# Power Set, Replacement or Infinity axioms unprovable in set theory

My question is related to the independence of the Power Set, Replacement, and Infinity Axioms. Can we show that Power Set, for example, is unprovable in $$ZFC-P$$? Is $$\neg$$Power Set unprovable in $$ZFC-P$$? (These questions go for Replacement and Infinity as well). I read that we cannot show that $$\text{Con}(ZFC -P) \rightarrow \text{Con}(ZFC)$$. Does this mean we can only show that either $$ZFC-P \vdash \text{Power Set}$$ or $$ZFC-P \vdash \neg\text{Power Set}$$ but not both?

Here's my intuition for these questions without looking at the model theory for it. It seems like the Power Set Axiom is unprovable from the other axioms because we can use the $$H(\kappa)$$ as a model for $$ZFC-P$$. But I don't think that $$\neg$$Power Set is provable from the others because that would imply that $$\text{Incon}(ZFC)$$. And if that were true, then we probably wouldn't work in $$ZFC$$. So I'm guessing that either $$\neg$$Power Set is unprovable or that we are unable to decide this finitistically because of some Second Incompleteness Theorem argument.

Is it informally believed that Power Set, as well as Replacement and Infinity, are independent from the others despite us potentially not being able to show it finitistically?

And finally, we know that the following axiomatic systems for set theory are all equivalently consistent: $$ZF^- \sim ZF \sim ZFC^- \sim ZFC$$ (where $$\Gamma^-$$ is $$\Gamma$$ minus the Axiom of Foundation). That is if we assume the consistency of one of these sets of axioms, then we can conclude the others are also consistent. How far can we reduce the assumption of consistency within the axioms, i.e. we have reduced accepting the consistency of $$ZFC$$ to accepting the consistency of $$ZF^-$$, can we do this any further with Power Set, Replacement and Infinity?

• In $\mathsf{ZFC}$ one can prove that $H(\omega_1)$, the set of hereditarily countable sets, is a model of $\mathsf{ZFC}-P+\forall X(x\text{ is countable})$ and hence of $\mathsf{ZFC}-P+\neg P$. And one can prove that $H(\omega)$, the set of hereditarily finite sets, is a model of $\mathsf{ZFC}-\text{Inf}+\neg\text{Inf}$. Feb 26, 2021 at 0:43

## 1 Answer

Expanding on Brian Scott's comment, the following are provable in $$\mathsf{ZF}^-$$:

• $$\mathsf{HF}\models\mathsf{ZFC-Inf}$$.

• $$(V_{\omega+\omega})^L\models\mathsf{ZC}$$.

• $$\mathsf{HC}^L\models\mathsf{ZFC-Pow}$$ (although we should be careful how we define that theory).

Consequently, $$\mathsf{ZF}^-$$ proves the consistency of each of those theories. So the situation with respect to powerset/replacement/infinity is very different from the situation with respect to choice/foundation.

It is generally believed that $$\mathsf{ZFC}$$ is consistent; consequently, it is generally believed e.g. that $$\mathsf{ZFC-Pow}$$ does not prove $$\neg\mathsf{Pow}$$ (since then $$\mathsf{ZFC}$$ would be obviously inconsistent) or $$\mathsf{Pow}$$ (since then $$\mathsf{ZFC}$$ would prove its own consistency and so be inconsistent per Godel), and similarly for infinity and replacement.

• Do you really need the relativizations to $L$? Apr 19, 2021 at 1:17
• @AndreasBlass Since I'm claiming the result in $\mathsf{ZF}^-$ and I want choice in the model produced, I do have to do some thinning. Apr 19, 2021 at 2:05