Could every Hausdorff space be induced by a total order relation [closed]
Let $(H,mathcal T)$ is a Hausdorff space then is there any total order relation on $H$ such that the topological space induced by the order relation be the same $(H,mathcal T)$?
Thanks a lots beforehand.
general-topology order-theory big-list well-orders
asked Nov 27 at 13:30
Rolling Stones
1
closed as off-topic by Brahadeesh, Saad, amWhy, Cesareo, Rebellos Nov 28 at 9:26
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – Brahadeesh, Saad, amWhy, Cesareo, Rebellos
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Let $(H,mathcal T)$ is a Hausdorff space then is there any total order relation on $H$ such that the topological space induced by the order relation be the same $(H,mathcal T)$?
Thanks a lots beforehand.
general-topology order-theory big-list well-orders
asked Nov 27 at 13:30
Rolling Stones
1
closed as off-topic by Brahadeesh, Saad, amWhy, Cesareo, Rebellos Nov 28 at 9:26
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – Brahadeesh, Saad, amWhy, Cesareo, Rebellos
If this question can be reworded to fit the rules in the help center, please edit the question.
What is your guess? Do you have any particular simple space for which you don't see any inducing total order?
– user87690
Nov 27 at 13:41
4
What happens if you remove a point from a connected space whose topology is induced by a total order? Can you think of a connected space where that doesn't happen?
– David Hartley
Nov 27 at 14:00
@user87690 no, I have no such space.
– Rolling Stones
Nov 27 at 14:26
1
@RollingStones David's example was a hint towards a counterexample - removing a point from an orderable space always disconnects it.
– Wojowu
Nov 27 at 18:55
1
@user87690 Woops, you are right. That's only two out of possibly infinitely many elements though.
– Wojowu
Nov 27 at 19:41
|
show 4 more comments
Let $(H,mathcal T)$ is a Hausdorff space then is there any total order relation on $H$ such that the topological space induced by the order relation be the same $(H,mathcal T)$?
Thanks a lots beforehand.
general-topology order-theory big-list well-orders
asked Nov 27 at 13:30
Rolling Stones
1
Let $(H,mathcal T)$ is a Hausdorff space then is there any total order relation on $H$ such that the topological space induced by the order relation be the same $(H,mathcal T)$?
Thanks a lots beforehand.
general-topology order-theory big-list well-orders
general-topology order-theory big-list well-orders
asked Nov 27 at 13:30
Rolling Stones
1
asked Nov 27 at 13:30
Rolling Stones
1
asked Nov 27 at 13:30
Rolling Stones
1
asked Nov 27 at 13:30
Rolling Stones
1
asked Nov 27 at 13:30
Rolling Stones
1
1
closed as off-topic by Brahadeesh, Saad, amWhy, Cesareo, Rebellos Nov 28 at 9:26
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – Brahadeesh, Saad, amWhy, Cesareo, Rebellos
If this question can be reworded to fit the rules in the help center, please edit the question.
closed as off-topic by Brahadeesh, Saad, amWhy, Cesareo, Rebellos Nov 28 at 9:26
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – Brahadeesh, Saad, amWhy, Cesareo, Rebellos
If this question can be reworded to fit the rules in the help center, please edit the question.
What is your guess? Do you have any particular simple space for which you don't see any inducing total order?
– user87690
Nov 27 at 13:41
4
What happens if you remove a point from a connected space whose topology is induced by a total order? Can you think of a connected space where that doesn't happen?
– David Hartley
Nov 27 at 14:00
@user87690 no, I have no such space.
– Rolling Stones
Nov 27 at 14:26
1
@RollingStones David's example was a hint towards a counterexample - removing a point from an orderable space always disconnects it.
– Wojowu
Nov 27 at 18:55
1
@user87690 Woops, you are right. That's only two out of possibly infinitely many elements though.
– Wojowu
Nov 27 at 19:41
|
show 4 more comments
What is your guess? Do you have any particular simple space for which you don't see any inducing total order?
– user87690
Nov 27 at 13:41
4
What happens if you remove a point from a connected space whose topology is induced by a total order? Can you think of a connected space where that doesn't happen?
– David Hartley
Nov 27 at 14:00
@user87690 no, I have no such space.
– Rolling Stones
Nov 27 at 14:26
1
@RollingStones David's example was a hint towards a counterexample - removing a point from an orderable space always disconnects it.
– Wojowu
Nov 27 at 18:55
1
@user87690 Woops, you are right. That's only two out of possibly infinitely many elements though.
– Wojowu
Nov 27 at 19:41
What is your guess? Do you have any particular simple space for which you don't see any inducing total order?
– user87690
Nov 27 at 13:41
What is your guess? Do you have any particular simple space for which you don't see any inducing total order?
– user87690
Nov 27 at 13:41
4
4
What happens if you remove a point from a connected space whose topology is induced by a total order? Can you think of a connected space where that doesn't happen?
– David Hartley
Nov 27 at 14:00
What happens if you remove a point from a connected space whose topology is induced by a total order? Can you think of a connected space where that doesn't happen?
– David Hartley
Nov 27 at 14:00
@user87690 no, I have no such space.
– Rolling Stones
Nov 27 at 14:26
@user87690 no, I have no such space.
– Rolling Stones
Nov 27 at 14:26
1
1
@RollingStones David's example was a hint towards a counterexample - removing a point from an orderable space always disconnects it.
– Wojowu
Nov 27 at 18:55
@RollingStones David's example was a hint towards a counterexample - removing a point from an orderable space always disconnects it.
– Wojowu
Nov 27 at 18:55
1
1
@user87690 Woops, you are right. That's only two out of possibly infinitely many elements though.
– Wojowu
Nov 27 at 19:41
@user87690 Woops, you are right. That's only two out of possibly infinitely many elements though.
– Wojowu
Nov 27 at 19:41
|
show 4 more comments
2 Answers
2
active
oldest
votes
A total order relation induces a normal topology, which means that if there is a Hausdorff space which is not normal as well, then it is not induced by an order relation.
Though I am also interested to see what counter-examples people will give you.
answered Nov 27 at 13:49
Keen-ameteur
1,257316
Thank you so much.
– Rolling Stones
Nov 27 at 14:20
add a comment |
One example is the Sorgenfrey line (i.e., $mathbb{R}$ with the lower-limit topology). See the following quetion and its answer for an outline showing why it is not orderable:
- Sorgenfrey line is not orderable
An interesting thing about this space is that it is "close" to being orderable. A topological space $X$ is called suborderable if it is homeomorphic to a subspace of an ordered space. We can show rather quickly that the Sorgenfrey line is suborderable.
Consider $mathbb{R} times { 0 , 1 }$ with the lexicographic order: $$(x,i) prec ( y , j ) Leftrightarrow begin{cases}
x < y, &text{or} \
x = y, i=0, j=1.
end{cases}$$
Then the Sorgenfrey line is homeomorphic to the subspace $mathbb{R} times { 1 }$ of the above ordered space.
answered Nov 27 at 14:26
1-3-7-Trimethylxanthine
4,597927
Thank you so much.
– Rolling Stones
Nov 27 at 14:29
add a comment |
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
A total order relation induces a normal topology, which means that if there is a Hausdorff space which is not normal as well, then it is not induced by an order relation.
Though I am also interested to see what counter-examples people will give you.
answered Nov 27 at 13:49
Keen-ameteur
1,257316
Thank you so much.
– Rolling Stones
Nov 27 at 14:20
add a comment |
A total order relation induces a normal topology, which means that if there is a Hausdorff space which is not normal as well, then it is not induced by an order relation.
Though I am also interested to see what counter-examples people will give you.
answered Nov 27 at 13:49
Keen-ameteur
1,257316
Thank you so much.
– Rolling Stones
Nov 27 at 14:20
add a comment |
A total order relation induces a normal topology, which means that if there is a Hausdorff space which is not normal as well, then it is not induced by an order relation.
Though I am also interested to see what counter-examples people will give you.
answered Nov 27 at 13:49
Keen-ameteur
1,257316
A total order relation induces a normal topology, which means that if there is a Hausdorff space which is not normal as well, then it is not induced by an order relation.
Though I am also interested to see what counter-examples people will give you.
answered Nov 27 at 13:49
Keen-ameteur
1,257316
answered Nov 27 at 13:49
Keen-ameteur
1,257316
answered Nov 27 at 13:49
Keen-ameteur
1,257316
answered Nov 27 at 13:49
Keen-ameteur
1,257316
1,257316
Thank you so much.
– Rolling Stones
Nov 27 at 14:20
add a comment |
Thank you so much.
– Rolling Stones
Nov 27 at 14:20
Thank you so much.
– Rolling Stones
Nov 27 at 14:20
Thank you so much.
– Rolling Stones
Nov 27 at 14:20
add a comment |
One example is the Sorgenfrey line (i.e., $mathbb{R}$ with the lower-limit topology). See the following quetion and its answer for an outline showing why it is not orderable:
- Sorgenfrey line is not orderable
An interesting thing about this space is that it is "close" to being orderable. A topological space $X$ is called suborderable if it is homeomorphic to a subspace of an ordered space. We can show rather quickly that the Sorgenfrey line is suborderable.
Consider $mathbb{R} times { 0 , 1 }$ with the lexicographic order: $$(x,i) prec ( y , j ) Leftrightarrow begin{cases}
x < y, &text{or} \
x = y, i=0, j=1.
end{cases}$$
Then the Sorgenfrey line is homeomorphic to the subspace $mathbb{R} times { 1 }$ of the above ordered space.
answered Nov 27 at 14:26
1-3-7-Trimethylxanthine
4,597927
Thank you so much.
– Rolling Stones
Nov 27 at 14:29
add a comment |
One example is the Sorgenfrey line (i.e., $mathbb{R}$ with the lower-limit topology). See the following quetion and its answer for an outline showing why it is not orderable:
- Sorgenfrey line is not orderable
An interesting thing about this space is that it is "close" to being orderable. A topological space $X$ is called suborderable if it is homeomorphic to a subspace of an ordered space. We can show rather quickly that the Sorgenfrey line is suborderable.
Consider $mathbb{R} times { 0 , 1 }$ with the lexicographic order: $$(x,i) prec ( y , j ) Leftrightarrow begin{cases}
x < y, &text{or} \
x = y, i=0, j=1.
end{cases}$$
Then the Sorgenfrey line is homeomorphic to the subspace $mathbb{R} times { 1 }$ of the above ordered space.
answered Nov 27 at 14:26
1-3-7-Trimethylxanthine
4,597927
Thank you so much.
– Rolling Stones
Nov 27 at 14:29
add a comment |
One example is the Sorgenfrey line (i.e., $mathbb{R}$ with the lower-limit topology). See the following quetion and its answer for an outline showing why it is not orderable:
- Sorgenfrey line is not orderable
An interesting thing about this space is that it is "close" to being orderable. A topological space $X$ is called suborderable if it is homeomorphic to a subspace of an ordered space. We can show rather quickly that the Sorgenfrey line is suborderable.
Consider $mathbb{R} times { 0 , 1 }$ with the lexicographic order: $$(x,i) prec ( y , j ) Leftrightarrow begin{cases}
x < y, &text{or} \
x = y, i=0, j=1.
end{cases}$$
Then the Sorgenfrey line is homeomorphic to the subspace $mathbb{R} times { 1 }$ of the above ordered space.
answered Nov 27 at 14:26
1-3-7-Trimethylxanthine
4,597927
One example is the Sorgenfrey line (i.e., $mathbb{R}$ with the lower-limit topology). See the following quetion and its answer for an outline showing why it is not orderable:
- Sorgenfrey line is not orderable
An interesting thing about this space is that it is "close" to being orderable. A topological space $X$ is called suborderable if it is homeomorphic to a subspace of an ordered space. We can show rather quickly that the Sorgenfrey line is suborderable.
Consider $mathbb{R} times { 0 , 1 }$ with the lexicographic order: $$(x,i) prec ( y , j ) Leftrightarrow begin{cases}
x < y, &text{or} \
x = y, i=0, j=1.
end{cases}$$
Then the Sorgenfrey line is homeomorphic to the subspace $mathbb{R} times { 1 }$ of the above ordered space.
answered Nov 27 at 14:26
1-3-7-Trimethylxanthine
4,597927
answered Nov 27 at 14:26
1-3-7-Trimethylxanthine
4,597927
answered Nov 27 at 14:26
1-3-7-Trimethylxanthine
4,597927
answered Nov 27 at 14:26
1-3-7-Trimethylxanthine
4,597927
4,597927
Thank you so much.
– Rolling Stones
Nov 27 at 14:29
add a comment |
Thank you so much.
– Rolling Stones
Nov 27 at 14:29
Thank you so much.
– Rolling Stones
Nov 27 at 14:29
Thank you so much.
– Rolling Stones
Nov 27 at 14:29
add a comment |
What is your guess? Do you have any particular simple space for which you don't see any inducing total order?
– user87690
Nov 27 at 13:41
4
What happens if you remove a point from a connected space whose topology is induced by a total order? Can you think of a connected space where that doesn't happen?
– David Hartley
Nov 27 at 14:00
@user87690 no, I have no such space.
– Rolling Stones
Nov 27 at 14:26
1
@RollingStones David's example was a hint towards a counterexample - removing a point from an orderable space always disconnects it.
– Wojowu
Nov 27 at 18:55
1
@user87690 Woops, you are right. That's only two out of possibly infinitely many elements though.
– Wojowu
Nov 27 at 19:41