Partition of interval is countable Consider the arbitrary partition of the real non-empty interval $[a,b)$, into the disjoint sets $[\alpha(i),\beta(i))$ where $i\in I$ and $I$ is some index set. Why is it necessary that $I$ be countable? Isn't it true that $[a,b)$ contains an uncountable set that could serve as a source of candidates for the endpoints of the partition intervals?

Hint: Each interval contains a rational.
Added: The following is probably more explicit tha necessary. Let $r_1,r_2, r_3,\dots$ be an enumeration of the rationals. For each $i\in I$, let $n_i$ be the least index such that $r_{n_i}$ is in $[\alpha(i),\beta(i))$. Then the set $\{n_i\}$, as $i$ ranges over $I$, is a subset $M$  of the natural numbers, and the mapping $i\to n_i$ gives a bijection between $I$ and $M$. Thus $I$ must be finite or countably infinite.
