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

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?

• Your attempted proof fails from the very beginning: $\lim_{n\to\infty}t_n$ does not need to exist. Sep 27, 2017 at 19:11
• You cannot assume the sequence converges. Sep 27, 2017 at 19:12
• You're both right, i assumed the sequence converged. Is there a way to fix my proof so it would work somehow? I don't know, maybe proving the theorem in two cases: one when $\{t_n \}$ does converge and one when it does not? Sep 27, 2017 at 19:23
• See also this question and this question. Dec 5, 2019 at 13:37

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 $$\inf_n \sup\{x_k:k \geq n\} \leq \inf_n \sup\{t_k:k \geq n\}$$