I am having a hard time following this (incomplete) proof that my analysis text offers as an example for finding the supremum of a subset of $\mathbb{R}$ using the epsilon definition of a supremum.
Proposition:
The supremum of a set $X := \{\frac{x}{1+x}: x \in \mathbb{R}, x>0\}$ is 1.
Proof:
for any $x$ in $\mathbb{R}$, $\frac{x}{1+x} < 1$, proving that 1 is indeed an upper bound.
for any $\epsilon>0$ there exists an $x'$ in $X$ such that $x'>1-\epsilon$ (where 1 is the lowest upper bound we are trying to prove), namely:
$x' = \frac{x}{1+x}$
$x' > 1-\epsilon$ if $x > \frac{1-\epsilon}{\epsilon}\qquad (\ast)$
The inequality is satisfied if, for example
$x = \frac{1-\epsilon}{\epsilon}+1\qquad (\dagger)$
Part 1 seems obvious since the elements of $X$ converge on 1. However, in part 2 the remaining proof in the marked sections is left as an exercise to the reader.
First, I wanted to know how we arrived at $(\ast)$ in the first place, so I tried setting
$\frac{x}{x+1}=1-\epsilon$
Solving for $x$ we get
$x = \frac{1-\epsilon}{\epsilon}$
But from this it does not immediately follow that $x$ must satisfy
$x > \frac{1-\epsilon}{\epsilon}$ so that $x' > 1-\epsilon$, or does it?
Regarding $(\dagger)$, the text refers to the following axioms from which the proposition should then follow:
- $ax+b=c \implies x=(c-b)/a$
- $x>0$ and $y>0 \implies x+y>0$
but I fail to see the connection here.
Instead of solving $$\frac{x}{1+x} = 1-\epsilon$$ and then arguing that $\frac{x}{1+x} > 1-\epsilon$ for $x > \frac{1-\epsilon}{\epsilon}$, rearrange $\frac{x}{1+x} > 1-\epsilon$ directly.
This is where the lemmas come in.