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cbarker1
Gold Member
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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
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
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