# Show that 1 is the supremum of $S = \{ x \epsilon R: x^2 < x \}$

I'm still new to proof writing so I was wondering if I could have a little help organizing my thoughts on this, I attempted this proof in a slightly oblong way that is probably not the standard, but I wanted to see if it still made sense. There is also a problem I was having at the end that I was hoping I could get help with.

So given that $S = \{x \epsilon R : x^2 < x\}$ we first need to find an upper bound. And then we have to show that the upper bound is the least upper bound, and moreover that it is equal to 1.

So the first thing I did, was observe that $(1)^2 = 1$ and that $(-1)^2 = 1$ by the positivity axioms, so they are not in the set. And as $1 > -1$, the set S must be contained in $(-1, 1)$. So therefore, 1 is an upper bound.

In order to show that it is the supremum, we need to show that $1 \leq b$ for all other upper bounds b. In my text it was pointed out that this is equivalent to showing that any other number less than sup S can't be an upper bound.

This is where I'm a little stuck. I know from experience that if a number is less than 1 but greater than 0, that it's square is less than itself. So I somehow need to say that if $x^2 < x$, there is another number $c$ that is larger than the value of $x$ that I chose that is in the set. But I get stuck here.

Your suggested argument, which is exactly what you need to proceed, can be framed as a contradiction: Suppose there is an upper bound $b < 1$ for $S$. Then, if you can show that $b$ is not an upper bound, i.e., that there is a number in $S$ larger than $b$, you'll have produced a contradiction; in other words, our assumption that there is an upper bound $b < 1$ cannot hold, or put another way, any upper bound is at least $1$. Then, since you've already shown that $1$ is an upper bound, it must be the least upper bound (supremum) of $S$.

$S=\{x\in \mathbb{R}\mid x^2<x\} \implies x<1$, $\forall x \in S$.

$\sup S=1 \impliedby \forall \epsilon>0$, $\exists s \in S \mid 1-\epsilon\leq s<1 \impliedby \forall \epsilon>0$, $1-\epsilon< 1-\dfrac{\epsilon}{2}<1$

$\left(1-\dfrac{\epsilon}{2}\right)^2<\left(1-\dfrac{\epsilon}{2}\right) \impliedby$ $\left(1-\dfrac{\epsilon}{2}\right)<1$

$\therefore\left(1-\dfrac{\epsilon}{2}\right) \in S$ $\forall \epsilon>0$

$\therefore \sup S=1$

@Rafael I'm assuming that you stick with the standard notation of $>$, which is a strict inequality. Therefore, your statement that $1^{2}> 1$ is incorrect.

You have to

1. first, establish 1 as an upper bound,
2. secondly, show that any upper bound $u$ of $S$ is greater or equal to 1.

One way to establish 1 as upper bound is to note that $x^{2}<x$ implies that $x(x-1)<0$. By what we know about quadratics, $0<x<1$. Therefore, $1$ must be an upper bound.

To show the second one, we assume that $u$ is any upper bound of $S$, we need to establish that $u\geq 1$. Well, suppose not, i.e $u<1$. Consider $x_{0}=\frac{1+u}{2}$, the "midpoint" of $u$ and $1$, we have that $u<x_{0}<1$ and furthermore, $x_{0}^{2}<x_{0}$. Thus, $x_{0}\in S$. So, $u$ cannot be an upper bound of $S$, a contradiction.

On hindsight, we could have just establish that $S=\{x\in \mathbb{R} \mid 0<x<1\}$ by using our knowledge about quadratics. The $\sup$ is then obvious.

• Yeah sorry I messed up, I think I meant to figure out to put a slash through the inequality sign but forgot to. Woops! – user1236 Oct 2 '14 at 5:27
• How did you go from $u < x_0 < 1$ to $x_0^2 < x_0$? – jaja Jan 24 '19 at 3:22
• just because $x_0 \in S$ does not mean it is not an upper bound. the upper bound can be in the set but need not be. – jaja Jan 24 '19 at 3:36