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

In summary, the conversation discusses whether all models of ZFC, specifically in a standard formulation, must be at least countable. The speaker argues that there are countably many instances of Replacement in a model, so it must have at least countably many satisfactions of it. The other speaker questions if this only applies to first-order formulations or if there are second-order formulations that can be finite. The first speaker responds that all models of ZFC are at least countable because they all have an empty set and sets containing it, and the argument using Replacement does not guarantee different sets.
  • #1
mpitluk
25
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Must all models of ZFC (in a standard formulation) be at least countable?

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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  • #2
Yes, they're all at least countable, because they all have an empty set, a set containing just the empty set, a set containing just the set containing just the empty set, etc. Your argument using Replacement doesn't work, since there's no guarantee that two different instances of Replacement generate different sets.
 

1. What is ZFC and why is it important for models to be countable?

ZFC stands for Zermelo-Fraenkel set theory with the Axiom of Choice, which is a foundational theory in mathematics. In order for models of ZFC to be well-defined and consistent, they must be countable. This means that there must be a way to assign a unique natural number to each element in the model.

2. Can models of ZFC be uncountable?

No, models of ZFC cannot be uncountable. This is because ZFC is a countable theory, meaning that it can only express statements about countable objects. Therefore, any model of ZFC must also be countable.

3. What happens if a model of ZFC is not countable?

If a model of ZFC is not countable, then it is not a valid model of ZFC. This means that it does not satisfy all of the axioms of ZFC and cannot be used as a foundation for mathematics. In other words, the model would be inconsistent and lead to contradictions.

4. Are there any alternative formulations of ZFC that allow for uncountable models?

Yes, there are alternative formulations of ZFC that allow for uncountable models. For example, there is the Multiverse Axiom of Set Theory, which allows for models of ZFC to exist in a larger "multiverse" of mathematical universes. However, this is a controversial topic in mathematics and is not widely accepted.

5. Are there any practical implications of models of ZFC being countable?

Yes, there are practical implications of models of ZFC being countable. For example, it means that we can effectively use computer simulations and other computational methods to study and explore properties of these models. Additionally, the countability of models makes it easier to compare and contrast different models of ZFC.

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