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Well, we know that for any set of axioms, it is necessary that no axiom or a conjunction of axioms contradict some other axiom. How do we know that this holds in the field, order and completeness axioms of the real number system?

Indrayudh Roy
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    There is an explicit construction of $\mathbb R$ where all these axioms can be proved. –  Oct 23 '13 at 05:46
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    As I think, we don't prove this fact within this theory (that is, second-order theory of the real numbers.) Because we interpret the Peano arithmetic within given theory, so by Gödel's second incompleteness theorem we can't prove the consistency of second-order theory of the real numbers within this theory. Of course, we can prove the consistency of this theory using the more stronger theory : set theory provides the model of the second-order theory of the reals. – Hanul Jeon Oct 23 '13 at 06:30
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    I have bearly no background in mathematical logic, but isn't that we can construct $\mathbb R$ from $\mathbb Q$ using only set theory (As in baby Rudin)? –  Oct 23 '13 at 06:42

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A general fact/insight in the foundations of mathematics is that one does not prove the consistency of a theory $T$ "out of thin air". If we want to prove that a theory $T$ is consistent (say, the set of axioms for a complete ordered field), we have to carry out that proof in some other axiom system, $S$.

In the case of the axioms for a complete ordered field, we can construct a model of those axioms in the theory $S$ of ZFC set theory, which is a standard system for carrying out mathematics. We can also prove in ZFC that if a set of axioms has a model, then it is syntactically consistent. That means there is no formula $\phi$ such that both $\phi$ and the negation of $\phi$ are provable in the theory.

In the case of the theory of complete ordered fields, we can prove the consistency in much weaker systems. In particular, we can prove the consistency in a theory known by the acronym $\mathsf{ACA}_0$. The nice thing about that is that there is a proof in a much weaker system that $\mathsf{ACA}_0$ is consistent if and only if Peano arithmetic is consistent. Peano arithmetic is a standard set of axioms for the natural numbers with induction.

So, in the end, the consistency of the axioms of complete ordered fields is no more "unreliable" than the consistency of Peano arithmetic.

Carl Mummert
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  • Please forgive my ignorance, but https://en.wikipedia.org/wiki/Reverse_mathematics#Arithmetical_comprehension_ACA0 says that $\mathsf{ACA}0$ is equivalent over $\mathsf{RCA}_0$ to the _sequential completeness of the real numbers. Is the distinction between completeness and sequential completeness relevant in this context? – Trevor Wilson Oct 24 '13 at 00:11
  • @Trevor Wilson: that is an excellent point. There are some issues with "arbitrary subsets" of reals in second-order arithmetic, so usually completeness in that context is taken to mean sequential completeness. So that is at least a slight difference from Dedekind completeness, although the two forms of completeness are closely related. – Carl Mummert Oct 24 '13 at 00:41
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    Oh, I see, (Dedekind) completeness would be a third-order statement. Is there a sense in which all of the second-order consequences of Dedekind completeness follow from sequential completeness? – Trevor Wilson Oct 24 '13 at 00:45
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    That is an excellent question (it can be phrased formally as a conservation result that some reasonable higher-order system with a Dedekind completeness axiom for $\mathbb{R}$ is conservative over $\mathsf{ACA}_0$). I don't know whether it's true. – Carl Mummert Oct 24 '13 at 00:51
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The reason that we know that is that the real numbers are a model of their axioms. And if a set of axioms has a model then it cannot prove a contradiction.

The reason is that if $T$ is any theory, and $T\vdash\varphi$, then in any model of $T$ we will have that $\varphi$ is true. This property is called soundness.

If $T$ is a first-order theory then the converse is true as well, if $\varphi$ is true in every model of $T$, then $T\vdash\varphi$. However note that the completeness axiom is not a first-order axiom (working in the language of fields), we quantify over all subsets of the real numbers.

But luckily, second-order logic is still sound, so if the axioms of the real numbers were inconsistent, it would prove a contradiction of the form $\varphi\land\lnot\varphi$. However in that case the real numbers themselves would satisfy both $\varphi$ and its negation. That is impossible, because of the way we define truth value for formulas in a structure guarantees that a structure cannot satisfy a statement and its negation.

Asaf Karagila
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  • The basic point - $\mathbb{R}$ is a model, so the axioms are consistent - is absolutely correct. But I have two small quibbles. First, there is not really such a thing as a "second order axiom" - if the axioms are consistent, then they are also consistent in two-sorted first-order semantics. And the axioms for real numbers are often written in first-order ZFC anyway. Second, if we really are thinking about full second order semantics, a theory could be semantically inconsistent (having no model) without also being syntactically inconsistent (proving a contradiction). – Carl Mummert Oct 23 '13 at 10:26
  • @Carl: When I say "second-order axiom", I mean that there is a quantification over subsets of the universe, rather than elements of the universe; in the language of ordered fields this is a second-order axiom. Secondly, I didn't claim that every consistent set of second-order axioms has a model. I even said that the relevant direction of the completeness theorem may fail (in some cases). However we still have soundness, so existence of a model implies consistency. – Asaf Karagila Oct 23 '13 at 10:29
  • #Asaf: the phrase that stands out is "if the axioms were inconsistent, it would prove a contradiction". That is true for syntactic consistency, but syntactic inconsistency is really a first-order type of inconsistency. If we are thinking of full second order semantics (i.e. the italicized all) then semantic inconsistency is more relevant. In any case the phrase is ambiguous, and maybe confusing for someone reading the answer. – Carl Mummert Oct 23 '13 at 10:36
  • Carl, I don't completely follow your logic. You took the sentence out of context. "the axioms" refer to the axioms of the real numbers, and we have a model for the real numbers. So semantic inconsistency is off the table, and I add that this implies that there cannot be syntactical inconsistency either. – Asaf Karagila Oct 23 '13 at 10:39
  • I'm sorry, Carl and Asaf, but I haven't studied up to that level of advanced mathematics. I just know a little bit of propositional logic and therefore, I couldn't get your point. I just know that for any set of axioms, it is necessary that the axioms do not contradict themselves. I was reading Apostol and that question props up in my mind. It would be helpful if you guide me as how to understand the concepts better. – Indrayudh Roy Oct 26 '13 at 18:03