Proof of Theorem 3.19 from Baby Rudin. Is this correct?

Question

Here is the theorem:

Let $\{s_n \}$ and $\{t_n \}$ be sequences of real numbers. If $s_n\leq t_n$ for $n\geq N$, where $N$ is fixed, then $\lim_{n\to\infty}\inf s_n\leq\lim_{n\to\infty}\inf t_n$

$\lim_{n\to\infty}\sup s_n\leq\lim_{n\to\infty}\sup t_n$

Here's my attempted proof:

By definition: $\lim_{n\to\infty}t_n\leq\lim_{n\to\infty}\sup t_n$

This means there exists a $M$ such that, for each $n\geq M$

$t_n\leq\lim_{n\to\infty}\sup t_n=t^*$

If we pick $n\geq\max\{N,M\}$ then we have

$t^*\geq t_n\geq s_n$

Therefore for each subsequence $s_{n(k)}$ we must have (for each $n(k)\geq\max\{N,M\}$)

$t^*\geq s_{n(k)}$

Therefore the set of all subsequential limit of $\{s_n \}$ is bounded above by $t^*$ which means that $s^*$ is not bigger than $t^*$ itself ($t^*\geq s^*$)

Is this correct?

Answer 1

Note that: $$\lim_n \sup x_n=\inf_n \sup\{x_k:k \geq n\}$$ $$\lim_n \inf x_n =\sup_n \inf \{x_k:k \geq n\}$$

Thus $$ \sup\{s_k:k \geq N\} \leq \sup\{t_k:k \geq N\}$$ so $$\inf_n \sup\{x_k:k \geq n\} \leq \inf_n \sup\{t_k:k \geq n\}$$

I leave the other case for you.