# Reining in the Axiom of Power Set in ZF

Given the powerset operator $$\mathit P$$, we have the following mapping

$$\tag 1 \mathcal \Phi: \mathbb N \to \Phi(\mathbb N)$$ $$\quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \, n \mapsto \mathit P^n(\mathbb N)$$

What happens if we take away the Axiom of Power Set in $$ZF\pm C$$ and replace it with $$\text{(1)}$$? Would this contradict the other axioms?

• Incidentally, there are some very interesting results on the "amount of powerset" (or really, "powerset-along-replacement") we need in various situations. Most famously, Harvey Friedman showed that more and more iterations of powerset are needed to prove determinacy further and further up the Borel hierarchy. – Noah Schweber Jan 11 at 22:08
• On formatting: You can use pairs of L& R dollar-signs to "display" a line instead of typing \quad \quad \quad ..., and to put un-formatted text (like "because") in a displayed line, enclose it within \text {...}, like \text { because }. – DanielWainfleet Jan 14 at 0:47
• @DanielWainfleet I've felt silly at times trying to (perfectly) align stuff since the final result might depend on the display device. But anyway, not exactly sure what you mean. Please share some links with useful mathstackexchange formatting technique so I can copy n paste it in the future. – CopyPasteIt Jan 14 at 1:27
• @NoahSchweber Another interesting paper, 'Large irredundant sets in operator algebras' arxiv.org/abs/1808.01511?context=math I don't know the formulation for the $\text{♢-Axiom}$ (stronger than the continuum hypothesis), but my guess is all modern day physics can be expressed/modeled in $V_{\omega+\omega} + \text{♢}$. I wonder how this holds up: "...every open set of Minkowski spacetime is associated with a C*-algebra" /wikipedia. – CopyPasteIt Jan 14 at 14:55
• Did something not work with the Ask Question page that you felt the need to ask a completely disjoint question on a page where you already accepted one answer? – Asaf Karagila Jan 14 at 15:44

I assume that by (1) you mean the existence of the operator $$\Phi$$ that you defined above. As you say, the existence of such a $$\Phi$$ is a consequence of the Powerset Axiom and of the other axioms of $$ZF$$, so if (1) contradicts the other axioms, the Powerset Axiom does as well.
Moreover, it seems to me that $$V_{\omega_1}$$ is a model of $$ZFC-P+(1)$$, so the consistency strength of this theory is strictly less that that of $$ZFC$$.
• Just $V_{\omega+\omega}$ is already enough. – Asaf Karagila Jan 9 at 12:18
• Correct, for Replacement we use $H(\kappa)$, the set of sets which are hereditarily smaller than $\kappa$. – Asaf Karagila Jan 9 at 13:18