# proving that the if $s_{n} \leq t_{n}$ for $n\geq N$ $\liminf_{n\to \infty} s_{n} \leq \liminf_{n \to \infty}t_{n}$ [duplicate]

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 $$\limsup s_{n} \leq \limsup t_{n}$$

• Is $N$ just one natural number or does this indicate that it holds for any natural $N$? – Snoop May 3 at 13:37
• $\sup$ and $\inf$ are dual of each other (by a change of sign). Proving one is enough. – Yves Daoust May 3 at 13:38
• @Snoop he says "for $n \geq N$ where $N$ is fixed" – learningmathematics May 3 at 13:38
• @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 $- \epsilon$ – learningmathematics May 3 at 13:41

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.

• 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. – learningmathematics May 4 at 8:21

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 $$\limsup s_{n} \leq \limsup t_{n}$$

• It should be $s_k \leq t_k \leq \sup_{m\geq n}t_m$. – Danny Pak-Keung Chan 2 days ago
• Then argue that $\sup_{m\geq n}$ is an upper bound of the set $\{s_n, s_{n+1},\ldots\}$. – Danny Pak-Keung Chan 2 days ago
• Also, $(S_n)$ and $(T_n)$ are decreasing rather than increasing. – Danny Pak-Keung Chan 2 days ago
• @DannyPak-KeungChan is this now correct? – learningmathematics yesterday
• 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. – Danny Pak-Keung Chan yesterday