I found the question Prove that $\sup(S)=1$ if $S=\{x \in \mathbb{R}| x^2 < x\}$, but I don't quite understand the answer.
This is my approach:
$S\equiv\{x\in\mathbb{R}\vert x^2<x\}\equiv\{x\in\mathbb{R}\vert 0<x<1\}\equiv\{x\in\mathbb{R}\vert 0<x^2<1\}$, by the definition of $S$, $S$ is bounded above and $\sup S = 1$. Now assume that $1$ is not the supremum but is $b$ where $b>1$ or $b<1$.
If $b>1$, then we can pick $\epsilon=\frac{b^2+1}{2b}$ such that $b-\epsilon<b$.
Then we can have $$(b-\epsilon)^2=b^2-2b\epsilon +\epsilon^2>b^2-2b\epsilon=1\Rightarrow b-\epsilon>1$$ This gives a contraction that $b$ is the least upper bound.
If $b<1$, let $\epsilon=\frac{1-b^2}{2b+1}$, then $$(b+\epsilon)^2=b^2+2b\epsilon+\epsilon^2<b^2+\epsilon(2b+1)=1\Rightarrow b+\epsilon<1$$
This contracts that $b$ is an upper bound, hence $\sup S =1$
I just follow the step from the text book. Could anyone give me a suggestion to write a better proof or explain more detail on the answer I linked? Thanks in advanced.
Step 1: Show that S is bounded above.
Show that if x >= 1 then x is not in S. Then 1 is greater than all elements of S so S is bounded above and 1 is an upper bound. (Although you don't know if it is the least upper bound yet.)
In your proof you just stated this without proving or showing anything.
Step 2: Show that if b < 1, then b is not an upper bound.
Show that if b < 1 then there is an x with b < x < 1 such that x is in S. In other words if b < 1 then there is an x; b < x < 1 such that $x^2 < x $. As b < x $\in$ S b can't be an upper bound. As b was arbitrary, 1 is the least upper bound.
(You won't need any epsilon proofs at all. In fact they'll just be too hard.)