6
$\begingroup$

I'm having trouble with the following exercise from the book Mathematical Logic by H.D. Ebbinghaus, J. Flum, and W. Thomas.

Show:

(a) The relation $<$ ("less-than") is elementarily definable in $(\mathbb{R},+,\cdot ,0)$, i.e, there is a formula $\varphi \in L_2^{(+,\cdot ,0)}$ such that for all $a,b\in \mathbb{R}$, $(\mathbb{R},+,\cdot ,0)\vDash \varphi [a,b]$ iff $a.

(b) The relation $<$ is not elementarily definable in $(\mathbb{R},+,0)$. (Hint: Work with a suitable automorphism of $(\mathbb{R},+,0)$, i.e, a suitable isomorphism of $(\mathbb{R},+,0)$ onto itself.)

It's the first time I see the term "elementarily definable". Still I was able to solve (a):

$\varphi \text{:=}\exists _x\left(x\neq 0\land v_0+x^2=v_1\right)$

But I have not been able to solve (b). Thanks.

  • 0
    Sorry got it. It could be negative2014-09-23

1 Answers 1

8

I gave it a little more thought and I figured it out.

Consider the automorphism $\pi (x)=-x$ (this would not be an automorphism if there were multiplication). If $<$ were elementarily definable, and if originally $a, then the isomorphism lemma would imply that $\pi (a)<\pi (b)$, which is a contradiction, since $a implies $-b<-a$.

  • 1
    @ becko: I'm glad you got it. It is perfectly acceptable for you to accept your own answer, as well.2011-07-27