Prove that it is possible to write $\mathbb{R}$ as a union $\mathbb{R} = \bigcup_{i\in I} A_{i}$ where $A_{i} \cap A_{j} = \phi$ if $i\neq j$, $i, j \in I$, and such that each $A_{i}$ and $I$ are uncountable sets.
It's fairly straight forward to find a union where either the $A_{i}$'s are countably infinite and $I$ is uncountably infinite or $I$ is countably infinite and the $A_{i}$'s are uncountably infinite.
One idea I had to get them to be both uncountable was to break the real line up into intervals $[i, i+\Delta i)$ and take the limit as $\Delta i$ goes to zero. But I don't think this works because either $\Delta i$ is zero or it's greater than zero. If it's zero then $I$ is uncountable and $A$ only contains one real number, and is therefore countable. If $\Delta i$ is greater than zero, then it's possible to enumerate the intervals, so $I$ is countable.
Another idea was to take the set of binary strings of infinite length and try to find a way to map each string to a uncountably infinite set of other binary strings of infinite length. I thought maybe this could be done by splitting up the strings into sets ($A_{i}$) depending on their first n digits, and then taking the limit as n goes to infinity.
The problem with this is that if we give infinite digits then we've specified the string itself and the set $A_{i}$ can only contain one string. Otherwise the first n digits can be mapped to a natural number and so $I$ is countable. Same problem.
Any help with this is appreciated.
Thanks
Best Answer
A proof (which, as yo' has pointed out, can be as contrustive as you like):
Since $|\mathbb{R}| = |\mathbb{R} \times \mathbb{R}| $, there exists a bijection $f$ from $\mathbb{R} \to \mathbb{R} \times \mathbb{R} $. Then $\mathbb{R} = \bigcup_{a \in \mathbb{R}} f^{-1}(\mathbb{R},a)$ where $f^{-1}(\mathbb{R},a) = \{b \in \mathbb{R}: f(b) = (c,a)$ for some $c \in \mathbb{R} \}$.