I am interested in proving Theorem 3.19 in Rudin only when $s^*$ and $t^*$ are infinite. (Many other posts on Math.SE prove the theorem when $s^*$ and $t^*$ are finite). While the proof for the infinite part might be trivial, I just want to make sure that I am not missing something. For brevity, I'll just prove the superior limits part of the theorem.
If $s_n\leqslant t_n$ for $n\geqslant N$, where $N$ is fixed, then $$\limsup_{n\to \infty} s_n\leqslant \limsup_{n\to \infty} t_n$$ (In alternative notation: $s^* \le t^*$). $$\liminf_{n\to \infty} s_n\leqslant \liminf_{n\to \infty} t_n$$
Okay, now for the proof. There are 4 possible cases when $s^*$ and $t^*$ are infinite:
- $s^* = t^* = +\infty \implies s^* \le t^*$
- $s^* = t^* = -\infty \implies s^* \le t^*$
- $s^* = +\infty, t^* = -\infty$
- $s^* = -\infty, t^* = +\infty \implies s^* \le t^*$
We argue that we can ignore Case 3 since that is never realized and contradicts our hypothesis of $s_n\leqslant t_n$. (Similarly for the inferior limits, we will be able to ignore Case 4). If $s^* = +\infty$, then we can find a subsequence $\{s_{n_k}\} \to +\infty$. This means that $\forall M \in \mathbb{R}, \exists K \in \mathbb{Z}$ such that
\begin{equation}\label{heck1}
k \ge K \implies s_{n_k} \ge M
\end{equation}
Similarly, we can find a $t_{n_p} \to -\infty$ which means that for some real-valued $Y < M$, $\exists P \in \mathbb{Z}$ such that
\begin{equation}\label{heck2}
p \ge P \implies t_{n_p} < Y
\end{equation}
Put $Z = \max \{N, K, P\}$. Then, $\forall n \ge Z, s_n \le t_n$ by the hypothesis. However, we also have that $ k, p \ge Z$ imply that $s_{n_k} \ge M > Y > t_{n_p}$, which is a contradiction.
Is this proof correct?
Best Answer
Your proof is correct, but there is a more direct way. Since $s_n\leq t_n$ then $$\sup_{i = 1,...,n} s_n \leq \sup_{i = 1,...,n} t_n$$ Now you can just pass the limit and use the property of order for the limit to conclude.