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I understand that to prove this, one considers that each open interval contains a rational number and since the open intervals are disjoint, there exists an injection from the collection of intervals to the rational numbers, which are countable, hence the collection of open intervals is countable.

I noticed, however, that this requires arbitrarily many simultaneous, non-defined, choices from the intersection of the rational numbers with each interval in the collection. So, my question is, does this proof require AC?

Killian Heanue
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  • Not really, because it’s easy to endow $\mathbb{Q}$ with a well-founded order (by smallest denominator first, then compare the numerators). – Aphelli Jan 03 '21 at 19:47

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No, we don't need the axiom of choice here. Take any bijection $f:\mathbb{Q}\to\mathbb{N}$, and for each interval in your collection choose the rational number $q$ in the interval for which $f(q)$ has the smallest value. This is a very specific choice function.

Mark
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