How to prove second part of this combinatorial lemma?
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Lemma is stated like this:
Lemma 1. Suppose that a finite number of points subdivides a closed interval
into smaller intervals. The left endpoint of the original interval is labeled by
0, the right one by 1, and each of the partitioning points inside the interval
is also labeled by either 0 or 1. Then there is an interval of the subdivision
whose endpoints are labeled by different numbers. Moreover, the number of
such intervals is odd.
I understand existence part, that is, that there is an interval of the subdivision
whose endpoints are labeled by different numbers. But I do not understand why the number of such intervals must be odd?
How would you prove that?
combinatorics
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up vote
1
down vote
favorite
Lemma is stated like this:
Lemma 1. Suppose that a finite number of points subdivides a closed interval
into smaller intervals. The left endpoint of the original interval is labeled by
0, the right one by 1, and each of the partitioning points inside the interval
is also labeled by either 0 or 1. Then there is an interval of the subdivision
whose endpoints are labeled by different numbers. Moreover, the number of
such intervals is odd.
I understand existence part, that is, that there is an interval of the subdivision
whose endpoints are labeled by different numbers. But I do not understand why the number of such intervals must be odd?
How would you prove that?
combinatorics
add a comment |Â
up vote
1
down vote
favorite
up vote
1
down vote
favorite
Lemma is stated like this:
Lemma 1. Suppose that a finite number of points subdivides a closed interval
into smaller intervals. The left endpoint of the original interval is labeled by
0, the right one by 1, and each of the partitioning points inside the interval
is also labeled by either 0 or 1. Then there is an interval of the subdivision
whose endpoints are labeled by different numbers. Moreover, the number of
such intervals is odd.
I understand existence part, that is, that there is an interval of the subdivision
whose endpoints are labeled by different numbers. But I do not understand why the number of such intervals must be odd?
How would you prove that?
combinatorics
Lemma is stated like this:
Lemma 1. Suppose that a finite number of points subdivides a closed interval
into smaller intervals. The left endpoint of the original interval is labeled by
0, the right one by 1, and each of the partitioning points inside the interval
is also labeled by either 0 or 1. Then there is an interval of the subdivision
whose endpoints are labeled by different numbers. Moreover, the number of
such intervals is odd.
I understand existence part, that is, that there is an interval of the subdivision
whose endpoints are labeled by different numbers. But I do not understand why the number of such intervals must be odd?
How would you prove that?
combinatorics
asked Aug 24 at 9:32
Right
1,026213
1,026213
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2 Answers
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up vote
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If you go from left to right and record the number of times 0 changed to 1 or vice versa, then this number is odd (because you start with 0 and end with 1).
But that number is precisely the number of intervals with different labels.
By the way this is the toy version of Sperner's lemma.
1
Perhaps insert or vice versa in the first clause for clarity?
â Peter Taylor
Aug 24 at 10:48
@PeterTaylor Good point. Thanks!
â Michal Adamaszek
Aug 24 at 19:34
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up vote
0
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You can use induction to prove that.
Extend the statement by adding that the number of such intervals is even if the points of the original interval are both labeled by $0$ or are both labeled by $1$.
This enforces your induction hypothese and makes things more easy to prove.
The statement is obviously true if $n=0$ points are added.
If $n>0$ points are added then pick out one of these (inner) points.
It splits up the original interval in two intervals with $<n$ points.
Treating them separatedly for both of them the induction hypothese informs you completely about the question whether the number of the special intervals are odd or even.
That leads to the expected conclusions for $n$.
add a comment |Â
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
1
down vote
If you go from left to right and record the number of times 0 changed to 1 or vice versa, then this number is odd (because you start with 0 and end with 1).
But that number is precisely the number of intervals with different labels.
By the way this is the toy version of Sperner's lemma.
1
Perhaps insert or vice versa in the first clause for clarity?
â Peter Taylor
Aug 24 at 10:48
@PeterTaylor Good point. Thanks!
â Michal Adamaszek
Aug 24 at 19:34
add a comment |Â
up vote
1
down vote
If you go from left to right and record the number of times 0 changed to 1 or vice versa, then this number is odd (because you start with 0 and end with 1).
But that number is precisely the number of intervals with different labels.
By the way this is the toy version of Sperner's lemma.
1
Perhaps insert or vice versa in the first clause for clarity?
â Peter Taylor
Aug 24 at 10:48
@PeterTaylor Good point. Thanks!
â Michal Adamaszek
Aug 24 at 19:34
add a comment |Â
up vote
1
down vote
up vote
1
down vote
If you go from left to right and record the number of times 0 changed to 1 or vice versa, then this number is odd (because you start with 0 and end with 1).
But that number is precisely the number of intervals with different labels.
By the way this is the toy version of Sperner's lemma.
If you go from left to right and record the number of times 0 changed to 1 or vice versa, then this number is odd (because you start with 0 and end with 1).
But that number is precisely the number of intervals with different labels.
By the way this is the toy version of Sperner's lemma.
edited Aug 24 at 19:33
answered Aug 24 at 9:54
Michal Adamaszek
1,69948
1,69948
1
Perhaps insert or vice versa in the first clause for clarity?
â Peter Taylor
Aug 24 at 10:48
@PeterTaylor Good point. Thanks!
â Michal Adamaszek
Aug 24 at 19:34
add a comment |Â
1
Perhaps insert or vice versa in the first clause for clarity?
â Peter Taylor
Aug 24 at 10:48
@PeterTaylor Good point. Thanks!
â Michal Adamaszek
Aug 24 at 19:34
1
1
Perhaps insert or vice versa in the first clause for clarity?
â Peter Taylor
Aug 24 at 10:48
Perhaps insert or vice versa in the first clause for clarity?
â Peter Taylor
Aug 24 at 10:48
@PeterTaylor Good point. Thanks!
â Michal Adamaszek
Aug 24 at 19:34
@PeterTaylor Good point. Thanks!
â Michal Adamaszek
Aug 24 at 19:34
add a comment |Â
up vote
0
down vote
You can use induction to prove that.
Extend the statement by adding that the number of such intervals is even if the points of the original interval are both labeled by $0$ or are both labeled by $1$.
This enforces your induction hypothese and makes things more easy to prove.
The statement is obviously true if $n=0$ points are added.
If $n>0$ points are added then pick out one of these (inner) points.
It splits up the original interval in two intervals with $<n$ points.
Treating them separatedly for both of them the induction hypothese informs you completely about the question whether the number of the special intervals are odd or even.
That leads to the expected conclusions for $n$.
add a comment |Â
up vote
0
down vote
You can use induction to prove that.
Extend the statement by adding that the number of such intervals is even if the points of the original interval are both labeled by $0$ or are both labeled by $1$.
This enforces your induction hypothese and makes things more easy to prove.
The statement is obviously true if $n=0$ points are added.
If $n>0$ points are added then pick out one of these (inner) points.
It splits up the original interval in two intervals with $<n$ points.
Treating them separatedly for both of them the induction hypothese informs you completely about the question whether the number of the special intervals are odd or even.
That leads to the expected conclusions for $n$.
add a comment |Â
up vote
0
down vote
up vote
0
down vote
You can use induction to prove that.
Extend the statement by adding that the number of such intervals is even if the points of the original interval are both labeled by $0$ or are both labeled by $1$.
This enforces your induction hypothese and makes things more easy to prove.
The statement is obviously true if $n=0$ points are added.
If $n>0$ points are added then pick out one of these (inner) points.
It splits up the original interval in two intervals with $<n$ points.
Treating them separatedly for both of them the induction hypothese informs you completely about the question whether the number of the special intervals are odd or even.
That leads to the expected conclusions for $n$.
You can use induction to prove that.
Extend the statement by adding that the number of such intervals is even if the points of the original interval are both labeled by $0$ or are both labeled by $1$.
This enforces your induction hypothese and makes things more easy to prove.
The statement is obviously true if $n=0$ points are added.
If $n>0$ points are added then pick out one of these (inner) points.
It splits up the original interval in two intervals with $<n$ points.
Treating them separatedly for both of them the induction hypothese informs you completely about the question whether the number of the special intervals are odd or even.
That leads to the expected conclusions for $n$.
edited Aug 24 at 13:06
answered Aug 24 at 10:01
drhab
88.3k541120
88.3k541120
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