the question is:

The relation $<$ is not definable in structure $\langle\mathbb R,+,0\rangle$ with a first order formula.

I asked my professor for help and he told me to consider $\langle\mathbb R,+,0\rangle$ as a vector space and work on its automorphism.

But even now I can't solve this problem...

Can someone please give me some help to understand the clue?

And also why I can't instead prove: the relation $<$ is not definable in the structure $\langle\mathbb N,+,0\rangle$ as a first order formula?

  • $\begingroup$ The case of $\Bbb N$ is a separate question, and I'm pretty sure that it was asked several times before. $\endgroup$ – Asaf Karagila Nov 20 '17 at 20:51
  • $\begingroup$ i know the proof for N , but i thought i could prove it for N as sub structure of <R,+,0> . but now i realized it's not possible because of quantifiers $\endgroup$ – user297564 Nov 20 '17 at 20:55
  • $\begingroup$ Could you prove it for $\{0\}$ as a substructure of $\langle\mathbb R,+,0\rangle$? $\endgroup$ – Asaf Karagila Nov 20 '17 at 20:56
  • $\begingroup$ no.. i don't get why should i do that? $\endgroup$ – user297564 Nov 20 '17 at 20:58
  • $\begingroup$ The point is that just because something holds for a structure does not mean it holds for a substructure. $\endgroup$ – Asaf Karagila Nov 20 '17 at 20:59

Note that $x\mapsto -x$ is both an automorphism and an order-reversing function.

But I guess your professor instead thought about taking a Hamel basis for $\Bbb R$ over $\Bbb Q$, then any permutation of this basis induces an automorphism of the structure, since it is an automorphism of the abelian group. So by switching just two basis elements, we necessarily don't preserve the usual ordering on $\Bbb R$.

(Or maybe he meant the first suggestion, by wanting you to note that $x\mapsto qx$ is an automorphism when $q\in\Bbb Q\setminus\{0\}$, and then consider $q=-1$, I can't quite know since I'm not your professor...)


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