Verification of proof concerning the supremum of a set.

Question

Given some set say $A$ = {$ x\over(x+7)$|$x \in \mathbb{R^+}$}

Where

$x\over(x+7)$ = 1 - $7\over(x+7)$ $\lt 1$

Hence $1$ is an upper bound of $A$.

Method $1$.

Consider $SupA = \alpha$ where $\alpha \lt 1$

$x \ge 0$ for all $x \in \mathbb{R^+}$

$\therefore$ $x+7\ge7$

$\therefore$ $(1-\alpha)(x+7)\gt7$

$\therefore 1-\alpha\gt$ $7\over(x+7)$

$\therefore -\alpha\gt$$-1 + $$7\over(x+7)$

$\therefore \alpha \lt 1 - $$7\over(x+7)$

Hence $\alpha \lt $$x\over(x+7)$

Therefore $\alpha$ is not an upper bound of $A$ and hence $SupA = 1$

Method $2.$

For $SupA = 1$ to be true the following must hold for any arbitrary $\epsilon \gt 0$

$ 1 - \epsilon \lt A^{'}$ for some $A^{'} \in A$

i.e. $1 - \epsilon \lt $$x\over(x+7)$ for some $x\in \mathbb{R^+}$

Take $x+7 \gt $$7\over \epsilon$

$\therefore \epsilon \gt $$7\over(x+7)$

$\therefore -\epsilon \lt $-$7\over(x+7)$

$\therefore 1-\epsilon \lt 1-$$7\over(x+7)$

Hence $1-\epsilon \lt $$x\over(x+7)$ for some $x \in \mathbb{R^+}$

Therefore $SupA = 1$

Are either of these methods of proving supremum correct? If both, which one is better? If neither, how do i do it?

Note: please excuse poor formatting, haven't learnt how to do it yet

Answer 1

Your first method has a big flaw: from $x+7\ge7$ it doesn't follow that $$ (1-\alpha)(x+7)\ge7 $$ For instance, if $\alpha=1/2$ and $x=1$, $$ (1-\alpha)(x+7)=\frac{1}{2}\cdot8=4<7 $$

Indeed, with this false statement, you conclude that $\alpha<x/(x+7)$ for every $x\in\mathbb{R}^+$, which is certainly false when $0<\alpha<1$.

What you want to prove is that $\alpha<x/(x+7)$ for (at least) one $x\in\mathbb{R}^+$. The inequality is equivalent to $\alpha<1-\frac{7}{x+7}$, that is $$ \frac{7}{x+7}<1-\alpha $$ or $$ x>\frac{7}{1-\alpha}-7=\frac{7\alpha}{1-\alpha} $$ which is certainly satisfied for some (actually infinitely many) $x\in\mathbb{R}^+$.

The second method is essentially the same: for $\varepsilon>0$ you want to see that there exists $x\in\mathbb{R}^+$ such that $$ 1-\varepsilon<\frac{x}{x+7} $$ that is $$ \varepsilon>\frac{7}{x+7} $$ that's equivalent to $$ x>\frac{7}{\varepsilon}-7 $$ Thus, taking $x=7/\varepsilon$ you are done.

There is also a third method, in this case.

Clearly, any $\alpha\le0$ is not an upper bound for the set $A$. Assume $0<\alpha<1$: we show that $$ \frac{1+\alpha}{2}\in A $$ Indeed $$ \frac{1+\alpha}{2}=\frac{x}{x+7} $$ becomes $$ (1+\alpha)x+7(1+\alpha)=2x $$ that is $$ x=\frac{7(1+\alpha)}{1-\alpha}\in\mathbb{R}^+ $$ Since $\alpha<\frac{1+\alpha}{2}$, we conclude $\alpha$ is not an upper bound for $A$.

Answer 2

The second method looks fine.

The first method is incorrect. You wrote $x+7\ge 7$ and then $$(1-\alpha)(x+7)>7$$ That's your mistake.

Remark: If you know how the $\sup$ is characterized by limits of sequences, then you can easily show your result using the fact that $$\lim_{n\to +\infty}\frac{n}{n+7}=1$$

Edit For your curiosity: If $A$ is an upper-bounded non-empty set and $s\in\mathbb{R}$, to show that $s=\sup A$ you can simply show that $s$ is an upper-bound of $A$ and that there exists a sequence $(a_n)_{n\in\mathbb{N}}$ of elements in $A$ (i.e $\forall n\in\mathbb{N},a_n\in A$) such that $\lim\limits_{n\to +\infty}a_n=s$. That's because if such a sequence exists, it means that $$\forall\varepsilon>0,\exists N\in\mathbb{N},\forall n\ge N,s-\varepsilon<a_n\le 1$$ and in particular $s-\varepsilon<a_N$ with $a_N\in A$. In fact, you can also prove that if $s=\sup{A}$ then such a sequence exists.