Is the theory for $\mathbb{R}$ categorical or not?

Question

The usual axioms of $\mathbb{R}$ consist of the ordered field axioms plus the least upperbound property. Because the least upperbound property is a statement that quantifies over sets of real numbers, people often say that it is a second-order logic statement, which makes the theory of real numbers categorical.

However, from the point of view of ZFC set theory, sets are simply objects of set theory, so quantification over sets (like in the least upperbound property) is a first-order statement. Then the theory of real numbers cannot be categorical.

The same kind of confusion can be posed when we're talking about natural numbers. The second-order logic Peano axioms include the induction axiom, which quantifies over properties of natural numbers. But we can recast this as a statement that quantifies over sets in the framework of set theory in first-order logic, and then it isn't clear whether or not (a certain version of) the Peano axioms leads to a categorical theory.

I would want someone with more understanding than me to clarify this situation. What is the right, straightforward way of thinking about this that steers clear of confusion?

Answer 1

In a given model of $\sf ZF$, the real numbers are unique up to isomorphism. But as soon as you reduced the problem to the first order theory of $\sf ZF$, you essentially have to contend with the fact that different models of $\sf ZF$ can exist, and they may have wildly different real numbers.

In some models the reals will be countable, maybe because the model itself is countable. In another model there will be many more reals, perhaps because we took an ultrapower and blew the (external size) of the reals of the model to be much larger than the continuum is supposed to be. And maybe some model contains all the reals.

All of these options are possible and plausible. But each model thinks that its real numbers are the unique Dedekind complete field. Even if we know the truth. The same goes for the second order Peano axioms.

Answer 2

Let $T$ be the first-order theory of $\mathbb{R}$. $T$ is not categorical. To see this, consider the theory $T$, but add in a new constant symbol $c$ and, for each $n \in \mathbb{N}$, the axiom $n < c$. This theory is clearly finitely satisfiable - given $T$ plus the axioms $n_1 < c, n_2 < c, \ldots, n_m < c$, we can take as a model $\mathbb{R}$ with $c = \max(n_1, \ldots, n_m) + 1$.

Ultimately, the notion of being “categorical” depends on which logic we are using and how we model this logic. If we work in first-order logic with ordinary semantics, many theories, such as those of $\mathbb{N}$ and $\mathbb{R}$, are not categorical. The same holds if we work in second-order logic with Henkin semantics, as this amounts to treating second-order logic as (many-sorted) first-order logic, so the compactness theorem still applies (so our above argument still shows that the theory of $\mathbb{R}$ isn’t categorical). In fact, using the compactness theorem, we can prove the upward Lowenheim-Skolem theorem, which proves that any theory with an infinite model isn’t categorical.

However, if we work in second-order logic with full semantics, we find that we can have plenty of consistent theories which don’t have models. Thus, unlike in first-order logic with ordinary semantics and unlike second-order logic with Henkin semantics, we can very often find theories with ana infinite model which is unique up to (potentially unique) isomorphism.