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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 $|{\frak P}_0| = 2^{\aleph_0}$?

Dominic van der Zypen
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    The complement ${\frak P}\setminus{\frak P}_0$ is at most countable because of the sigma-finiteness. So the question is equivalent to ask if $|{\frak P}| = 2^{\aleph_0}$. But this is true, for we can make a continuous surjective function $\mathbb{R}^n\to[0,1]$ which is constant on each $F\in{\frak P} $. – Pietro Majer Jun 08 '19 at 06:26
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    Just a small remark: actually all elements of $\mathfrak{P}$ are closed, thus Borel measurable, hence Lebesgue measurable, right? – Jochen Glueck Jun 08 '19 at 06:29
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    @PietroMajer how do we make such a function? – Fedor Petrov Jun 08 '19 at 07:49
  • Thanks @JochenGlueck , I adapted the post – Dominic van der Zypen Jun 08 '19 at 08:24
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    @FedorPetrov for the initial question it is sufficient to assume $n=1$, since one can consider the traces on a line meeting more than one member of $\frak P$; I think it is analogous for any n. For $n=1$ a function is defined assigning inductively a dyadic value to any member of a given countable subfamily $\frak P'$ of $\frak P$, so as to get a continuous function on the union of $\frak P'$, and deduce $\frak P\neq \frak P'$. (Defining a function taking at most countably many values on each closed set is also ok). Actually this construction only proves that $\frak P$ is uncountable! – Pietro Majer Jun 08 '19 at 09:04
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    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 http://www.math.wisc.edu/~miller/res/cov.pdf – Ashutosh Jun 08 '19 at 09:25
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    A better argument: replace $\mathbb{R}^n$ with a closed ball $K$, and $\frak P$ with ${F\cap K:F\in\frak{P}}$. One can consider the quotient topology on $K/\frak{P}$: a connected $T_2$ compact separable space with more than one point: so it's cardinality is $2^{\mathbf c}$, right? – Pietro Majer Jun 08 '19 at 09:33
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    I think several of these comments should be posted as answers. – Joel David Hamkins Jun 08 '19 at 09:52
  • @ Ashutosh I'm confused. If $\frak P$ is a partition of a compact connected metric space $K$ into closed sets, the quotient $K/\frak P$ is a compact connected metric space (it's a continuous image of $K$). But a compact connected metric space with more than one point has cardinality $2^{\bf c}$ no? – Pietro Majer Jun 08 '19 at 09:58
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    I see, the quotient is not necessarily $T_2$! – Pietro Majer Jun 08 '19 at 10:02
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    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..." – Todd Trimble Jun 08 '19 at 12:02
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    See my answer and Taras Banakh's comments at https://mathoverflow.net/questions/48970 . My answer shows that a partition of $[0,1]$ into closed sets must have cardinality at least the covering number for Baire category, and Taras's comments improve the lower bound to the dominating number. – Andreas Blass Jun 08 '19 at 13:36

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