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Let $n$ be a positive integer. It is well-known that $\mathbb{R}^n$ cannot be non-trivially partitioned into open sets, since it is connected.

Let $\frak P$ be a partition of $\mathbb{R}^n$ into closed sets and assume $\mathbb{R}^n\notin{\frak P}$ (that is, ${|\frak P|}>1$). Let ${\frak P}_0\subseteq {\frak P}$ consist of the elements of ${\frak P}$ that have Lebesgue measure $0$. Is it necessarily true that $|{\frak P}_0| = 2^{\aleph_0}$?

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    $\begingroup$ Just a small remark: actually all elements of $\mathfrak{P}$ are closed, thus Borel measurable, hence Lebesgue measurable, right? $\endgroup$
    – Jochen Glueck
    Commented Jun 8, 2019 at 6:29
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    $\begingroup$ @PietroMajer how do we make such a function? $\endgroup$
    – Fedor Petrov
    Commented Jun 8, 2019 at 7:49
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    $\begingroup$ Not necessarily true. It is consistent that $2^{\aleph_0}$ is arbitrarily large and every Polish space can be partitioned into $\aleph_1$ non empty closed sets. See Theorems 3 and 4 in A. Miller, Covering $2^{\omega}$ with $\omega_1$ disjoint closed sets, here math.wisc.edu/~miller/res/cov.pdf $\endgroup$
    – Ashutosh
    Commented Jun 8, 2019 at 9:25
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    $\begingroup$ I think several of these comments should be posted as answers. $\endgroup$
    – Joel David Hamkins
    Commented Jun 8, 2019 at 9:52
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    $\begingroup$ Among the interesting things in the Arnold Miller paper cited by Ashutosh is: "Tall remarks that Booth (1968, unpublished) proved that MA [Martin's Axiom -- TT] implies the closed unit interval is not the union of less than $|2^\omega|$ disjoint nonempty closed sets..." $\endgroup$
    – Todd Trimble
    Commented Jun 8, 2019 at 12:02

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