is $\lim_{n\to \infty} \sup_{i \in I}(x^n_i) =\sup_{i \in I}(\lim_{n\to\infty}(x^n_i)$? for each i, $x_i^n$ is a monotone increasing sequence.

Question

Is $\displaystyle{\lim_ {n\to \infty} \sup_ {i \in I} (x^n_i) =\sup_ {i \in I} \ \left(\lim_{n\to\infty}(x^n_i)\right)}$ ?
For each $i \in I$, $(x^n_i)_{n \in N}$ is a monotone increasing sequence. We can write $\lim_{n\to\infty}(x^n_i) = x_i$. I don't know how to start approaching this problem. Any help would be appreciated.

Answer 1

A lot of time has passed, so I'll answer this fully now.

Since $\left(x_i^{(n)}\right)_n$ is increasing, we have, for all $i \in I$ and all $n$: $$x_i^{(n+1)} \geq x_i^{(n)}$$ By taking the $\sup$ on the left side, we get, for all $i$: $$\sup_{i \in I} x_i^{(n+1)} \geq x_i^{(n)}$$ Since the left side is constant compared to $i$, we can take the $\sup$ on the right side, which grants: $$\sup_{i \in I} x_i^{(n+1)} \geq \sup_{i \in I} x_i^{(n)}$$ Therefore, the sequence $\left(\sup_{i \in I} x_i^{(n)}\right)_n$ is also increasing, which implies that all $\lim$s in the starting equation exist and were all $\sup$s, meaning that what we wanted to prove is equivalent to: $$\sup_{n \in \mathbb{N}} \sup_{i \in I} x_i^{(n)} = \sup_{i \in I} \sup_{n \in \mathbb{N}} x_i^{(n)}\,\,\,?$$

Let $i \in I$ and $n \in \mathbb{N}$. We have, quite trivially: $$x_i^{(n)} \leq \min\left(\sup_{i \in I}\sup_{n \in \mathbb{N}} x_i^{(n)}, \sup_{n \in \mathbb{N}}\sup_{i \in I} x_i^{(n)}\right)$$ By then taking the $\sup$s in two different ways, we obtain: $$\max\left(\sup_{i \in I}\sup_{n \in \mathbb{N}} x_i^{(n)}, \sup_{n \in \mathbb{N}}\sup_{i \in I} x_i^{(n)}\right) \leq \min\left(\sup_{i \in I}\sup_{n \in \mathbb{N}} x_i^{(n)}, \sup_{n \in \mathbb{N}}\sup_{i \in I} x_i^{(n)}\right)$$ Hence: $$\max\left(\sup_{i \in I}\sup_{n \in \mathbb{N}} x_i^{(n)}, \sup_{n \in \mathbb{N}}\sup_{i \in I} x_i^{(n)}\right) = \min\left(\sup_{i \in I}\sup_{n \in \mathbb{N}} x_i^{(n)}, \sup_{n \in \mathbb{N}}\sup_{i \in I} x_i^{(n)}\right)$$ which finally yields: $$\sup_{i \in I}\sup_{n \in \mathbb{N}} x_i^{(n)} = \sup_{n \in \mathbb{N}}\sup_{i \in I} x_i^{(n)}$$