Definition of Limit Superior Bartle & Sherbert Theorem $3.4.11$

Question

I'm trying to understand the proof of equivalent definitions of the limit superior.

Given the definition that $X=(x_n)$ is a bounded sequence of real numbers and that the limit superior of $(x_n)$ is the infimum of the set $V$ of $v\in \mathbb{R}$ such that $v<x_n$ for at most a finite number $n\in\mathbb{N}$, I want to show the following implication, $(b)\implies (c)$, about a real number $x^*$:

$(b)$ If $\epsilon>0$, there are at most a finite number of $n\in\mathbb{N}$ such that $x^*+\epsilon < x_n$, but an infinite number of $n\in\mathbb{N}$ such that $x^*-\epsilon<x_n$.

$(c)$ If $u_m=\sup\{x_n:n\geq m\}$, then $x^*=\inf\{u_m:m\in\mathbb{N}\}=\lim(u_m)$.

The proof starts as follows:

Given $\epsilon>0$, $(b)$ implies for all sufficiently large $m$ that $u_m<x^*+\epsilon$. Therefore, $\inf\{u_m:m\in\mathbb{N}\}\leq x^*+\epsilon$.

I'm trying to understand why this start of the proof is correct rigorously. Here's my attempt to derive that statement:

From part $(b)$, we know that for all $n$ greater than some $m\in\mathbb{N}$, $x_n\leq x^*+\epsilon$. Since $u_m$ is the supremum of $\{x_n:n\geq m\}$, for any given $\epsilon>0$, we have $u_m-\epsilon<x_n\leq u_m$ for some $n\geq m$. I feel like these are the two pieces of data I need to combine into "$(b)$ implies for 'all sufficiently large $m$' that $u_m<x^*+\epsilon$" but I'm not sure how.

How does that imply anything about the set $\{u_m:m\in\mathbb{N}\}$ and its infimum, in which $u_m$'s index ranges over all natural numbers unlike in the previous "sufficiently large $m$" statement?

Answer 1

For all $n\ge m$, $x_n\le x^*+\epsilon$. $x^*+\epsilon$ is an upper bound for the set $\{x_n: n\ge m\}$. And supremum is the smallest bound so $u_m:=\sup\{x_n: n\ge m\}\le x^*+\epsilon$. Note that $u_m$ exists because of completeness of $\mathbb R$.

Note that $(u_n)_{n\ge m}$ is a monotonically decreasing sequence and is bounded so must converge to $u:=\inf \{u_n:n\ge m\}$.

Note that $u\le u_m$ by definition of a lower bound. It follows that $u\le x^*+\epsilon.$ Since $\epsilon$ is arbitrary, it follows that $u\le x^*$.

Suppose on the contrary that $u<x^*$, then $\color{blue}{u<x_n}$ for infinitely many $n$. Since $u_n$ converges to $u$, for all large $n$, $u_n\le \frac{x^*+u}2$. That is, $x_n\le \frac{x^*+u}2<x^*$ for all large $n$. This contradicts condition in blue color. So $u\ge x^*$. This alongwith last para gives $u=x^*$.

Answer 2

Wait, is it this simple?

From part $(b)$, we know that given $\epsilon=\frac{\epsilon_1}{2}>0$ with $\epsilon_1>0,$ then for all $n$ greater than some $m\in\mathbb{N}$, $x_n\leq x^*+\epsilon=x^*+\frac{\epsilon_1}{2}$. We also know that since $u_m$ is the supremum of $\{x_n:n\geq m\}$, for any $\epsilon>0$, $u_m-\epsilon<x_n\leq u_m$ for some $n\geq m$. Thus, combining the two inequalities gives $$u_m-\epsilon=u_m-\frac{\epsilon_1}{2}<x_n\leq x^*+\epsilon=x^*+\frac{\epsilon_1}{2}$$ or $$u_m-\frac{\epsilon_1}{2}<x^*+\frac{\epsilon_1}{2}$$

after adding $\frac{\epsilon_1}{2}$ to both sides $$u_m<x^*+\epsilon_1$$ for arbitrary $\epsilon_1>0.$

Now since $u_m<x^*+\epsilon_1$ for all $m$ sufficiently large, we know $\inf\{u_m:m\in\mathbb{N}\}\leq x^*+\epsilon_1$ since only a finite number of terms $u_m$ are such that $u_m> x^*+\epsilon_1$, and we'd need all terms of $u_m$ to be greater than $x^*+\epsilon_1$ for $\inf\{u_m:m\in\mathbb{N}\}\geq x^* +\epsilon_1$ to hold.