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.



Does there exist an uncountable partition of a Polish space so that the union of any collection of blocks is Borel?
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.


