# Proving that $\sup(E) = 1$, where $E := \{x \in \mathbb{Q}: x < 1\}$

I'm reviewing basic analysis for tutoring. This is my friend's first day taking an analysis course, so trying to keep things at an intuitive level is needed as much as possible (i.e., I don't want to use fancy tricks that I can't explain where they come from).

Suppose $$E := \{x \in \mathbb{Q}: x < 1\}\text{.}$$ I wish to find $\sup(E)$. I have already explained the intuition that $\sup(E) = 1$, but right now, it's a matter of proving it.

To prove that $\sup(E) = 1$, I need to show:

1. $1$ is an upper bound of $E$, i.e., $\forall e \in E$, $e \leq 1$.
2. If $y \in \mathbb{R}$ is any upper bound of $E$, then $1 \leq y$.

Statement 1 is easy to show, by the definition of $E$: since if $e \in E$, $e < 1 \implies e \leq 1$.

I'm not sure how to show statement 2. If we let $y$ be an upper bound of $E$, if $e \in E$, $e \leq y$. But how do I show $1 \leq y$? I have a feeling there's some proof-by-contradiction involved here.

The only definitions and assumptions available so far are:

1. Set theory vocabulary, including DeMorgan's laws;
2. Well-ordering property of $\mathbb{N}$;
3. Induction, strong induction;
4. Images and inverse images of functions;
5. Cardinality, and Cantor's proof that there isn't a surjection $A \to \mathcal{P}(A)$.
• If you suppose $y<1$ then you can find $e\in E$ with $e\in(y,1)$ which contradicts that $y$ is an upper bound. – John Martin May 17 '16 at 20:30
• You haven't included any properties of the rational numbers $\Bbb{Q}$ or or of the real numbers $\Bbb{R}$ in your list of definitions and assumptions (even though $\Bbb{Q}$ appears in your problem). Bear in mind also that the notion of supremum depends on the ordered set you are working in. – Rob Arthan May 17 '16 at 20:46

## 3 Answers

You could use a proof by contradiction, or more directly, you could prove the contrapositive:
if $y<1$, then $y$ is not an upper bound of $E$.

Let $y<1$. The condition for $y$ to be an upper bound is that for all $z\in E$, $z\leq y$. Hence for $y$ not to be an upper bound, there should exist some $z\in E$ such that $z>y$. Can you cook up such a value for $z$?

You have to use the fact that given two distinct real numbers $a<b$ there is a rational number $q$ with $a<q<b$.

Suppose $y$ is an upper bound of $E$, $y<1$; then there exists a rational $q$ with $y<q<1$; this is a contradiction, because $q\in E$.

I assume that with the things you've listed, the person you're trying to help is familiar with the archimedian property of the reals, that is : $$\forall x > 0, \exists n \in \mathbb N \quad \mathrm{s.t.} \quad nx \ge 1.$$ You can reverse that and claim $x \ge 1/n$, or by taking $n$ a bit larger, $x > 1/n$.

You also need to use the following trichotomy : for two real numbers $x,y$, either $x < y, x=y$ or $x>y$. Now you want to show that all upperbounds of $E$ satisfy $y \ge 1$, so it suffices to show they can't satisfy $y < 1$. Suppose this is the case, and let $z = 1-y$ (the distance between the two). Pick $n$ such that $0 < 1/n < z$. Then $y < 1-1/n < 1$ and $w = 1-1/n \in E$, so $y$ is not an upper bound for $E$.

I assume you had an intuition (i.e. that there would be a rational in between two real numbers), but this is the proof. The numbers $y$ and $1$ played no particular role ; this is the proof that any real interval contains a rational number.

Hope that helps,