Proofs with limit superior and limit inferior: $\liminf a_n \leq \limsup a_n$

Question

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...

Answer 1

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

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

Answer 2

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 $\inf \{a_k \textrm{ : } k \geq n\}$ ?

Answer 3

To make @Pedro's answer more complete, here's how things work in the unbounded case. Now we're given any sequence $\{a_n\}$, and would like to show $\liminf a_n \leq \limsup a_n$. I use the more general definition: $\liminf a_n := \sup_N a_N^- = \sup_N \inf_{n\geq N} a_n$, and $\limsup a_n := \inf_N a_N^+ = \inf_N \sup_{n\geq N} a_n$. I also use the standard definition of $\sup$ for a subset $E$ of the extended reals $\mathbb{R}^* = \mathbb{R} \cup{ \{\infty, -\infty \}}$ (see definition 6.2.6 in Analysis I by Terrance Tao).

Suppose $\{a_n\}$ does not have an upper bound. Then $\forall n, a_n^+ = \infty$, so $\limsup a_n = \inf \{\infty\} = \inf \{\} = \infty$; the claim immediately follows from the standard arithmetic on $\mathbb{R}^*$.

Similarly, if $\{a_n\}$ does not have a lower bound, we have $\liminf a_n =\sup \{-\infty\} = \sup \{\} = -\infty \leq \limsup a_n$.

We're now left with the case where $\{a_n\}$ is bounded, which is handled by @Pedro's answer.