MHB Real Analysis, liminf/limsup inequality

[joypav]

I am working a bunch of problems for my Real Analysis course.. so I am sure there are more to come. I feel like I may have made this proof too complicated. Is it correct? And if so, is there a simpler method?

Problem:
Show that $liminfa_n \leq limsupa_n$.

Proof:
Consider a sequence of real numbers, $(a_n)$.
By the definitions of inf and sup, we know,

$\forall n \in N, infa_m \leq supa_m, m \in N$

Now consider the following sequences:

$(inf_{m \ge n}a_m)$ and $(sup_{m \ge n}a_m)$

We know that,

(1.) $inf_{m \ge n}a_m \le sup_{m \ge n}a_m$ for all $n > N$, some $N \in \Bbb{N}$

Claim 1: $lim_{n \rightarrow \infty}inf_{m \ge n}a_m \le lim_{n \rightarrow \infty}sup_{m \ge n}a_m $
Proof of Claim:
Let $\underline{a'}=lim_{n \rightarrow \infty}inf_{m \ge n}a_m$ and $\overline{a'}=lim_{n \rightarrow \infty}sup_{m \ge n}a_m$
Suppose the claim is false, then $\underline{a'}>\overline{a'}$.
Given $\epsilon>0, \exists N_1 \in \Bbb{N}, \vert inf_{m \ge n}a_m - \underline{a'} \vert < \frac{\epsilon}{2}$ and
$\exists N_2 \in \Bbb{N}, \vert sup_{m \ge n}a_m - \overline{a'} \vert < \frac{\epsilon}{2}$

Choose $\epsilon = \frac{\underline{a'}-\overline{a'}}{2} > 0$. (Because $\underline{a'} > \overline{a'} \implies \underline{a'}-\overline{a'} > 0$).
If $n > max{N,N_1,N_2}$, then
$inf_{m \ge n}a_m > \underline{a'} - \epsilon = \underline{a'} - \frac{\underline{a'}-\overline{a'}}{2} = \overline{a'} + \frac{\underline{a'}-\overline{a'}}{2} = \overline{a'} + \epsilon > sup_{m \ge n}a_m$
$\implies inf_{m \ge n}a_m > sup_{m \ge n}a_m$, a contradiction of (1.).
$\implies$ Claim 1 is true.

Then,
$lim_{n \rightarrow \infty}inf_{m \ge n}a_m \le lim_{n \rightarrow \infty}sup_{m \ge n}a_m$
$\implies lim_{n \rightarrow \infty}infa_n \le lim_{n \rightarrow \infty}supa_n$

 

[Euge]

Three things:

1. Based on your argument, it looks like there’s an implicit assumption that $(a_n)$ is bounded. Is that an assumption in the problem?

2. The inequality $\inf_{m\ge n} a_m \le \sup_{m\ge n}a_m$ is true for all positive integers $n$, not just all $n$ greater than some positive integer $N$.

3. In your argument, if you choose $N_1$ such that $\lvert \inf_{m\ge n} a_m - \underline{a’}\rvert < \epsilon$ for all $n > N_1$ and $\lvert \sup_{m\ge n} a_m - \overline{a’}\rvert < \epsilon$ for all $n > N_2$ (where $\epsilon = \frac{\underline{a’} - \overline{a’}}{2}$), then the string of inequalities you have at the end make sense for $n > \max\{N_1,N_2\}$.

The theorem itself can be proven directly as follows. First, suppose $(a_n)$ is unbounded. If $\limsup a_n =+\infty$, then the inequality is clear. If $\limsup a_n = -\infty$, given $\epsilon > 0$, $a_k < -\epsilon$ for all sufficiently large $k$. Thus $\liminf a_n \le -\epsilon$. Since $\epsilon$ was arbitrary, $\liminf a_n = -\infty$ and the inequality is satisfied.

Now suppose $(a_n)$ is bounded. For each $n$ and $k$ with $k > n$, $a_k \le \sup_{m \ge n} a_m$. Thus, for each $n$, $\inf_{k \ge n} a_k \le \sup_{m \ge n} a_m$. Taking limits as $n \to \infty$ results in $\liminf_{n \to \infty} a_n \le \limsup_{n\to \infty} a_n$.

 

[joypav]

That is... much better. Thanks!