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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?

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  • $\begingroup$ 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? $\endgroup$ – Wojowu Nov 9 at 12:32
  • $\begingroup$ Ok, corrected. Thanks! $\endgroup$ – Zuhair Al-Johar Nov 9 at 12:53
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    $\begingroup$ If $\kappa$ is inaccessible, then it is a $\beth$-fixed point, and so $V_\kappa$ has size $\kappa$. $\endgroup$ – Joel David Hamkins Nov 9 at 13:12
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    $\begingroup$ 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. $\endgroup$ – Joel David Hamkins Nov 10 at 13:04
  • $\begingroup$ @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. $\endgroup$ – Zuhair Al-Johar Nov 11 at 5:24

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