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Andrej Bauer
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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?
  4. $\mathbf{G}$ is two-sided complete
  5. $\langle G,\cdot \rangle$ is abelian
  6. Countable Choice
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