# 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
Would it be interesting to come up with a model of ZF and a topological group there that is not completely regular. – Gerald Edgar Mar 3 '15 at 15:02

I'm not sure about the proof I gave, But as I checked, it didn't use full AC, but as Asaf mentioned in a comment it uses DC. The following theorem is due to Pontrjagin. See Book by Montgomery and Zippin(Page 29). I will give a sketch of proof.

Note: Maybe it's needed to add some separation axiom to the following theorem. please edit it, if needed.

### Theorem: Every $T_{0}$ topological group is completely regular.

Proof. It's enough to prove that a given topological group $(G,*)$ is completely regular at $e$. Let $F$ be a closed set not containing $e$. Put $O=F^{c}$. Choose symmetric open neighborhoods of $e$, $U_{n}$ By continuity of $*$, such that $U_{0}=O$(w.l.o.g assume $O$ is symmetric) $U_{n+1}^2 \subseteq U_{n} \cap O ~~~~n=0,1,2...$.

Now for rational numbers of the form $r=\frac{k}{2^n}~~k \in \{1,2,3,...2^n \},~~ n \in \{0,1,2,...\}$ inductively define open neighborhood $V_{r}$ of $e$ such that:

1) $V_{\frac{1}{2^n}}=U_{n}~~~~~\forall n$

2) $V_{\frac{2k}{2^{n+1}}}=V_{\frac{k}{2^n}}$

3) $V_{\frac{2k+1}{2^{n+1}}}=V_{\frac{1}{2^{n+1}}}V_{\frac{k}{2^n}}$

The definition does not depend to the representation of $r$ and the family $V_{r}$ has the following properties:

4) $V_{\frac{1}{2^n}}V_{\frac{m}{2^{n}}} \subseteq V_{\frac{m+1}{2^n}}~~~~~~~~~m+1 \leq 2^n$

5) $V_{r} \subseteq V_{s}~~~~~$if $~r <s<1$ Now define the function $f:G \longrightarrow [0, 1]$ as follows:

$f(x)=0~$ if $x \in \bigcap \limits_{r} V_{r}$

$f(x)=1$ if $~~x \notin V_{1}$, and in other cases define:

$f(x)= \sup \{r \leq 1 : x \notin V_{r}\}$

### Claim: $f$ is as required.

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Defining a sequence by induction is the quintessential use of dependent choice. – Asaf Karagila Mar 3 '15 at 10:31
@AsafKaragila Thank you. I don't know much about AC and related things, Is it an answer of question 2. – Rahman. M Mar 3 '15 at 11:19
No, it just answers (part of) question 1. – Todd Trimble Mar 3 '15 at 12:47
About" maybe closedness of $\{e\}$ is sufficient": In a topological group, if $\{e\}$ is closed, then, by homogeneity, all singletons are closed, and so the topology satisfies $T_1$. – Andreas Blass Mar 3 '15 at 13:19
Although $T_1$ is stronger than $T_0$ in general, they're equivalent in topological groups. – Andreas Blass Mar 3 '15 at 15:19