n\geq 0$

With $(1)$ you should show $a_n^-\leq a_n^+$ for each $n\in \Bbb N$. Monotone convergence says both $\{a_n^+\}$ and $\{a_n^-\}$ converge, since they are bounded (above/below) and are monotone (increasing/decreasing)$^{(*)}$. But $a_n^+- a_n^-\geq 0$

for each $n\in \Bbb N$, so use $(2)$ to show

$\lim a_n^+-\lim a_n^-\geq 0$

that is:

$\liminf a_n\leq \limsup a_n$

$(*)$ To prove this, you need to show that if $A\subseteq B$, then $\sup A\leq \sup B$ $\inf A\geq \inf B$ Then, observe that

$\{a_{n+1},a_{n+2},\cdots\}\subseteq \{a_n,a_{n+1},a_{n+2},\cdots\}$

  • 0
    Yes, thanks. Check out this one: http://math.stackexchange.com/questions/213324/suppose-a-n-is-a-sequence-with-l-1-l-2-subsequential-limits-supp2012-10-14
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Hint: Think about what the definitions mean. We have $\limsup a_n = \lim_n \sup \{ a_k \textrm{ : } k \geq n\}$ and $\liminf a_n = \lim_n \inf \{ a_k \textrm{ : } k \geq n\}$

What can you say about the individual terms $\sup \{a_k \textrm{ : } k \geq n\}$ and $\inf \{a_k \textrm{ : } k \geq n\}$ ?

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    Yes, $b_n$ is set to the supremum of the set of values $a_k$, for $k \geq n$. This is a single number, so that $b_n$ is a sequence (one value for each $n$).2012-10-14