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Is it consistent that there exists a partition $P$ of the real number line $\mathbb{R}$ such that $|P|>\aleph_{0}$ but where $\bigcup R$ is Borel whenever $R\subseteq P$?

If $2^{\aleph_{0}}<2^{\aleph_{1}}$, then the answer to this question is $\textbf{no}$ since there would be at least $2^{\aleph_{1}}$ subsets of each uncountable partition $P$ but there are only $2^{\aleph_{0}}$ Borel sets. We therefore need to use a model such that $2^{\aleph_{0}}=2^{\aleph_{1}}$ to construct our counterexample.

Under $MA+\neg CH$ does there exist a partition $P$ of the real number line $\mathbb{R}$ such that $|P|>\aleph_{0}$ but where $\bigcup R$ is Borel whenever $R\subseteq P$?

Perhaps such a partition $P$ is impossible to construct by some fact obvious to descriptive set theorists that I probably have overlooked.

I call a partition $p$ of a Boolean algebra $B$ such that $\bigvee R$ exists whenever $R\subseteq p$ a subcomplete partition and subcomplete partitions came up all the time in my work on Boolean partition algebras, so I would be interested in which partitions of a given Boolean algebra are subcomplete and which ones are not.

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    $\begingroup$ Trivial observation: all but countably many $A \in P$ must be uncountable (and hence have $|A| = \mathfrak{c}$). For suppose there is an uncountable $P_c \subset P$ with every $A \in P_c$ countable. Since we are assuming $\lnot \mathsf{CH}$, we can find $P_0 \subset P_c$ with $\aleph_0 < |P_0| < \mathfrak{c}$. Then $B = \bigcup P_0$ has cardinality $|P_0|$, which cannot be if $B$ is Borel. $\endgroup$
    – Nate Eldredge
    Commented Aug 18, 2015 at 4:21
  • $\begingroup$ Another trivial observation: if by "consistent" we mean "consistent with ZF", then it is, because we can work in a model of ZF in which every subset of $\mathbb{R}$ is Borel. (For instance, a model in which $\mathbb{R}$ is a countable union of countable sets.) In that case we can take the trivial partition of $\mathbb{R}$ into singletons. Ashutosh's answer below shows it is not consistent with ZFC. $\endgroup$
    – Nate Eldredge
    Commented Sep 8, 2015 at 16:36

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Suppose $\{P_i : i < \kappa\}$ is such a partition. Let $f: R \to R$ be a function satisfying $|f[P_i]| = 1$ and for all $i < j < \kappa$, $f[P_i] \cap f[P_j] = \phi$. Then $f$ is Borel so its image is an uncountable analytic set of size less than continuum: Contradiction.

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    $\begingroup$ It took me a moment to understand $f$. Another way to describe it: Using AC, choose one element $x_i$ from each $P_i$, and set $f(x) = x_i$ if $x \in P_i$. For any Borel set $B$, we have $f^{-1}(B) = \bigcup_{i : x_i \in B} P_i$ which by assumption is Borel. So $f$ is Borel. $\endgroup$
    – Nate Eldredge
    Commented Aug 18, 2015 at 4:54
  • $\begingroup$ @NateEldredge Does this only show that the partition must be size $2^{\aleph_0}$? $\endgroup$
    – William
    Commented Aug 18, 2015 at 4:59
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    $\begingroup$ @William: If the partition has size $2^{\aleph_0}$ then by choosing subsets of the partition we get $2^{2^{\aleph_0}}$ distinct Borel sets, which is impossible. $\endgroup$
    – Nate Eldredge
    Commented Aug 18, 2015 at 5:04
  • $\begingroup$ So, long story short, given any uncountable partition of $\Bbb R$ into Borel sets, the partition is a refinement of a partition into non-Borel sets. $\endgroup$
    – Asaf Karagila ♦
    Commented Aug 18, 2015 at 5:34

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