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Profinite groups are usually characterized as compact, totally disconnected, Hausdorff groups. However, as shown here, every totally disconnected topological group already has the Hausdorff property.

Still, every textbook I've come across explicitly mentions (and even proves) the Hausdorff property in the characterization. Is there any reason whatsoever for this emphasis?

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Yes it is quite funny that every totally disconnected topological group is automatically Hausdorff.
There are more places in mathematics where this superfluousness is present. For example the definition of a (left) module $M$ over a ring $R$ with identity $1$. In every textbook $M$ is required to be an abelian group, but it is implied by the axioms: $r(m_1+m_2)=rm_1+rm_2$ en $(r_1+r_2)m=r_1m+r_2m$. For these yield $(1+1)(m_1+m_2)=m_1+m_1+m_2+m_2$ and also $(1+1)(m_1+m_2)=m_1+m_2+m_1+m_2$, whence $m_1+m_2=m_2+m_1$!
So I guess it is laziness or a matter of ease.

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For a topological group it's equivalent to be $T_0$, $T_1$, $T_2$ or even completely regular ($T_{3\frac12}$). So we just want to exclude the case where we have a large indiscrete subgroup like $\overline{\{e\}}$, even though the totally disconnected condition already implies this.

It's pretty routine to demand compact spaces to be also Hausdorff, and all topological groups to have non-trivial separation axioms too. So it's probably force of habit.

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