Proving order topology is normal without going through complete-normality
1
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I've been looking extensively for a proof that any toset equipped with the order topology is normal. Every proof I've seen actually shows that the space is in fact completely normal, but they use complicated arguments about convex sets. I would like to know if there exists a simple way to prove directly that the order topology is normal, without going through complete-normality.
general-topology order-theory
asked Dec 1 '18 at 6:47
RlosRlos
1337
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add a comment |
1
$begingroup$
I've been looking extensively for a proof that any toset equipped with the order topology is normal. Every proof I've seen actually shows that the space is in fact completely normal, but they use complicated arguments about convex sets. I would like to know if there exists a simple way to prove directly that the order topology is normal, without going through complete-normality.
general-topology order-theory
asked Dec 1 '18 at 6:47
RlosRlos
1337
$endgroup$
$begingroup$
completely normal means that every subspace is normal, so it's not a real complication. In fact you can prove that the fact that an arbitrary ordered topological space is normal needs the axiom of choice so is always non-constructive, e.g. The standard proof I know shows that an ordered space is monotonically normal (stronger than completely normal, and has all sorts of other extra consequences) using an extra well-order on the set (so making the AC necessity obvious). It's a bit of case checking but not very hard.
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– Henno Brandsma
Dec 1 '18 at 6:57
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toset? You mean linearly ordered set?
$endgroup$
– Henno Brandsma
Dec 1 '18 at 7:01
$begingroup$
Yes, I mean linearly ordered. So there is just no constructive proof of this fact? I find that amazing and rather odd, because it's usually not too hard to show that a given space is normal...
$endgroup$
– Rlos
Dec 1 '18 at 7:08
$begingroup$
In fact, normality is a hard thing to show for general spaces. Only metric spaces as a class are easy.
$endgroup$
– Henno Brandsma
Dec 1 '18 at 7:25
add a comment |
1
1
1
$begingroup$
I've been looking extensively for a proof that any toset equipped with the order topology is normal. Every proof I've seen actually shows that the space is in fact completely normal, but they use complicated arguments about convex sets. I would like to know if there exists a simple way to prove directly that the order topology is normal, without going through complete-normality.
general-topology order-theory
asked Dec 1 '18 at 6:47
RlosRlos
1337
$endgroup$
I've been looking extensively for a proof that any toset equipped with the order topology is normal. Every proof I've seen actually shows that the space is in fact completely normal, but they use complicated arguments about convex sets. I would like to know if there exists a simple way to prove directly that the order topology is normal, without going through complete-normality.
general-topology order-theory
general-topology order-theory
asked Dec 1 '18 at 6:47
RlosRlos
1337
asked Dec 1 '18 at 6:47
RlosRlos
1337
asked Dec 1 '18 at 6:47
RlosRlos
1337
asked Dec 1 '18 at 6:47
RlosRlos
1337
asked Dec 1 '18 at 6:47
RlosRlos
1337
1337
$begingroup$
completely normal means that every subspace is normal, so it's not a real complication. In fact you can prove that the fact that an arbitrary ordered topological space is normal needs the axiom of choice so is always non-constructive, e.g. The standard proof I know shows that an ordered space is monotonically normal (stronger than completely normal, and has all sorts of other extra consequences) using an extra well-order on the set (so making the AC necessity obvious). It's a bit of case checking but not very hard.
$endgroup$
– Henno Brandsma
Dec 1 '18 at 6:57
$begingroup$
toset? You mean linearly ordered set?
$endgroup$
– Henno Brandsma
Dec 1 '18 at 7:01
$begingroup$
Yes, I mean linearly ordered. So there is just no constructive proof of this fact? I find that amazing and rather odd, because it's usually not too hard to show that a given space is normal...
$endgroup$
– Rlos
Dec 1 '18 at 7:08
$begingroup$
In fact, normality is a hard thing to show for general spaces. Only metric spaces as a class are easy.
$endgroup$
– Henno Brandsma
Dec 1 '18 at 7:25
add a comment |
$begingroup$
completely normal means that every subspace is normal, so it's not a real complication. In fact you can prove that the fact that an arbitrary ordered topological space is normal needs the axiom of choice so is always non-constructive, e.g. The standard proof I know shows that an ordered space is monotonically normal (stronger than completely normal, and has all sorts of other extra consequences) using an extra well-order on the set (so making the AC necessity obvious). It's a bit of case checking but not very hard.
$endgroup$
– Henno Brandsma
Dec 1 '18 at 6:57
$begingroup$
toset? You mean linearly ordered set?
$endgroup$
– Henno Brandsma
Dec 1 '18 at 7:01
$begingroup$
Yes, I mean linearly ordered. So there is just no constructive proof of this fact? I find that amazing and rather odd, because it's usually not too hard to show that a given space is normal...
$endgroup$
– Rlos
Dec 1 '18 at 7:08
$begingroup$
In fact, normality is a hard thing to show for general spaces. Only metric spaces as a class are easy.
$endgroup$
– Henno Brandsma
Dec 1 '18 at 7:25
$begingroup$
completely normal means that every subspace is normal, so it's not a real complication. In fact you can prove that the fact that an arbitrary ordered topological space is normal needs the axiom of choice so is always non-constructive, e.g. The standard proof I know shows that an ordered space is monotonically normal (stronger than completely normal, and has all sorts of other extra consequences) using an extra well-order on the set (so making the AC necessity obvious). It's a bit of case checking but not very hard.
$endgroup$
– Henno Brandsma
Dec 1 '18 at 6:57
$begingroup$
completely normal means that every subspace is normal, so it's not a real complication. In fact you can prove that the fact that an arbitrary ordered topological space is normal needs the axiom of choice so is always non-constructive, e.g. The standard proof I know shows that an ordered space is monotonically normal (stronger than completely normal, and has all sorts of other extra consequences) using an extra well-order on the set (so making the AC necessity obvious). It's a bit of case checking but not very hard.
$endgroup$
– Henno Brandsma
Dec 1 '18 at 6:57
$begingroup$
toset? You mean linearly ordered set?
$endgroup$
– Henno Brandsma
Dec 1 '18 at 7:01
$begingroup$
toset? You mean linearly ordered set?
$endgroup$
– Henno Brandsma
Dec 1 '18 at 7:01
$begingroup$
Yes, I mean linearly ordered. So there is just no constructive proof of this fact? I find that amazing and rather odd, because it's usually not too hard to show that a given space is normal...
$endgroup$
– Rlos
Dec 1 '18 at 7:08
$begingroup$
Yes, I mean linearly ordered. So there is just no constructive proof of this fact? I find that amazing and rather odd, because it's usually not too hard to show that a given space is normal...
$endgroup$
– Rlos
Dec 1 '18 at 7:08
$begingroup$
In fact, normality is a hard thing to show for general spaces. Only metric spaces as a class are easy.
$endgroup$
– Henno Brandsma
Dec 1 '18 at 7:25
$begingroup$
In fact, normality is a hard thing to show for general spaces. Only metric spaces as a class are easy.
$endgroup$
– Henno Brandsma
Dec 1 '18 at 7:25
add a comment |
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completely normal means that every subspace is normal, so it's not a real complication. In fact you can prove that the fact that an arbitrary ordered topological space is normal needs the axiom of choice so is always non-constructive, e.g. The standard proof I know shows that an ordered space is monotonically normal (stronger than completely normal, and has all sorts of other extra consequences) using an extra well-order on the set (so making the AC necessity obvious). It's a bit of case checking but not very hard.
$endgroup$
– Henno Brandsma
Dec 1 '18 at 6:57
$begingroup$
toset? You mean linearly ordered set?
$endgroup$
– Henno Brandsma
Dec 1 '18 at 7:01
$begingroup$
Yes, I mean linearly ordered. So there is just no constructive proof of this fact? I find that amazing and rather odd, because it's usually not too hard to show that a given space is normal...
$endgroup$
– Rlos
Dec 1 '18 at 7:08
$begingroup$
In fact, normality is a hard thing to show for general spaces. Only metric spaces as a class are easy.
$endgroup$
– Henno Brandsma
Dec 1 '18 at 7:25