Timeline for Why does the category of definable sets of $T^\mathrm{eq}$ have coproducts?
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| Mar 27, 2022 at 12:28 | comment | added | user478652 | Ah, I see. Thanks for the clarification. So your argument is a bit more general because your assumption ("for every model there is a sort with at least two elements") is strictly weaker than that of proper-ness ("there is a sort which has at least two elements in every model")! | |
| Mar 26, 2022 at 20:44 | comment | added | James E Hanson | @user478652 Ah, you are right. I misread the definition of proper. The compactness argument is just this: Assume that for any finite set of sorts there is a model in which each of those sorts has one element. Then by compactness, there is a model in which all sorts have one element. Since we assumed this can't happen, there must be a finite collection $O_0,\dots,O_{n-1}$ of sorts with the property that any model has at least two elements in one of those sorts. The theory is $T$ together with sentences for each sort saying that that sort has one element. | |
| Mar 26, 2022 at 19:14 | comment | added | user478652 | As I said, I don't understand that. By the definition goedelian gave, a theory is proper if there is a sort which has at least two elements in every model. Just work with that sort. I don't know why you work with a different definition ("for every model there is a sort with at least two elements" - the quantifiers are the other way around). Also, even if you use that definition, I don't get how your compactness argument works. The compactness theorem says that if each finite subset of a theory has a model, then the theory has a model. To which theory do you apply that? | |
| Mar 26, 2022 at 18:44 | comment | added | James E Hanson | @user478652 In order for this to work, we need to assume that the theory $T$ is proper, i.e., that every model has a sort with at least two elements. If we assume this, then a compactness tells us that there must be a finite list of sorts $O_0,\dots,O_{n-1}$ such that for any model $M\models T$, there is an $i<n$ such that $|O_i^M| \geq 2$. That's where $n$ comes from. | |
| Mar 26, 2022 at 18:43 | comment | added | James E Hanson | @user478652 I'm interpreting the overall question being relative to a possibly incomplete theory rather than relative to a particular structure, and furthermore, I'm establishing that the coproduct can be defined in a uniform way across all models of that theory, rather than just in each model individually. For an incomplete theory, there isn't necessarily a single sort that has at least two elements for every model, even if every model has a sort with at least two elements. | |
| Mar 26, 2022 at 12:04 | comment | added | user478652 | Just fix any sort $O$ with at least two elements. What are $O_0, \dots, O_{n-1}$, what is $n$? I still don't get that. | |
| Mar 26, 2022 at 3:20 | comment | added | James E Hanson | @user478652 Maybe a better way to phrase my argument would be to break it up into two steps. The first would be to show that any proper theory has an imaginary sort that always has at least two elements (namely the product $O_0 \times \dots \times O_{n-1}$), and the second would be what you're suggesting, using that product sort. | |
| Mar 26, 2022 at 3:18 | comment | added | James E Hanson | @user478652 If the theory is incomplete, it may not be the case that $O$ has at least two elements in all models of the theory. For a complete theory, though, we can just use a single $O$ like you say. | |
| Mar 25, 2022 at 15:57 | comment | added | user478652 | I don't really get why you are working with a sequence $O_0, \dots, O_{n-1}$. Can't one realize $A+B$ as a subquotient of $A\times B\times O$ where $O$ is any sort with at least two elements? | |
| Mar 20, 2022 at 22:43 | history | answered | James E Hanson | CC BY-SA 4.0 |