Is there a proof of axiom supremum for real number?

Question

If A$\subset R$ bounded above then the supremum exists.. I cant find the proof anywhere

Answer 1

Of course you can prove that if a subset $A\subset\mathbb{R}$ is bounded above then it has a supremum, according to the definition of $\mathbb{R}$ you are dealing with: the fact that taken $A\subset \mathbb{R}$, $B\subset \mathbb{R}$, where $A,B\neq\emptyset$ such that for every $a\in A$, $b\in B$, $a\leq b$, there exists $\xi \in \mathbb{R}$ such that $\forall a\in A:\forall b\in B: a\leq\xi\leq b$, can be put as an axiom, making $\mathbb{R}$ a (the one and only up to isomorphism) complete ordered field that we want it to be.

In this setup, let $\emptyset\neq E\subset\mathbb{R}$ bounded above; $E$ and $K=\{\text{upper bounds of}\ E\}$ are both nonempty sets satisfying the statement above. This of course means that the supremum exists; remembering the definition of supremum ($\operatorname{min}$ of something), you can check unicity.

Answer 2

This is an expansion of my comment (to the question).

---

Following Dedekind we say that

A real number is a subset $A$ of the set of rationals (ie $A\subseteq\mathbb{Q} $) with the following properties:

• $A$ as well as $\mathbb {Q} - A$ is non-empty ie $\emptyset \neq A\neq\mathbb {Q} $.
• If a rational number lies in $A$ then all the smaller rationals also lie in $A$.
• If a rational number lies in $A$ then there is a greater rational also lying in $A$ ie $A$ has no greatest member.

Without going into details we focus on the part needed here. A real number $A$ is said to be less than another real number $B$, written as $A<B$, if $A\subset B$. In what follows usual capital Roman letters like $A, B, C$ denote real numbers ie sets of rationals with the specific properties listed above and letters like $\mathcal{A, B, C} $ represent sets of real numbers.

Let's come back to your axiom:

Completeness axiom: Let $\mathcal{A} $ be a non-empty set of real numbers which is bounded above ie there is a real number $K$ such that $A\leq K, \forall A\in \mathcal{A} $. Then there is a real number $M$ such that no member of $\mathcal{A} $ exceeds $M$ and any real number less than $M$ is exceeded by some member of $\mathcal{A}$. This number $M$ is called the supremum of $\mathcal{A} $ and denoted by $M=\sup\, \mathcal{A} $.

Outline of proof: Consider the set $M\subseteq \mathbb {Q} $ defined by $$M=\bigcup_{A\in\mathcal{A}} A$$ ie $M$ is the union of all the members of $\mathcal{A} $ (note that by definition the members of $\mathcal{A} $ are sets of rationals and hence their union makes sense). Show that the set $M$ satisfies all the three properties of a real number mentioned earlier. This part is easy and you will need the fact that $\mathcal{A}$ is bounded above.

Next one needs to show that $A\leq M$ ie $A\subseteq M$ for all members $A\in\mathcal {A} $. This is obvious as $M$ is the union of all such $A$.

And finally let $N$ be any real number less than $M$ ie $N\subset M$. And then we have to show that there is some member $A\in\mathcal{A} $ such that $N<A$ ie $N\subset A$. This is not so difficult to show and one can do it via contradiction.