Timeline for Is the theory of the adjunction operation, used for algebraization of hereditarily finite set theory, algebraizable?
Current License: CC BY-SA 4.0
12 events
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| Nov 17, 2021 at 9:15 | comment | added | Ioachim Drugus | My earlier question here math.stackexchange.com/questions/4306165/… requests a more or less "direct answer". However, this theory having directly to do with algebraization sounds to me also an interesting subject for research. | |
| Nov 17, 2021 at 6:56 | comment | added | Andrej Bauer | Ok, thanks. So you more or less randomly suggested $\triangleright$, it's not like you want to do something with it, you're just trying to understand what "algebraic" means in the context of set theory? Well, it would have been easier to answer that question directly :-) | |
| Nov 16, 2021 at 20:22 | comment | added | Ioachim Drugus | However, I hope, that by giving up some conditions required to be an algebraic theory, one can get a theory close to 'algebraic theory'. What one must give up? I suspect this is the condition that the axioms must be identities. Thus, maybe we must look for quasi-idetities which axiomatize this theory. I chose specifically this theory to learn about algebraization of set theory, because adjunction is the best known operation used for algebraization purposes. | |
| Nov 16, 2021 at 20:21 | comment | added | Ioachim Drugus | I want to understand what is to 'algebraize set theory' in terms of universal algebra. I learnt from reading 'categorical set theory' that this is not what is called 'algebraic theory' since the algebraic structures used by this theory have infinitary operations. However, this is an algebraic set theory in a wider sense. Thus, the authors had to give up something (finiteness) in order to still algebraize ZF close to 'algebraic theory'. I suspect, that in terms of universal algebra one also cannot algebraize set theory to obtain an 'algebraic theory'... | |
| Nov 16, 2021 at 17:31 | comment | added | Andrej Bauer | @IoachimDrugus: it would help to know what the larger context of this question is. What are you trying to accomplish? What do you want to do with the theory, once you have it? | |
| Nov 16, 2021 at 15:57 | comment | added | Emil Jeřábek | Well neither $x\notin\emptyset$ nor $z\in x\cup\{y\}\to z\in x\lor z=y$ can be expressed by quasi-identities, as they are not preserved by products (including the empty product, for the former). | |
| Nov 16, 2021 at 15:07 | comment | added | Ioachim Drugus | It doesn't sound probable that this theory is algebraizable in the sense that it is algebraic theory. It could be algebraizable in another sense: this theory is the theory of a quasi-variety of algebras. I would call such a theory ``quasi-algebraic''. | |
| Nov 16, 2021 at 12:18 | comment | added | Emil Jeřábek | Yeah, it’s not very clear to me either what the OP is after. | |
| Nov 16, 2021 at 11:17 | comment | added | Andrej Bauer | Thanks, I removed the offending axiom. I did say "to get things started"... I am a bit confused as to what the OP is after. We can't hope an algebraic theory to describe an initial algebra (which is what the hereditary finite sets are), we would need also a suitable induction principle. But would that count as "algebraic"? | |
| Nov 16, 2021 at 11:14 | history | edited | Andrej Bauer | CC BY-SA 4.0 |
deleted 82 characters in body
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| Nov 16, 2021 at 10:48 | comment | added | Emil Jeřábek | Your axiom $\emptyset\triangleright x\triangleright y=x\triangleright y$ is not valid. Also, you are missing the adjunction axiom $z\in x\cup\{y\}\to z\in x\lor z=y$, which is even worse than $x\notin\emptyset$; cf. mathoverflow.net/questions/408327/…. | |
| Nov 16, 2021 at 9:44 | history | answered | Andrej Bauer | CC BY-SA 4.0 |