Must all models of ZFC (in a standard formulation) be at least countable?(adsbygoogle = window.adsbygoogle || []).push({});

Why I think this: there are countably many instances of Replacement, and so, if a model is to satisfy Replacement, it must have at least countably many satisfactions of it.

Does my question only apply to first-order formulations of ZFC, or are there second-order formulations of ZFC that can be finite?

Thanks.

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# Must all models of ZFC (in a standard formulation) be at least countable?

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