# Partitioning $\mathbb{R}^n$ into closed sets

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

• Just a small remark: actually all elements of $\mathfrak{P}$ are closed, thus Borel measurable, hence Lebesgue measurable, right? – Jochen Glueck Jun 8 at 6:29
• @PietroMajer how do we make such a function? – Fedor Petrov Jun 8 at 7:49
• 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 – Ashutosh Jun 8 at 9:25
• I think several of these comments should be posted as answers. – Joel David Hamkins Jun 8 at 9:52
• I see, the quotient is not necessarily $T_2$! – Pietro Majer Jun 8 at 10:02