1
$\begingroup$

In answering a previous post of mine, Henning Makholm showed that the theory, $K$, of the real number system* has a consistent extension, $K'$, with a function symbol, $^{-1}$, satisfying $$ \begin{alignat}{3} &\vdash_{K'} x &&\neq 0 \implies x\cdot x^{-1} = 1 \tag{1}\label{a} \\ &\vdash_{K'} 0^{-1} &&= 1 \tag{2}\label{b} \end{alignat} $$

My question now is this: Is this extension conservative? My feeling is that it is, but I don't know how to prove it.

If $K'$ was not stipulated to satisfy \eqref{b}, but only \eqref{a}, then $K'$'s conservativity would follow from The Conservativity Theorem (for a proof, see [2] Proposition 2.28, pp. 102-103 (link)). It is the added stipulation \eqref{b} that throws me for the loop.

Footnotes

* $K$ is a second-order theory with equality with constants $0,1$, with binary function symbols $+,\cdot$, and with unary predicate symbol $\mathbb{R}_+$, that has for axioms: (1) the field axioms, (2) the order axioms, and (3) the completeness axiom (it is this axiom that requires second-order rather than first-order logic), as described on pp. 13 and 14 of [1] (link).

References

[1] Dipak Chatterjee (2012). Real Analysis (2nd ed.) PHI Learning.

[2] Elliott Mendelson (2015). Introduction to Mathematical Logic (6th ed.) CRC Press.

  • 1
    Careful with the terminology. If your $K$ is "_the_ theory of the reals", then that phrasing usually means the set of all sentences that are true in $(\mathbb R,0,1,{+},{\cdot})$. This theory is trivially _complete_, and a consistent extension of a complete theory is trivially conservative. – 2017-02-03
  • 0
    @HenningMakholm: Thanks for your reply. I'm afraid I took Logic a long time ago, and am very rusty with it. To me the fact you mentioned is not trivial. Could you please explain why it holds, or refer me to a source where I can read about it? – 2017-02-03
  • 1
    x @Evan: That a consistent extension of a complete theory is conservative? If the extension proves some $\phi$ in the original language that the original theory didn't already prove, then the original theory proved $\neg \phi$ (because it was complete), and therefore the extended theory proves both $\phi$ and $\neg\phi$ and so is not consistent, contradicting the assumption that it is. – 2017-02-03
  • 0
    @HenningMakholm: Thanks. This is much clearer. – 2017-02-03
  • 0
    @HenningMakholm: Just to make sure I understand why the axiomatic system that defines $\mathbb{R}$ is complete: it can be shown that if a theory has a model, then the theory is complete, correct? – 2017-02-03
  • 2
    x @Evan, no, no. If a theory has a model then it is _consistent_. "The theory of $\mathbb R$" is _complete_ because, by definition, it has _everything that is actually true about $(\mathbb R,0,1,{+},\cdot)$_ as axioms. My original point here was that I don't think you actually wanted to say "the theory of $\mathbb R$" in the question. – 2017-02-03
  • 0
    @HenningMakholm: I was thinking about the theory of $\mathbb{R}$ as being a first order theory with constants $0$ and $1$ and with function symbols $+$ and $\cdot$ that has as axioms (1) the field axioms, (2) the order axioms, and (3) the completeness axiom, as described on pp. 13 and 14 [here](https://books.google.co.il/books?id=FK97dlYL4GQC&lpg=PP25&dq=real%20number%20axiom&pg=PP25#v=onepage&q&f=false). – 2017-02-03
  • 2
    The textbook axioms for real numbers are in fact _also_ a complete theory, because they imply that $\mathbb R$ is a [real closed field](https://en.wikipedia.org/wiki/Real_closed_field) -- but it is not trivial to prove this completeness. – 2017-02-03

1 Answers 1

1

In order not to make the question trivial, let's assume that $K$ is merely some (not necessarily complete) conventional set of axioms for $(\mathbb R,0,1,{+},{\cdot})$, such as you might find in a first-year university text.

In that case, $K$ hopefully proves $$ (\forall x)(\exists_1 y)((x=0\land y=1)\lor(x\ne 0 \land x\cdot y =1))$$ and then you can justify your $^{-1}$ as a conservative extension by the general principle of defining new function letters. (In the 4th edition of Mendelson this is section 2.9, proposition 2.28).

  • 0
    I've realized that the axiomatic theory of the real line that I referred to in my question is actually based on second-order logic, due to the completeness axiom. Does this change your answer to this post, or to [this related post](http://math.stackexchange.com/a/2124407/37058)? Note that the "extension by definition" principle that you referred to in your answer is stated in Mendelson's textbook for theories in first-order languages. – 2017-02-04
  • 1
    @EvanAad: The principle actually holds for higher-order logics too; Mendelson just doesn't treat such logics in much detail. In second-order logic you can prove $$ (\exists F)((F(0)=1) \land (\forall x)(x\ne 0\to x\cdot F(x)=1))$$ and therefore it is conservative to introduce a name for a thing with this property. – 2017-02-04
  • 1
    @Evan Aad This shows that, in some sense, the situation is *easier* in second-order logic with full semantics. We don't need to extend the model, just add a constant function symbol for a function that was already in the model. Moreoever, the usual theory of the reals is already not only complete but also categorical in second-order logic with full semantics. So there would be no harm in going ahead and adding a constant function symbol for *every* function from $\mathbb{R}$ to $\mathbb{R}$. – 2017-02-04