Post #2264410
2026-05-06 12:26 UTC
@johncarlosbaez@mathstodon.xyz @typeswitch@gamedev.lgbt I must admit that I am not an expert and might have misunderstood the assumptions of the quote. In particular, I read the ZFC part in the quote more as an example rather than an assumption. Though I also don't know prior to googling whether "non-standard model of set theory" implicitly excludes IST?
Either way, I would enjoy to learn more about any of these things and am happy to accept if my answer was missing the point, if that is the case?
Ignoring any nitty gritty interpretations of the original quote. You wrote "in a non-standard model of ZFA". What actually does this include? I would assume anything with ZFA axioms plus whatever other non-standard axioms goes?
EDIT: Ah, I think I get it.. Sorry.. IST is a different theory right? Not a different model?
Replies (1)
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@papalex@mathstodon.xyz 2026-05-06 13:47
@johncarlosbaez@mathstodon.xyz @typeswitch@gamedev.lgbt but still, what is wrong with the following reasoning: Since IST is a conservative extension of ZF(C), we have that the reduct of IST models (forgetting the additional structure on these models) gives us ZF(C) models. That means: via IST we get non standard models of ZF(C). In these, we cannot prove via ZF(C) machinery that the nonstandard elements are there, but we know it from our meta view knowing that this is a reduct coming from an IST model.