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Hello friends of mathematics :) I have some problems with the topic "Is something definable in a structure". I can solve some problems for example the following questions:

  • Is the relation definable in the structure $(\Bbb{Q},+,0,1)$
  • Is the function "sin" definable in the structure $(\Bbb{R},<,+,\cdot,0,1)$
  • Is $\Bbb{N}$ definable subset of $(\Bbb{Z},<,+,\cdot,0,1)$

The answers are No, No, Yes resp.

But now i want to solve the following problems:

  • Is $\Bbb{Z}$ definable subset of $(\Bbb{R},<,+,\cdot,0,1)$
  • Is $\Bbb{Q}$ definable subset of $(\Bbb{R},<,+,\cdot,0,1)$
  • There is no qeuntifier-free formula that defines the set $2\Bbb{Z}$ in the structure $(\Bbb{Z},+,<,S,0)$ (with $S$ the sucessorfunction)

For the first problem i thought this: If we can define $\Bbb{Z}$ then we could find a polynomial with zeros in all integers. But such a polynomial doesn't exist. I don't know how to solve it on another way. Can someone help me? Thank you beforehand :)

Mathematic Research
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1 Answers1

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For the non-definability of $\mathbb{Z}$ in the reals, there is good structure theory (keyword: o-minimal) for the definable subsets of $\mathbb{R}$. One can also argue using decidability/undecidability.

It is an old result of Tarski that there is a decision procedure for the elementary theory of real-closed fields. If $Z$ were definable, that decision procedure could be used to produce a decision procedure that would determine, for every sentence $\phi$, whether or not $\phi$ is true in the natural numbers.

For definability of $\mathbb{Q}$, again structural information about definable subsets of $\mathbb{R}$ will do it. Another approach is to use the old, and not easy, result of Julia Robinson that the natural numbers are definable in $\mathbb{Q}$. (Please See Definability and Decision Problems in Arithmetic Journal of Symbolic Logic, Vol. 14, No. 2 (June 1949).)

Hence if $\mathbb{Q}$ were definable in $\mathbb{R}$, then $\mathbb{Z}$ would be. This, as shown above, is not true.

André Nicolas
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  • In the lecture we do not use the keyword o-minimal. Oh yes but such a procedure doesn't exist because of uncompleteness theorem thus $\Bbb{Z}$ is not definable, true?! Wat do you mean with the last sentence? – Mathematic Research Dec 29 '12 at 21:40
  • @MathematicResearch: I have added a sentence at the end which I hope clarifies things. – André Nicolas Dec 29 '12 at 21:44
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    Dear André, I am currently doing a little undergraduate research project in o-minimal structures. Thank you for showing me that this is actually something more people know about :P – akkkk Dec 29 '12 at 21:48
  • @André Nicolas: Thank you :) i think i understand the argument ... but is the argument with uncompleteness theorem correct? – Mathematic Research Dec 29 '12 at 21:48
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    The "we could find a polynomial $\dots$" argument is, at least, highly incomplete. And for example $\mathbb{Z}$ is definable in $\mathbb{Q}$, so an argument based on roots of polynomials is unlikely to be enough. For a good understanding of definability in $\mathbb{R}$, the fact that the definable subsets are simple combinations of semi-algebraic sets is the key result, and in a sense undecidability is peripheral. – André Nicolas Dec 29 '12 at 22:01
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    @AndréNicolas would you please explain how do you prove the following? "definable subsets of $\mathbb{R}$ are simple combinations of semi-algebraic sets $\Rightarrow \mathbb{Z}$ is not definable over $\mathbb{R}$" Thanks – aerdna91 Jun 29 '14 at 13:40
  • PS: I already know that by "combinations of semi-algebraic sets" you mean "the smallest class containing semi-algebraic sets and closed under union, intersection and complement" – aerdna91 Jun 29 '14 at 13:48
  • The argument is geometric: there are only finitely many connected components. – André Nicolas Jun 29 '14 at 15:42