I wanted to give some of the new undergrad analysis students the following problem: given the real numbers (with the standard topology, as they'd expect) one cannot have an uncountable set such that every point is an isolated point. A few of my fellow grad students attempted solutions first.
A sketch of a potential proof was given as follows: take each point and create as large a ball as is possible without it containing any other point of the set, so say $(a-\alpha, a+\alpha)$. Then construct corresponding balls with a smaller radius: $(a-\frac{\alpha}{2}, a+\frac{\alpha}{2})$. These new balls will intersect each other trivially. Then choose a rational point from each of these new balls. This set will be in one-to-one correspondence with the balls which are in one-to-one correspondence with the points of the original set.
After presenting this proof, it was argued that it requires the (non-finite) axiom of choice (as we are picking one point from a potentially uncountable set). We modified the proof by using the density of the rationals to pick rationals $p_{a},q_{a}$ such that $(a-p_{a},a+q_{a})\subseteq (a-\frac{\alpha}{2}, a+\frac{\alpha}{2})$, and then have our function pick the "left-most" end-point. It was still argued that this used the axiom of choice.
Because I am not entirely familiar with what constitutes use of the axiom of choice, I wanted to open the question up to all of you. Do these proofs require the (non-finite) axiom of choice? If so, is there a proof you know of which does not require it?
Best Answer
You don't need the Axiom of Choice here to pick a rational in each $(a-\frac{\alpha}{2}, a+\frac{\alpha}{2})$. You can fix an enumeration $(q_r)$ of the rationals and just pick the rational $q_i \in (a-\frac{\alpha}{2}, a+\frac{\alpha}{2})$ with the smallest index $i$.