# Why is no AOC needed when proving Baire's theorem for a separable complete metric space?

Wikipedia states

"A restricted form of the Baire category theorem, in which the complete metric space is also assumed to be separable, is provable in ZF with no additional choice principles."

Could someone please either explain why this is the case or point me to some appropriate (undergrad level) reference?

• This is because the space has a base (basis) that can be well-ordered. – DanielWainfleet Aug 14 at 16:10
• I wrote a small review work on analysis without choice where the proof of Baire's theorem is given in ZF for separable spaces. You can find it here. – Asaf Karagila Aug 20 at 13:18