The theorem:
I am trying to prove:
Let $\{s_n\}$ and $\{t_n\}$ be real-valued sequences. If $$\exists I\in \mathbb N, \forall n \geq I, s_n \leq t_n$$ Then $$\lim_{n\to \infty} sup (s_n) \leq \lim_{n\to \infty} sup (t_n) $$
Here, $$\lim_{n\to \infty} sup (u_n)$$ is the $sup$ of the set $E$ made of all the (real) subsequential limits of $\{u_n\}$, plus $\{+\infty\}$ and $\{-\infty\}$ added in case one (or more) subsequence goes to $\pm \infty$.
I will note it (according to W. Rudin's notation) $u^*$.
The proof:
Here is my attempt of proof (inspired from this proof). The main difficulty, and the reason I'm submitting it, is in handling the indices $I, I', I''$.
Given:
- $\exists I\in \mathbb N, \forall n \geq I, s_n \leq t_n$
Suppose $$s^* > t^*$$
Let $\epsilon = \frac{s^* - t^*}{2}$ and $x= s^* - \epsilon$
- Since $x> t^*$ we have (Rudin $3.17 (b)$) the existence of $I'\in \mathbb N$ such as $$\forall n \geq I', t_n <x$$
- Rudin $3.17 (a)$ gives: $s^* \in E$ (the $E$ of $\{s_n\}$). Therefore, $$\exists I'' \in \mathbb N , \forall k \geq I'', s_{n_k} \in (s^* - \epsilon, s^* + \epsilon)$$ $$ \exists I'' \in \mathbb N , \forall k \geq I'', s_{n_k} > s^* - \epsilon = x$$
Since $I''$ is a subsequence index let's call it $i''$.
$**$ Here is the part I'm not sure of $**$:
$n_k$ is a subsequence ($n_0 < n_1 < n_2 $ ...).
Let $i'$ be the smallest integer $k$ such as $n_k \geq I'$.
Let $i$ be the smallest integer $k$ such as $n_k \geq I$.
Then, if $i_{max} = max (i,i',i'')$ we have: $$\forall k \geq i_{max}, s_{n_k} > x > t_{n_k}$$ $$\forall k \geq i_{max}, s_{n_k} \leq t_{n_k}$$
Contradiction.