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Does ZF prove that topological groups are completely regular?

Let $\; \mathbf{G} = \langle G,\cdot,\cal{T}\hspace{.06 in}\rangle \;$ be a topological group. $\;\;$ Let $\mathbf{e}$ be the identity element of $\langle G,\cdot \rangle$.
Assume $\{\mathbf{e}\}$ is closed in $\langle G,\cal{T}\hspace{.06 in}\rangle$. $\;\;$ Then, I have manged to convince myself that:

a. $\;\;$ ZF proves that $\langle G,\cal{T}\hspace{.06 in}\rangle$ is regular Haudorff.
b. $\;\;$ ZF + (Dependent Choice) $\;$ proves that $\langle G,\cal{T}\hspace{.06 in}\rangle$ is completely regular.



  1. $\;$ Are those right?


  2. $\;$ Does ZF prove that $\langle G,\cal{T}\hspace{.06 in}\rangle$ is completely regular?


  3. $\;$ If no to 2,
    does assuming one or more of following the suffice for ZF to conclude that $\langle G,\cal{T}\hspace{.06 in}\rangle$ is completely regular?

    (i) $\;$ $\mathbf{G}$ is two-sided complete
    (ii) $\;$ $\langle G,\cdot \rangle$ is abelian
    (iii) $\;$ Countable Choice
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