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It is well known that the original proof of Urysohn's lemma uses a choice principle. (DC)

What are weak forms of the Urysohn's lemma that don't require choice? For example, one is in the link. ($\text{ Second Countable } + \text{ Regular } \Rightarrow \text{ Urysohn's lemma }$)

I also saw somewhere (I don't remember) that Urysohn's Lemma is provable without choice if it is metric space. (Edit; Now I remember. It was an exercise on PMA, using Hausdorff distance)

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I am fairly sure that the arguments I've seen for 2nd countable + regular => urysohn's lemma required choice. – gnometorule Jan 17 '13 at 0:37
@gnometorule Are you even sure for the argument in the link? Otherwise i'm going to spend my whole day to check it. – Katlus Jan 17 '13 at 3:08
For metric spaces and disjoint non-empty closed subsets $A$ and $B$ you can just write $f(x) = \frac{d(x,A)}{d(x,A)+d(x,B)}$ to get a function $f$ which is $0$ on $A$ and $1$ on $B$. Is that what you mean by alluding to Hausdorff distance (which is something else)? – Martin Jan 17 '13 at 7:22
@Martin Yes, exactly. How do i call $d(x,A)$ then? – Katlus Jan 17 '13 at 7:53
It's usually called the distance of $x$ to $A$ or something to that effect. – Martin Jan 17 '13 at 7:56

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