# Does ZF prove that topological groups are completely regular?

Let $\mathbf{G} = \langle G,\cdot,\mathcal{T}\;\rangle$ be a topological group. Let $\mathbf{e}$ be the identity element of $\langle G,\cdot \rangle$. Assume $\{\mathbf{e}\}$ is closed in $\langle G,\cal{T}\;\rangle$. Then, I have managed to convince myself that:

1. ZF proves that $\langle G,\cal{T}\;\rangle$ is regular Haudorff.
2. ZF + (Dependent Choice) proves that $\langle G,\cal{T}\;\rangle$ is completely regular.

My questions are:

1. Are those right?
2. Does ZF prove that $\langle G,\cal{T}\;\rangle$ is completely regular?
3. If no to question 2, does assuming one or more of following suffice for ZF to conclude that $\langle G,\cal{T}\,\rangle$ is completely regular?
1. $\mathbf{G}$ is two-sided complete
2. $\langle G,\cdot \rangle$ is abelian
3. Countable Choice
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I think I can confirm that ZF+DC proves that a Hausdorff topological group is completely regular. Dependent choice appears in Urysohn's proof that a uniform space is completely regular, and I do not see how to remove it. In fact, the whole question hangs on the following: does ZF prove that a uniform space is completely regular? My instinct tells me that the answer is negative, even with Countable choice. Of course, my instincts are often wrong. –  Andrej Bauer Aug 27 '11 at 10:08
I'll go on a limb here and claim that intuitionistic logic with Dependent Choice proves every Hausdorff group to be regular. –  Andrej Bauer Aug 27 '11 at 12:59
Ricky just posted a new version of this question in terms of proximities - mathoverflow.net/questions/73952/… –  François G. Dorais Aug 29 '11 at 12:36