Why is the Hausdorff property mentioned in the characterization of profinite groups?
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?
group-theory topological-groups profinite-groups
asked Jan 4 at 12:37
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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?
group-theory topological-groups profinite-groups
asked Jan 4 at 12:37
user631620user631620
213
New contributor
user631620 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
add a comment |
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?
group-theory topological-groups profinite-groups
asked Jan 4 at 12:37
user631620user631620
213
New contributor
user631620 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
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?
group-theory topological-groups profinite-groups
group-theory topological-groups profinite-groups
asked Jan 4 at 12:37
user631620user631620
213
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user631620 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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asked Jan 4 at 12:37
user631620user631620
213
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asked Jan 4 at 12:37
user631620user631620
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asked Jan 4 at 12:37
user631620user631620
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asked Jan 4 at 12:37
user631620user631620
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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.
answered Jan 4 at 13:43
Nicky HeksterNicky Hekster
28.3k53456
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For a topological group it's equivalent to be $T_0$, $T_1$, $T_2$ or even completely regular ($T_{3frac12}$). 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.
answered Jan 4 at 23:54
Henno BrandsmaHenno Brandsma
105k347114
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2 Answers
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2 Answers
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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.
answered Jan 4 at 13:43
Nicky HeksterNicky Hekster
28.3k53456
add a comment |
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.
answered Jan 4 at 13:43
Nicky HeksterNicky Hekster
28.3k53456
add a comment |
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.
answered Jan 4 at 13:43
Nicky HeksterNicky Hekster
28.3k53456
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.
answered Jan 4 at 13:43
Nicky HeksterNicky Hekster
28.3k53456
edited Jan 4 at 14:10
answered Jan 4 at 13:43
Nicky HeksterNicky Hekster
28.3k53456
answered Jan 4 at 13:43
Nicky HeksterNicky Hekster
28.3k53456
answered Jan 4 at 13:43
Nicky HeksterNicky Hekster
28.3k53456
28.3k53456
add a comment |
add a comment |
For a topological group it's equivalent to be $T_0$, $T_1$, $T_2$ or even completely regular ($T_{3frac12}$). 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.
answered Jan 4 at 23:54
Henno BrandsmaHenno Brandsma
105k347114
add a comment |
For a topological group it's equivalent to be $T_0$, $T_1$, $T_2$ or even completely regular ($T_{3frac12}$). 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.
answered Jan 4 at 23:54
Henno BrandsmaHenno Brandsma
105k347114
add a comment |
For a topological group it's equivalent to be $T_0$, $T_1$, $T_2$ or even completely regular ($T_{3frac12}$). 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.
answered Jan 4 at 23:54
Henno BrandsmaHenno Brandsma
105k347114
For a topological group it's equivalent to be $T_0$, $T_1$, $T_2$ or even completely regular ($T_{3frac12}$). 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.
answered Jan 4 at 23:54
Henno BrandsmaHenno Brandsma
105k347114
edited Jan 5 at 9:45
answered Jan 4 at 23:54
Henno BrandsmaHenno Brandsma
105k347114
answered Jan 4 at 23:54
Henno BrandsmaHenno Brandsma
105k347114
answered Jan 4 at 23:54
Henno BrandsmaHenno Brandsma
105k347114
105k347114
add a comment |
add a comment |
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