Can $mathbb R$ be written as the disjoint union of (uncountably many) closed intervals?
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From this post: http://terrytao.wordpress.com/2010/10/04/covering-a-non-closed-interval-by-disjoint-closed-intervals/
We know that $mathbb R$ can't be written as the countable union of disjoint closed intervals. Can we do it if we allow uncountably many intervals? There doesn't seem to be a nice way to construct such a cover, since if we choose a closed interval, we split the line into two remaining pieces to cover that re still homeomorphic to the original real line. But on the other hand, I feel as if it should be possible, since with the possibility of uncountably many intervals, you should be able to find a way to cover everything.
real-analysis
asked Oct 17 '14 at 21:57
Nishant
5,7342931
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up vote
3
down vote
favorite
From this post: http://terrytao.wordpress.com/2010/10/04/covering-a-non-closed-interval-by-disjoint-closed-intervals/
We know that $mathbb R$ can't be written as the countable union of disjoint closed intervals. Can we do it if we allow uncountably many intervals? There doesn't seem to be a nice way to construct such a cover, since if we choose a closed interval, we split the line into two remaining pieces to cover that re still homeomorphic to the original real line. But on the other hand, I feel as if it should be possible, since with the possibility of uncountably many intervals, you should be able to find a way to cover everything.
real-analysis
asked Oct 17 '14 at 21:57
Nishant
5,7342931
4
It is possible if you allow degenerate intervals $[a,a]$, of course.
– Henning Makholm
Oct 17 '14 at 22:02
It is possible with $1$ closed interval.
– Jonas Meyer
Oct 17 '14 at 22:02
Okay, $[a, a]$ and unbounded intervals don't count.
– Nishant
Oct 17 '14 at 22:03
add a comment |Â
up vote
3
down vote
favorite
up vote
3
down vote
favorite
From this post: http://terrytao.wordpress.com/2010/10/04/covering-a-non-closed-interval-by-disjoint-closed-intervals/
We know that $mathbb R$ can't be written as the countable union of disjoint closed intervals. Can we do it if we allow uncountably many intervals? There doesn't seem to be a nice way to construct such a cover, since if we choose a closed interval, we split the line into two remaining pieces to cover that re still homeomorphic to the original real line. But on the other hand, I feel as if it should be possible, since with the possibility of uncountably many intervals, you should be able to find a way to cover everything.
real-analysis
asked Oct 17 '14 at 21:57
Nishant
5,7342931
From this post: http://terrytao.wordpress.com/2010/10/04/covering-a-non-closed-interval-by-disjoint-closed-intervals/
We know that $mathbb R$ can't be written as the countable union of disjoint closed intervals. Can we do it if we allow uncountably many intervals? There doesn't seem to be a nice way to construct such a cover, since if we choose a closed interval, we split the line into two remaining pieces to cover that re still homeomorphic to the original real line. But on the other hand, I feel as if it should be possible, since with the possibility of uncountably many intervals, you should be able to find a way to cover everything.
real-analysis
asked Oct 17 '14 at 21:57
Nishant
5,7342931
asked Oct 17 '14 at 21:57
Nishant
5,7342931
asked Oct 17 '14 at 21:57
Nishant
5,7342931
asked Oct 17 '14 at 21:57
Nishant
5,7342931
5,7342931
4
It is possible if you allow degenerate intervals $[a,a]$, of course.
– Henning Makholm
Oct 17 '14 at 22:02
It is possible with $1$ closed interval.
– Jonas Meyer
Oct 17 '14 at 22:02
Okay, $[a, a]$ and unbounded intervals don't count.
– Nishant
Oct 17 '14 at 22:03
add a comment |Â
4
It is possible if you allow degenerate intervals $[a,a]$, of course.
– Henning Makholm
Oct 17 '14 at 22:02
It is possible with $1$ closed interval.
– Jonas Meyer
Oct 17 '14 at 22:02
Okay, $[a, a]$ and unbounded intervals don't count.
– Nishant
Oct 17 '14 at 22:03
4
4
It is possible if you allow degenerate intervals $[a,a]$, of course.
– Henning Makholm
Oct 17 '14 at 22:02
It is possible if you allow degenerate intervals $[a,a]$, of course.
– Henning Makholm
Oct 17 '14 at 22:02
It is possible with $1$ closed interval.
– Jonas Meyer
Oct 17 '14 at 22:02
It is possible with $1$ closed interval.
– Jonas Meyer
Oct 17 '14 at 22:02
Okay, $[a, a]$ and unbounded intervals don't count.
– Nishant
Oct 17 '14 at 22:03
Okay, $[a, a]$ and unbounded intervals don't count.
– Nishant
Oct 17 '14 at 22:03
add a comment |Â
1 Answer
1
active
oldest
votes
up vote
11
down vote
accepted
No.
Since $mathbbQ$ is dense in $mathbbR$, within each interval $[a_i, b_i]$, we can select a single $q_i in mathbbQ$ with which
to "label" the interval. The set of $q_i$'s forms a subset of $mathbbQ$, which is countable.
answered Oct 17 '14 at 22:01
Kaj Hansen
27k43679
Short and sweet. Thanks!
– Nishant
Oct 17 '14 at 22:04
Glad I could help!
– Kaj Hansen
Oct 17 '14 at 22:04
1
This shows more: it is not possible to have any uncountable family of pairwise disjoint closed subintervals in $mathbbR$. The same holds with closed balls in any separable metric space. :)
– Pete L. Clark
Oct 19 '14 at 5:22
1
@PeteL.Clark, indeed it does! Basically the same problem came up for $mathbbR^n$ in class on Friday. :)
– Kaj Hansen
Oct 19 '14 at 5:25
add a comment |Â
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
11
down vote
accepted
No.
Since $mathbbQ$ is dense in $mathbbR$, within each interval $[a_i, b_i]$, we can select a single $q_i in mathbbQ$ with which
to "label" the interval. The set of $q_i$'s forms a subset of $mathbbQ$, which is countable.
answered Oct 17 '14 at 22:01
Kaj Hansen
27k43679
Short and sweet. Thanks!
– Nishant
Oct 17 '14 at 22:04
Glad I could help!
– Kaj Hansen
Oct 17 '14 at 22:04
1
This shows more: it is not possible to have any uncountable family of pairwise disjoint closed subintervals in $mathbbR$. The same holds with closed balls in any separable metric space. :)
– Pete L. Clark
Oct 19 '14 at 5:22
1
@PeteL.Clark, indeed it does! Basically the same problem came up for $mathbbR^n$ in class on Friday. :)
– Kaj Hansen
Oct 19 '14 at 5:25
add a comment |Â
up vote
11
down vote
accepted
No.
Since $mathbbQ$ is dense in $mathbbR$, within each interval $[a_i, b_i]$, we can select a single $q_i in mathbbQ$ with which
to "label" the interval. The set of $q_i$'s forms a subset of $mathbbQ$, which is countable.
answered Oct 17 '14 at 22:01
Kaj Hansen
27k43679
Short and sweet. Thanks!
– Nishant
Oct 17 '14 at 22:04
Glad I could help!
– Kaj Hansen
Oct 17 '14 at 22:04
1
This shows more: it is not possible to have any uncountable family of pairwise disjoint closed subintervals in $mathbbR$. The same holds with closed balls in any separable metric space. :)
– Pete L. Clark
Oct 19 '14 at 5:22
1
@PeteL.Clark, indeed it does! Basically the same problem came up for $mathbbR^n$ in class on Friday. :)
– Kaj Hansen
Oct 19 '14 at 5:25
add a comment |Â
up vote
11
down vote
accepted
up vote
11
down vote
accepted
No.
Since $mathbbQ$ is dense in $mathbbR$, within each interval $[a_i, b_i]$, we can select a single $q_i in mathbbQ$ with which
to "label" the interval. The set of $q_i$'s forms a subset of $mathbbQ$, which is countable.
answered Oct 17 '14 at 22:01
Kaj Hansen
27k43679
No.
Since $mathbbQ$ is dense in $mathbbR$, within each interval $[a_i, b_i]$, we can select a single $q_i in mathbbQ$ with which
to "label" the interval. The set of $q_i$'s forms a subset of $mathbbQ$, which is countable.
answered Oct 17 '14 at 22:01
Kaj Hansen
27k43679
edited Aug 12 at 21:24
answered Oct 17 '14 at 22:01
Kaj Hansen
27k43679
answered Oct 17 '14 at 22:01
Kaj Hansen
27k43679
answered Oct 17 '14 at 22:01
Kaj Hansen
27k43679
27k43679
Short and sweet. Thanks!
– Nishant
Oct 17 '14 at 22:04
Glad I could help!
– Kaj Hansen
Oct 17 '14 at 22:04
1
This shows more: it is not possible to have any uncountable family of pairwise disjoint closed subintervals in $mathbbR$. The same holds with closed balls in any separable metric space. :)
– Pete L. Clark
Oct 19 '14 at 5:22
1
@PeteL.Clark, indeed it does! Basically the same problem came up for $mathbbR^n$ in class on Friday. :)
– Kaj Hansen
Oct 19 '14 at 5:25
add a comment |Â
Short and sweet. Thanks!
– Nishant
Oct 17 '14 at 22:04
Glad I could help!
– Kaj Hansen
Oct 17 '14 at 22:04
1
This shows more: it is not possible to have any uncountable family of pairwise disjoint closed subintervals in $mathbbR$. The same holds with closed balls in any separable metric space. :)
– Pete L. Clark
Oct 19 '14 at 5:22
1
@PeteL.Clark, indeed it does! Basically the same problem came up for $mathbbR^n$ in class on Friday. :)
– Kaj Hansen
Oct 19 '14 at 5:25
Short and sweet. Thanks!
– Nishant
Oct 17 '14 at 22:04
Short and sweet. Thanks!
– Nishant
Oct 17 '14 at 22:04
Glad I could help!
– Kaj Hansen
Oct 17 '14 at 22:04
Glad I could help!
– Kaj Hansen
Oct 17 '14 at 22:04
1
1
This shows more: it is not possible to have any uncountable family of pairwise disjoint closed subintervals in $mathbbR$. The same holds with closed balls in any separable metric space. :)
– Pete L. Clark
Oct 19 '14 at 5:22
This shows more: it is not possible to have any uncountable family of pairwise disjoint closed subintervals in $mathbbR$. The same holds with closed balls in any separable metric space. :)
– Pete L. Clark
Oct 19 '14 at 5:22
1
1
@PeteL.Clark, indeed it does! Basically the same problem came up for $mathbbR^n$ in class on Friday. :)
– Kaj Hansen
Oct 19 '14 at 5:25
@PeteL.Clark, indeed it does! Basically the same problem came up for $mathbbR^n$ in class on Friday. :)
– Kaj Hansen
Oct 19 '14 at 5:25
add a comment |Â
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4
It is possible if you allow degenerate intervals $[a,a]$, of course.
– Henning Makholm
Oct 17 '14 at 22:02
It is possible with $1$ closed interval.
– Jonas Meyer
Oct 17 '14 at 22:02
Okay, $[a, a]$ and unbounded intervals don't count.
– Nishant
Oct 17 '14 at 22:03