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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...

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    Actually, `\liminf` and `\limsup` is a LaTeX command.2012-10-13
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    I've changed (sequences) to (sequences-and-series). From [FAQ about tags](http://meta.math.stackexchange.com/questions/107/faq-for-math-stackexchange/128#128): *Try to avoid creating new tags. Instead, check if there is some synonym that already has a popular tag.* It's not easy to keep balance between too specific tags and not having enough tags, but it is always good to search first and to ask yourself, whether newly created tag is not too specific. (Of course, you can disagree with the removal of the tag you've created, and there is possibility for further discussion, if needed.)2012-10-14
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    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

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

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    In 2) Why must $a_n \geq 0$?2012-10-14
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    It is the hypothesis. "**If** $a_n\geq0$ **then** $\lim a_n\geq 0$"2012-10-14
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    I don't see how to use this fact...2012-10-14
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    nevermind... hahah2012-10-14
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    @CodeKingPlusPlus Did you get it?2012-10-14
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    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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    What do you mean by the individual terms $\sup\{a_k: k \geq n\}$2012-10-14
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    My understanding of $\limsup$ is the supremum of subsequential limits and $\liminf$ is the infimum of subsequential limits. Where subsequential limits are limits of subsequences.2012-10-14
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    For instance $\limsup a_n$ is by definition the limit as $n$ goes to infinity of the sequence $b_n$, where $b_n = \sup \{a_k \textrm{ : } k \geq n\}$. Each $b_n$ is an `individual term' in the sequence.2012-10-14
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    Your definition is equivalent to mine (under the assumption that the sequence is bounded -think about why). But let's use your definition. What you are trying to prove is that the supremum of the subsequential limits must be bigger than (or equal to) the infimum of subsesquential limits. Tell me why that's true :)2012-10-14
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    What is the $n$ in $k \geq n$? Also, I don't understand how $b_n$ is a sequence. If you take the supremum of a set of numbers greater than $k$ Isn't $b_n$ just set to that supremum?2012-10-14
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    Given the sequence $a_n$, you may define a new sequence by the rule $$b_n = \sup\{a_k \textrm{ : } k \geq n\}$$. Then $\limsup a_n = \lim_{n \to \infty} b_n$. That is where the $n$ comes from.2012-10-14
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    let us [continue this discussion in chat](http://chat.stackexchange.com/rooms/6117/discussion-between-codekingplusplus-and-treble)2012-10-14
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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