Prove this proposition 2.1.13 in Induction to Real Analysis by Jiri Lebel

[cbarker1]

Dear Everybody,

I need some help with seeing if there any logical leaps or any errors in this proves.

Corollary 1.2.8 to Proposition 1.2.8 states:
if $S\subset\Bbb{R}$ is a non-empty set, bounded from below, then for every $\varepsilon>0$ there exists a $y\in S$ such that $\inf S+\varepsilon>y\ge \inf S$.

Proof: Let $-S\subseteq\Bbb{R}$ and $-S\ne\emptyset$. Since $-S$ is bounded above by the least-upper-bound property of real numbers, there exists a $\sup (-S)$. By Proposition 1.2.8, for all $\varepsilon>0$ there exists a $y\in-S$ such that $\sup(-S)-\varepsilon<y\le\sup(-S).$ By Proposition 1.2.6 v, which states if $y<0$ and $S$ is bounded below, then $\sup(yS)=y(\inf S)$, $-\inf S-\varepsilon<y\le-\inf S$. We factored out the -1 in the inequality, then $\inf S +\varepsilon>y\ge\inf S$. QED

Proposition 2.1.13
Let $S\subseteq \Bbb{R}$ be a nonempty bounded set. Then, there exist monotone sequences $\left\{{x}_{n}:n\in\Bbb{N} \right\}$ and $\left\{{y}_{n}:n\in\Bbb{N} \right\}$ such that ${x}_{n},{y}_{n}\in S$ and
$\lim_{{n}\to{\infty}}{x}_{n}=\sup S$ and $\lim_{{n}\to{\infty}}{y}_{n}=\inf S$.Proof:
For $\lim_{{n}\to{\infty}} {x}_{n}=\sup S$

Since $S\subseteq \Bbb{R}$ and $S$ is a nonempty set and $S$ is bounded, there exists the $\sup S$ due to least-upper-bound property for real numbers. Then, to construct a montone increasing sequences, we need ${x}_{1},{x}_{2},\cdots {x}_{n}$, and we need to formulate an ${\varepsilon}_{n}$. The formula is ${\varepsilon}_{n}=\frac{1}{n}{x}_{n-1}$ for $n\ge 2$ where ${\varepsilon}_{1}=1$.
Since $\sup S$ exists, we can apply proposition 1.2.8 (if $S\subset\Bbb{R}$ is a nonempty set, bounded from above, then for every $\varepsilon>0$ there exists a $x\in S$ such that $\sup S-\varepsilon<x\le \sup S$), let ${\varepsilon}_{1}=1$, there exists a ${x}_{1}\in S$ such that $\sup S -1<{x}_{1}\le\sup S$.
By applying proposition 1.2.8, let ${\varepsilon}_{2}=\sup S-{x}_{1}$ there exists a ${x}_{2}\in S$ such that $\sup S - {\varepsilon}_{2}<{x}_{2}\le\sup S$. So if we substitute ${\varepsilon}_{2}$ into $\sup S - {\varepsilon}_{2}<{x}_{2}\le\sup S$ , then ${x}_{1}<{x}_{2}\le \sup S$.
By apply the prosition 1.2.8, let ${\varepsilon}_{3}=\sup S-{x}_{2}$, there exists a ${x}_{3}\in S$ such that $\sup S - {\varepsilon}_{3}<{x}_{3}\le\sup S$. So if we substitute ${\varepsilon}_{3}$ into $\sup S - {\varepsilon}_{3}<{x}_{3}\le\sup S$ , then ${x}_{2}<{x}_{3}\le \sup S$.

By reapplying the proposition 1.2.8 for n times, let ${\varepsilon}_{n}=\sup S -{x}_{n-1}$, there exists ${x}_{n}\in S$ such that $\sup S - {\varepsilon}_{n}<{x}_{n}\le\sup S$. So if we substitute ${\varepsilon}_{n}$ into $\sup S - {\varepsilon}_{n}<{x}_{n}\le\sup S$ , then ${x}_{n-1}<{x}_{n}\le \sup S$.

Therefore ${{x}_{n}}$ is a monotone increasing and is bounded. By the Monotone Convergence Theorem, the sequence ${{x}_{n}}$ converges to $\sup S$.

For $\lim_{{n}\to{\infty}}{y}_{n}=\inf S$

Let $S\subseteq\Bbb{R}$ and $S\ne\emptyset$. Since $S$ is bounded below by the greatest-lower-bound property of real numbers, there exists a $\inf (S)$. Then, to construct a monotone decreasing sequence, we need ${y}_{1},{y}_{2},\cdots {y}_{n}$, and we need to formulate ${\varepsilon}_{n}$. The formula is ${\varepsilon}_{n}={y}_{n-1}-\inf S$ for $n\ge2$ where ${\varepsilon}_{1}=1$.
Since $\inf (S)$ exists, we apply the corollary 1.2.8; let${\varepsilon}_{1}=1$, there exists a ${y}_{1}\in S$ such that $\inf S+{\varepsilon}_{1}>{y}_{1}\ge \inf S$.
By applying corollary 1.2.8, let ${\varepsilon}_{2}={y}_{1}-\inf S$ there exists a ${y}_{2}\in S$ such that $\inf S + {\varepsilon}_{2}>{y}_{2}\ge\inf S$. So if we substitute ${\varepsilon}_{2}$ into $\inf S + {\varepsilon}_{2}>{y}_{2}\ge\inf S$ , then ${y}_{1}>{y}_{2}\ge \inf S$.
By apply the corollary 1.2.8, let ${\varepsilon}_{3}={y}_{2}-\inf S$, there exists a ${y}_{3}\in S$ such that $\inf S + {\varepsilon}_{3}>{y}_{3}\ge\inf S$. So if we substitute ${\varepsilon}_{3}$ into $\inf S + {\varepsilon}_{3}>{y}_{3}\ge\inf S$ , then ${y}_{2}>{y}_{3}\ge \inf S$.
By reapplying the corollary 1.2.8 for n times, let ${\varepsilon}_{n}= {y}_{n-1}-\inf S$, there exists a ${y}_{n}\in S$ such that $\inf S + {\varepsilon}_{n}>{y}_{n}\ge\inf S$. So if we substitute ${\varepsilon}_{n}$ into $\inf S + {\varepsilon}_{n}>{y}_{n}\ge\inf S$ , then ${y}_{n-1}>{y}_{n}\ge \inf S$.

Therefore ${{y}_{n}}$ is monotone decreasing and bounded. By the Monotone Convergence Sequence, the sequence ${{y}_{n}}$ converges to $\inf S$. QED

 

[Jhoneine]

There are no logical leaps or errors in this proof. It is clear and well-structured, with each step of the proof properly justified in accordance with the given propositions.

 

[math771]

Your proofs for both Corollary 1.2.8 and Proposition 2.1.13 seem to be logically sound. However, there are a few things that could be clarified or expanded upon for better understanding.

For Corollary 1.2.8, it may be helpful to explain why you chose to use the set $-S$ instead of just using $S$ directly. Additionally, you could expand on the reasoning behind using Proposition 1.2.6 v and how it relates to the overall proof.

For Proposition 2.1.13, it would be beneficial to provide some context or explanation for the formula used to construct the monotone sequences. Also, it may be helpful to clarify how the sequences are related to the supremum and infimum of $S$ and why they converge to those values.

Overall, your proofs seem to be well-structured and logically sound. However, providing some additional explanations and clarifications could make them easier to follow for those who may not be as familiar with the concepts being discussed. Great job!