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It is a famous that the consistency strength of the statement "every set is measurable" $+ZF+DC$ is the same as the existence of an inaccessible cardinal. Does Shelah's argument also answer the following question?

Question: Does the consistency of $ZF+DC+\mbox{ every non-null set has a perfect subset}$ imply the consistency of the existence of an inaccessible cardinal?

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    $\begingroup$ @JasonZeshengChen In view of the context and standard usage in set theory, I assume "non-null set" just means a set that doesn't have Lebesgue measure zero. So it might be a non-measurable set. $\endgroup$
    – Andreas Blass
    Commented Aug 8, 2021 at 3:34
  • $\begingroup$ Should "set" be interpreted as "subset of the reals" in this question? $\endgroup$
    – YCor
    Commented Aug 8, 2021 at 7:24
  • $\begingroup$ @AndreasBlass right, that was a silly reading I made. Thanks $\endgroup$
    – Jason Zesheng Chen
    Commented Aug 8, 2021 at 7:36
  • $\begingroup$ I understand that there are many typos and misleading statements, but I will not correct them. $\endgroup$
    – 喻 良
    Commented Aug 8, 2021 at 10:20
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    $\begingroup$ @喻良 Why not? $\,$ $\endgroup$
    – Jeremy Rickard
    Commented Aug 8, 2021 at 13:47

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