1. ## Supremum proof

Define S≡{x in ℝ | x2 < x}. Prove that sup S=1.

2. Originally Posted by freddyg
Define S≡{x in ℝ | x2 < x}. Prove that sup S=1.

note that the set of x's that satisfy this condition is -1 < x < 1

now what can you say about the sup?

3. Originally Posted by freddyg
Define S≡{x in ℝ | x2 < x}. Prove that sup S=1.

Proving a Sup is (usually) accomplished in two steps.
(1) Show that $\displaystyle x \in S$ implies $\displaystyle x \le 1$, and
(2) Show that if $\displaystyle \alpha$ is an upper bound, then $\displaystyle 1 \le \alpha$.

For the first one, pick any $\displaystyle x \in S$, then $\displaystyle x^2 < x$. Since we know $\displaystyle 0 \notin S$ because $\displaystyle 0^2 = 0$, we can safely divide by $\displaystyle x$ to get the inequality $\displaystyle x < 1$.

So we know that 1 is an upper bound of S. The next step is to prove that there are no upper bounds of the set that are less than 1.
First it should be clear that everything in S is positive. Negative numbers are never larger than their square, which are always positive, and we showed 0 is also not a member.

Suppose for contradiction that 1 were not the Sup. Then there is some $\displaystyle \alpha$ which is an upper bound of S, but also so $\displaystyle \alpha < 1$. First note that $\displaystyle \alpha$ is a positive number because everything in S is positive, and $\displaystyle \alpha$ is an upper bound of the set. So we can multiply the inequality by $\displaystyle \alpha$ without changing the sign. So $\displaystyle \alpha < 1$ implies $\displaystyle \alpha^2 < \alpha$. So it follows that $\displaystyle \alpha \in S$.
Now consider the number $\displaystyle \alpha + \frac{1-\alpha}{2}$, which is the number lying halfway between $\displaystyle \alpha$ and 1. Now you can easily prove that its square is smaller than itself -- that is, $\displaystyle (\alpha + \frac{1-\alpha}{2})^2 < \alpha + \frac{1-\alpha}{2}$, which means that creature is also in S.
But here's the contradiction. We said $\displaystyle \alpha$ was an upper bound, which means it's greater-than-or-equal-to everything in S. But we found an element, which lies between $\displaystyle \alpha$ and 1, that lies in S but is bigger than $\displaystyle \alpha$. Therefore, no such $\displaystyle \alpha$ exists, so 1 must be the least upper bound, i.e., the Sup of S.

4. This is somewhat shorter.
Clearly 1 is an upper bound for S because $\displaystyle x > 1\quad \Rightarrow \quad x^2 > 1$.
If $\displaystyle 1 > \varepsilon > 0$ then $\displaystyle 1 > 1 - \varepsilon > 0$.
Consider $\displaystyle \left( {1 - \frac{\varepsilon }{2}} \right)^2 = 1 - \varepsilon + \frac{{\varepsilon ^2 }}{4}$ observe that $\displaystyle 1 - \varepsilon < 1 - \varepsilon + \frac{{\varepsilon ^2 }}{4} < 1$.

That shows that no number less that 1 is an upper bound for S.
Thus 1 is sup(S).