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We are given a language $L$ and a structure $M$ (model). Definable sets in this model are subsets of $M^n$ definable by a formula of $L$. The Grothendieck semiring of the structure is defined in the following way:

  1. First one forms the free semigroup on symbols [X] for $X$ a definable subset of $M^n$ for some $n$.
  2. Then one quotient the obtained semigroup by relations $$ [X] =[Y] $$ iff $X$ is isomorphic to $Y$ by a definable bijection, i.e. a bijection whose graph is a definable set. $$ [X \setminus Y] +[Y] =[X]$$
  3. Next one considers the product defined by $[X]\cdot [Y] = [X \times Y]$ in the same way.

If for some reason, I need to consider only compact definable subsets of $M^n$ (for some topology) at first then I obtain a Grothendieck semiring $R^0$. Secondly I consider all definable sets and I get a Grothendieck semiring $R$. Then my question is: What is the relation between $R$ and $R^0$?

More precisely: If any definable set ($\subset M^n$ say) is an infinite increasing union of compact definable sets, can we say that $R$ is in some sense a completion to $R^0$?

Also, if I know $R^0 \neq 0$ does it follow $R \neq 0$?

Edit: The language $L$ is Macintyre's language and the model $M$ is $\mathbb{Q}_p$. Definable sets in Macintyre language are semi-algebraic sets. I usually require that definable bijections are also (Haar) measure preserving. My original question does not depend on these details, though.

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  • $\begingroup$ I retagged it, because it seems more like a Model Theory question. $\endgroup$
    – Mahdi Majidi-Zolbanin
    Commented Jan 13, 2012 at 17:26
  • $\begingroup$ I haven't thought about defining the right $L$, but can't we find a counterexample to "if $R^0 \neq 0$ then $R \neq 0$" by taking $M=\mathbb Z$ and the discrete topology (so that compact = finite)? $\endgroup$
    – darij grinberg
    Commented Jan 13, 2012 at 17:27
  • $\begingroup$ @darij: I added some details, but as I said in the edit, any counterexample or proof does not need to follow these details. $\endgroup$
    – user16974
    Commented Jan 13, 2012 at 19:29
  • $\begingroup$ We can define a neighborhood of a set $X$ to consist of all the sets isomorphic to a definable subset of $X$ which intersects some finite collection of open definable subsets of $X$. This should make addition and multiplication continuous, and would make $R^0$ dense in $R$. However I know nothing about model theory so this is probably wrong. $\endgroup$
    – Will Sawin
    Commented Jan 13, 2012 at 22:23

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