- #1

joypav

- 151

- 0

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?

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,

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

$\implies$ Claim 1 is true.

Then,

$lim_{n \rightarrow \infty}inf_{m \ge n}a_m \le lim_{n \rightarrow \infty}sup_{m \ge n}a_m$

$\implies lim_{n \rightarrow \infty}infa_n \le lim_{n \rightarrow \infty}supa_n$

**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,

$lim_{n \rightarrow \infty}inf_{m \ge n}a_m \le lim_{n \rightarrow \infty}sup_{m \ge n}a_m$

$\implies lim_{n \rightarrow \infty}infa_n \le lim_{n \rightarrow \infty}supa_n$

Last edited: