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Using Danny Pak-Keung Chan's answer I've tried to do the supremum case.

We want to show if $s_{n} \leq t_{n}$ then $\limsup s_{n} \leq \limsup t_{n}$

Let $n\geq N$ be arbitrary and let $k \geq N$ be arbitrary. Then $s_{k} \leq t_{k} \leq \sup_{m \geq n}t_{m}$. Thus $\sup_{m \geq n}t_{m}$ is an upper bound of the set $\{s_{n}, s_{n+1}, \dots ,\}$ therefore $\sup_{m \geq n}t_{m} \geq \sup_{m \geq n}s_{m}$.

Define $S_{n}:= \sup_{m \geq n}s_{m}$ and $T_{n}:= \sup_{m \geq n}t_{m}$. Then we have $S_{n}$ and $T_{n}$ are decreasing. Hence $S_{n} \leq T_{n}$. Therefore

$\lim_{n \to \infty}S_{n} \leq \lim_{n \to \infty}T_{n}$. Hence $\limsup s_{n} \leq \limsup t_{n}$

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  • $\begingroup$ Is $N$ just one natural number or does this indicate that it holds for any natural $N$? $\endgroup$ – Snoop May 3 at 13:37
  • $\begingroup$ $\sup$ and $\inf$ are dual of each other (by a change of sign). Proving one is enough. $\endgroup$ – Yves Daoust May 3 at 13:38
  • $\begingroup$ @Snoop he says "for $n \geq N$ where $N$ is fixed" $\endgroup$ – learningmathematics May 3 at 13:38
  • $\begingroup$ @YvesDaoust Right, i mean it's only really the final case that is much work. In that case I guess I can just do $s_{n} \leq t_{n} < t^{*} + \epsilon$ (for the supremum case) rather than $- \epsilon$ $\endgroup$ – learningmathematics May 3 at 13:41
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Let $n\geq N$ be arbitrary. Let $k\geq n$ be arbitrary. We have that $\inf_{m\geq n}s_{m}\leq s_{k}\leq t_{k}$. Since $k$ is arbitrary, we conclude that $\inf_{m\geq n}s_{m}$ is a lower bound of the set $\{t_{n},t_{n+1},\ldots\}$. Therefore, $\inf_{m\geq n}s_{m}$ is smaller than or equal to the greatest lower bound of $\{t_{n},t_{n+1},\ldots\}$, i.e, $\inf_{m\geq n}s_{m}\leq\inf_{m\geq n}t_{m}$. Denote $S_{n}=\inf_{m\geq n}s_{m}$ and $T_{n}=\inf_{m\geq n}t_{m}$. Note that $(S_{n})_{n\geq N}$ and $(T_{n})_{n\geq N}$ are increasing and $S_{n}\leq T_{n}$, so $\lim_{n}S_{n}\leq\lim_{n}T_{n}$. Hence, $\liminf s_{n}\leq\liminf t_{n}$. The proof for limsup is similar.

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  • $\begingroup$ I've tried to use your answer as a template for the supremum case, (answer below) but I think my answer is incorrect? I've used $\geq$ as the supremum is an upper bound. $\endgroup$ – learningmathematics May 4 at 8:21
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Using Danny Pak-Keung Chan's answer I've tried to do the supremum case.

We want to show if $s_{n} \leq t_{n}$ then $\limsup s_{n} \leq \limsup t_{n}$

Let $n\geq N$ be arbitrary and let $k \geq N$ be arbitrary. Then $s_{k} \leq t_{k} \leq \sup_{m \geq n}t_{m}$. Thus $\sup_{m \geq n}t_{m}$ is an upper bound of the set $\{s_{n}, s_{n+1}, \dots ,\}$ therefore $\sup_{m \geq n}t_{m} \geq \sup_{m \geq n}s_{m}$.

Define $S_{n}:= \sup_{m \geq n}s_{m}$ and $T_{n}:= \sup_{m \geq n}t_{m}$. Then we have $S_{n}$ and $T_{n}$ are decreasing. Hence $S_{n} \leq T_{n}$. Therefore

$\lim_{n \to \infty}S_{n} \leq \lim_{n \to \infty}T_{n}$. Hence $\limsup s_{n} \leq \limsup t_{n}$

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  • $\begingroup$ It should be $s_k \leq t_k \leq \sup_{m\geq n}t_m$. $\endgroup$ – Danny Pak-Keung Chan May 4 at 14:23
  • $\begingroup$ Then argue that $\sup_{m\geq n}$ is an upper bound of the set $\{s_n, s_{n+1},\ldots\}$. $\endgroup$ – Danny Pak-Keung Chan May 4 at 14:25
  • $\begingroup$ Also, $(S_n)$ and $(T_n)$ are decreasing rather than increasing. $\endgroup$ – Danny Pak-Keung Chan May 4 at 14:26
  • $\begingroup$ @DannyPak-KeungChan is this now correct? $\endgroup$ – learningmathematics May 5 at 17:24
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    $\begingroup$ Remark: In the first part, we have already proved that $S_n \leq T_n$. We mention that $(S_n)$ and $(T_n)$ are monotone sequences because we need to make sure that limits $\lim S_n$ and $\lim T_n$ exist. $\endgroup$ – Danny Pak-Keung Chan May 5 at 17:26

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