# Proving the supremum of a set in the general case

The question is as follows:

Prove that for every $\beta \in \mathbb{R}$, $\sup(-\infty, \beta) = \beta$.

The goal of the problem is to prove this without using the $\epsilon-\delta$ method. My professor gave us an idea in class, but for some reason, it isn't really making any sense to me.

Call the set $S.$ We need to show that:

1) $\beta$ is an upper bound of $S.$

2) $\beta \leq B \ \forall$ upper bounds $B$ of $S.$

The first condition is clearly satisfied by the fact that $x \leq \beta \ \forall x \in S$.

The second condition is what we did in class and the part that I am most confused about.

If $B$ is an upper bound of $S$, then $\beta \leq B.$

Now assume that $B < \beta.$

If $B < \beta, \exists x$ such that $B < x < \beta$. But then $x < \beta$ and $B<x$, contradicing the assumption that $B$ is an upper bound of $S$, thus proving that $\sup(-\infty, \beta) = \beta$.

Q.E.D.

This proof does not make much sense to me, so I am asking for a clarification. If this prove is indeed wrong, I would appreciate if it someone could provide me with a clear and concise way of proving this.

Thank you so much!

• What passage(s) do you want to be clarified, exactly? – Saucy O'Path Sep 1 '18 at 18:32
• @SaucyO'Path Well, firstly I would just like to know if this proof is correct! – MathIsLife12 Sep 1 '18 at 18:36

## 2 Answers

The proof is correct. You are using the fact that for any two distinct elements of $\mathbb{R},$ say $x, y \in \mathbb{R}$ with $x < y,$ there exists $z \in \mathbb{R}$ such that $x < z < y.$ This tells you that since $B < \beta$ you have this element $x$ such that $B < x < \beta.$ Hence, $x \in (-\infty, \beta)$ as $x < \beta.$ Thus, $B$ cannot be an upper bound as there is some element within the set, $x,$ such that $B < x.$ Hope this helps.

Let $S=\{x\in\mathbb{R}:x<\beta\}$. You've already shown (1): $\beta$ is an upper bound of $S$.
You want to show (2): $\forall B,B\ \text{is an upper bound of}\ S\implies \beta\le B$.
The proof is by contradiction:

Suppose (2) is false, then there is some upper bound $B$ of $S$ with $\beta\not\le B$, i.e. $B<\beta$. Clearly $B\in S$. We use that fact that between any two reals $c$ and $d$ is a real $e$ s.t. $c<e<d$, so in particular there is a real $e$ s.t. $B<e<\beta$. But then $e<\beta$ so $e\in S$, and $B<e$, i.e. $B$ is not an upper bound of $S$, contradiction.