# How much high iterative partitioning of the first inaccessible stage can be?

Working in $$ZF$$ + existence of an inaccessible ordinal.

Let $$\kappa$$ be the first inaccessible ordinal (i.e. the first regular ordinal that is a limit of regular ordinals).

Let subpartition of $$X$$ stand for a "subset of a partition of $$X$$".

Question 1: Can we have a $$V_{\kappa}$$ sized subpartition of $$\mathcal P(V_{\kappa})$$ whose elements are $$V_{\kappa}$$ sized subpartitions of $$V_{\kappa}$$ whose elements are $$V_{\kappa}$$ sized?

Question 2:If so, can we go up? i.e. have a $$V_{\kappa}$$ sized subpartition of $$\mathcal P^2(V_{\kappa})$$ whose elements are $$V_{\kappa}$$ sized subpartitions of $$\mathcal P(V_{\kappa})$$ whose elements are $$V_{\kappa}$$ sized subpartitions of $$V_{\kappa}$$ whose elements are $$V_{\kappa}$$ sized?

Question 3:If so, then how long can we go up further? i.e. up to which finite ordinal $$i$$ we can have have a $$V_{\kappa}$$ sized subpartition of $$\mathcal P^i(V_{\kappa})$$ whose elements are $$V_{\kappa}$$ sized subpartitions of $$\mathcal P^{i-1}(V_{\kappa})$$,..., whose elements are $$V_{\kappa}$$ sized subpartitions of $$V_{\kappa}$$ whose elements are $$V_{\kappa}$$ sized?

• How do you define an inaccessible? The definitions equivalent in ZFC are not equivalent in ZF. Also, do you have to use a notation which is widely used for the entire cumulative hierarchy for one of its stages? – Wojowu Nov 9 at 12:32
• Ok, corrected. Thanks! – Zuhair Al-Johar Nov 9 at 12:53
• If $\kappa$ is inaccessible, then it is a $\beth$-fixed point, and so $V_\kappa$ has size $\kappa$. – Joel David Hamkins Nov 9 at 13:12
• Well, you are referring throughout to "size $V_\kappa$", but this just means "size $\kappa$", and so your way of stating the problem is needlessly cumbersome. – Joel David Hamkins Nov 10 at 13:04
• @JoelDavidHamkins, yes I know. That doesn't essentially change the question in any substantial manner. Anyhow I want to keep it as such, because I want the same question to work under absence of choice, for whatever the definition of inaccessible ordinal is. – Zuhair Al-Johar Nov 11 at 5:24