# What can be written instead of the supremum axioms to get same structure in the set of real numbers

Question goes as follows : "What can be written instead of the supremum axioms, to get the same structure in $\mathbb{R}$." Now for the supremum axioms I have : If $K_a \neq Ø \Rightarrow \exists$ a smallest upper bound $\exists min$ $K_a \iff \exists \alpha \in K_a$ $\forall k \in K_a : \alpha \leq k$

$\alpha = sup(A)$ ergo the least upper bound.

1. $\alpha$ is upper bound, $\forall a \in A : a \leq \alpha$
2. $\alpha$ is the smallest/least upper bound, $\forall \epsilon \gt 0$ $\exists a \in A : a \gt \alpha - \epsilon$

Now my question is what can I write instead of these axioms. I was thinking some kind of isomorphism to preserve the structure, but I am not sure.

• Jun 9, 2016 at 21:03

If you're hoping for an alternate axiom which is "purely algebraic" - that is, avoids talking about arbitrary sets of reals, and is closer to (say) the field axioms - then there is a sense in which no such axiom exists: specifically, any first-order axioms will fail to capture the real numbers, in the following sense. If $T$ is a set of first-order sentences which are true about the structure $(\mathbb{R}; +, \times, <)$, then

• There will be models of $T$ which are "too big", e.g. have cardinality much greater than that of $\mathbb{R}$. In particular, there will be models which are non-Archimedean (an ordered field is Archimedean if for any $x>0$, there is some natural number $n$ such that $1+1+...+1$ ($n$ times) is $>x$).

• There will be models of $T$ which are "too small," e.g. are countable. Some of these will be non-Archimedean, but we'll also have proper substructures of the reals satisfying $T$.

This is due to the compactness and Lowenheim-Skolem theorems of first-order logic, respectively.

That said, we can certainly avoid sups and infs! $\mathbb{R}$ is a maximal Archimedean ordered field; and any two maximal Archimedean ordered fields are isomorphic. Note, however, that this is non-first-order in two ways:

• We have to say that $\mathbb{R}$ is Archimedean.

• We have to say that any proper ordered field extension of $\mathbb{R}$ is not Archimedean.

There is a precise sense in which this second sentence is "even less first-order" than the first; but that's getting a bit off-track. If you're interested in these sorts of things, you might look at abstract model theory.