Suppose we delete the axiom of infinity in ZFC, which states that there exists a set of all natural numbers, and instead put in the axiom that there exists a complete ordered field. Would we still be able to derive the axiom of infinity? What I am basically asking is whether the axiom of infinity is equivalent to the existence of a complete ordered field, modulo (ZFC – Infinity).
Is the axiom of infinity in ZFC equivalent to the existence of a complete ordered field
set-theory
Related Solutions
The corollary from the incompleteness theorems is that you cannot prove the consistency of $\sf ZFC$ from $\sf ZFC$ itself. You have to have a stronger theory.
For example, in $\sf ZFC+\text{There exists an inaccessible cardinal}$, you can in fact prove the consistency of $\sf ZFC$ because this is a stronger theory.
Similarly this is the case of $\sf ZF_{fin}$ ($\sf ZF$ without infinity). The theory itself cannot prove its own consistency. However $\sf ZFC$ is a strictly stronger theory, and it proves the consistency of $\sf ZF_{fin}$. It does so by exhibiting a set which is a model of the theory, $V_\omega$ - the set of the hereditarily finite sets.
Large cardinal axioms are often called "strong infinity axioms" because they mimic the axiom of infinity, in the sense that they make a stronger theory by describing that a certain set of ordinals exists.
BrianO's answer is spot-on, but it seems to me you may not be too familiar with models and consistency proofs, so I'll try to provide a more complete explanation. If anything it may better steer you towards what you need to study, as admittedly I'm about to gloss over a lot of material.
Why do we need the axiom of infinity? Because we know (and can prove) that the other axioms of ZFC cannot prove that any infinite set exists. The way this is done is roughly by the following steps:
- Remember a set of axioms $\Sigma$ is inconsistent if for any sentence $A$ the axioms lead to a proof of $A \land \neg A$. This can be written as $\Sigma \vdash A \land \neg A \to \neg Con(\Sigma)$
- If $Inf$ is the statement "an infinite set exists", then $\neg Inf$ is the statement "no infinite sets exist".
- The axiom of infinity is essentially the assumption that $Inf$ is true and hence $\neg Inf$ is false.
- If we don't need the axiom of infinity, then with the other axioms $ZFC^* = ZFC - Inf$, we should be able to prove $Inf$ as a theorem, in other words we'll posit that $ZFC^* \vdash Inf$
- We assume that $ZFC$, and hence the subset $ZFC^*$, are consistent.
- We then add $\neg Inf$ as an axiom to $ZFC^*$, which we'll call $ZFC^+$
- By showing that $(ZFC - Inf) + \neg Inf$ has a model (a set in which all the axioms are true when quantifiers range only over the elements of the set), we can prove the relative consistency $Con(ZFC) \to Con(ZFC^+)$. In other words we're basically just proving $ZFC^+$ is consistent, but we need to be explicit that this proof assumes $ZFC$ is consistent.
- The model we want is $HF$, the set of all hereditarily finite sets. I'll leave it you to verify all the axioms of $ZFC^+$ hold in this set. But the important point is $HF \models ZFC^+$, and our relative consistency is proven. (This follows from Godel's completeness theorem)
- We are assuming that $ZFC^* \vdash Inf$, but because $ZFC^+$ is an extension of $ZFC^*$ it must also be the case that $ZFC^+ \vdash Inf$. But then we have $ZFC^+ \vdash Inf \land \neg Inf$ and is thus inconsistent, a contradiction.
Thus we must conclude that our hypothesis $ZFC^* \vdash Inf$ is false and there is no proof of $Inf$ from the other axioms of ZFC. $Inf$ must be taken as an axiom to be able to prove that any infinite set exists.
Best Answer
Of course, if we have a complete ordered field $F$ we can define a copy of $\Bbb N$ inside it: $0$ and $1$ are the ones from the field, and we can define $\Bbb Z$ as the smallest subring that contains $1$ and $\Bbb N$ are just the elements $\ge 0$ in that $\Bbb Z$. I don't see any issue with that and it's well-known that from $\Bbb N$ we can construct a complete ordered field. So assuming one or the other is equivalent. But $\Bbb N$ is more in line with a minimality ideal: assume the simplest object as an axiom (Occam's razor like). And we have the tradition of starting with Peano's axioms in foundational theory as well. So it's never done in "your" order.