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

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(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 $a<b$.

(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.

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For your solution to (a), you also need to specificy $x\neq 0$, lest $v_{1} = v_{2}$. –  Quinn Culver Jul 26 '11 at 21:10
    
@Quinn: Thanks. Fixed it. –  becko Jul 26 '11 at 23:33
    
Will $\varphi \text{:=}\exists _x\left(x\neq 0\land v_0+x=v_1\right)$ work for (a)? why do we need '.' here? Please explain –  vinothkr Sep 23 at 15:53
    
Sorry got it. It could be negative –  vinothkr Sep 23 at 15:55

1 Answer 1

up vote 8 down vote accepted

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 $a<b$, then the isomorphism lemma would imply that $\pi (a)<\pi (b)$, which is a contradiction, since $a<b$ implies $-b<-a$.

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@ becko: I'm glad you got it. It is perfectly acceptable for you to accept your own answer, as well. –  Carl Mummert Jul 27 '11 at 1:12

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