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Every Hausdorff topological group is regular (completely regular, in fact). Is this true if I replace topological group with homogeneous space?

This is not obvious to me because there are Hausdorff homogeneous spaces which are not topological groups, such as $S^2$. Also, there are $T_1$ spaces which are homogeneous but not regular, such as $\omega$ with the cofinite topology.

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An example of a homogeneous Hausdorff space which is not regular is obtained by giving $\mathbb R$ the topology where the open sets are those of the form $U \setminus A$ where $U \subseteq \mathbb R$ is open in the usual metric/order topology, and $A \subseteq \mathbb R$ is countable.

  • It is Hausdorff since the topology is finer than the usual topology which is itself Hausdorff.
  • Any autohomeomorphism of the usual topology on $\mathbb R$ is also an autohomeomorphism of this space, so the homogeneity of the usual topology implies the homogeneity of this space.
  • It is not regular because $U = (-1,1) \setminus \{ \frac 1n : n \geq 1 \}$ is an open neighborhood of $0$, but there is no neighborhood $V$ of $0$ with $\overline V \subseteq U$.
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