Mixing
A sequence of $rs + 1$ real numbers has an increasing subsequence of length $r + 1$ or a decreasing...
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Problem: Prove the following: a sequence of $rs + 1$ real numbers has an increasing subsequence of length $r + 1$ or a decreasing subsequence of length $s + 1$.
Solution: Define a partial ordering on the sequence $a _ { 1 } , ldots , a _ { r s + 1 }$ by $a _ { i } preceq a _ { j }$ iff, $a _ { i } leq a _ { j }$ and $i leq j$ A chain is an increasing subsequence, an antichain is a decreasing subsequence. Now I would like to apply Dilworth theorem. Suppose that the maximum size of a chain is $r+1$, the poset can be partitioned into $r+1$ antichain. However from there I don't now how to continue, what would be size of those antichains?
sequences-and-series combinatorics discrete-mathematics elementary-set-theory
asked Jan 19 at 8:33
NotAbelianGroupNotAbelianGroup
18211
$endgroup$
add a comment |
$begingroup$
Problem: Prove the following: a sequence of $rs + 1$ real numbers has an increasing subsequence of length $r + 1$ or a decreasing subsequence of length $s + 1$.
Solution: Define a partial ordering on the sequence $a _ { 1 } , ldots , a _ { r s + 1 }$ by $a _ { i } preceq a _ { j }$ iff, $a _ { i } leq a _ { j }$ and $i leq j$ A chain is an increasing subsequence, an antichain is a decreasing subsequence. Now I would like to apply Dilworth theorem. Suppose that the maximum size of a chain is $r+1$, the poset can be partitioned into $r+1$ antichain. However from there I don't now how to continue, what would be size of those antichains?
sequences-and-series combinatorics discrete-mathematics elementary-set-theory
asked Jan 19 at 8:33
NotAbelianGroupNotAbelianGroup
18211
$endgroup$
$begingroup$
The proof of Dilworth's theorem is short and easy but a little bit tricky. With chains and antichains interchanged, it's completely obvious: if the maximum size of a chain is $r$, then the poset can be partitioned into $r$ antichains. This poor relation of Dilworth's theorem is just as good for proving that theorem about monotone subsequences.
$endgroup$
– bof
Jan 19 at 10:08
add a comment |
$begingroup$
Problem: Prove the following: a sequence of $rs + 1$ real numbers has an increasing subsequence of length $r + 1$ or a decreasing subsequence of length $s + 1$.
Solution: Define a partial ordering on the sequence $a _ { 1 } , ldots , a _ { r s + 1 }$ by $a _ { i } preceq a _ { j }$ iff, $a _ { i } leq a _ { j }$ and $i leq j$ A chain is an increasing subsequence, an antichain is a decreasing subsequence. Now I would like to apply Dilworth theorem. Suppose that the maximum size of a chain is $r+1$, the poset can be partitioned into $r+1$ antichain. However from there I don't now how to continue, what would be size of those antichains?
sequences-and-series combinatorics discrete-mathematics elementary-set-theory
asked Jan 19 at 8:33
NotAbelianGroupNotAbelianGroup
18211
$endgroup$
Problem: Prove the following: a sequence of $rs + 1$ real numbers has an increasing subsequence of length $r + 1$ or a decreasing subsequence of length $s + 1$.
Solution: Define a partial ordering on the sequence $a _ { 1 } , ldots , a _ { r s + 1 }$ by $a _ { i } preceq a _ { j }$ iff, $a _ { i } leq a _ { j }$ and $i leq j$ A chain is an increasing subsequence, an antichain is a decreasing subsequence. Now I would like to apply Dilworth theorem. Suppose that the maximum size of a chain is $r+1$, the poset can be partitioned into $r+1$ antichain. However from there I don't now how to continue, what would be size of those antichains?
sequences-and-series combinatorics discrete-mathematics elementary-set-theory
sequences-and-series combinatorics discrete-mathematics elementary-set-theory
asked Jan 19 at 8:33
NotAbelianGroupNotAbelianGroup
18211
asked Jan 19 at 8:33
NotAbelianGroupNotAbelianGroup
18211
asked Jan 19 at 8:33
NotAbelianGroupNotAbelianGroup
18211
asked Jan 19 at 8:33
NotAbelianGroupNotAbelianGroup
18211
asked Jan 19 at 8:33
NotAbelianGroupNotAbelianGroup
18211
18211
$begingroup$
The proof of Dilworth's theorem is short and easy but a little bit tricky. With chains and antichains interchanged, it's completely obvious: if the maximum size of a chain is $r$, then the poset can be partitioned into $r$ antichains. This poor relation of Dilworth's theorem is just as good for proving that theorem about monotone subsequences.
$endgroup$
– bof
Jan 19 at 10:08
add a comment |
$begingroup$
The proof of Dilworth's theorem is short and easy but a little bit tricky. With chains and antichains interchanged, it's completely obvious: if the maximum size of a chain is $r$, then the poset can be partitioned into $r$ antichains. This poor relation of Dilworth's theorem is just as good for proving that theorem about monotone subsequences.
$endgroup$
– bof
Jan 19 at 10:08
$begingroup$
The proof of Dilworth's theorem is short and easy but a little bit tricky. With chains and antichains interchanged, it's completely obvious: if the maximum size of a chain is $r$, then the poset can be partitioned into $r$ antichains. This poor relation of Dilworth's theorem is just as good for proving that theorem about monotone subsequences.
$endgroup$
– bof
Jan 19 at 10:08
$begingroup$
The proof of Dilworth's theorem is short and easy but a little bit tricky. With chains and antichains interchanged, it's completely obvious: if the maximum size of a chain is $r$, then the poset can be partitioned into $r$ antichains. This poor relation of Dilworth's theorem is just as good for proving that theorem about monotone subsequences.
$endgroup$
– bof
Jan 19 at 10:08
add a comment |
1 Answer
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$begingroup$
Assume that the maximum size of an antichain is $le s$. Then the poset can be partitioned into at most $s$ chains. By pigeonhole principle, there is at least one chain of size $ge lceil frac{rs+1}{s}rceil= r+1. $ Thus either there is a chain of size $ge r+1$ or an antichain of size $ge s+1$.
answered Jan 19 at 8:55
SongSong
16.2k1739
$endgroup$
$begingroup$
Thank you I was stuck on finding the size of those chains.
$endgroup$
– NotAbelianGroup
Jan 19 at 9:02
add a comment |
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1 Answer
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$begingroup$
Assume that the maximum size of an antichain is $le s$. Then the poset can be partitioned into at most $s$ chains. By pigeonhole principle, there is at least one chain of size $ge lceil frac{rs+1}{s}rceil= r+1. $ Thus either there is a chain of size $ge r+1$ or an antichain of size $ge s+1$.
answered Jan 19 at 8:55
SongSong
16.2k1739
$endgroup$
$begingroup$
Thank you I was stuck on finding the size of those chains.
$endgroup$
– NotAbelianGroup
Jan 19 at 9:02
add a comment |
$begingroup$
Assume that the maximum size of an antichain is $le s$. Then the poset can be partitioned into at most $s$ chains. By pigeonhole principle, there is at least one chain of size $ge lceil frac{rs+1}{s}rceil= r+1. $ Thus either there is a chain of size $ge r+1$ or an antichain of size $ge s+1$.
answered Jan 19 at 8:55
SongSong
16.2k1739
$endgroup$
$begingroup$
Thank you I was stuck on finding the size of those chains.
$endgroup$
– NotAbelianGroup
Jan 19 at 9:02
add a comment |
$begingroup$
Assume that the maximum size of an antichain is $le s$. Then the poset can be partitioned into at most $s$ chains. By pigeonhole principle, there is at least one chain of size $ge lceil frac{rs+1}{s}rceil= r+1. $ Thus either there is a chain of size $ge r+1$ or an antichain of size $ge s+1$.
answered Jan 19 at 8:55
SongSong
16.2k1739
$endgroup$
Assume that the maximum size of an antichain is $le s$. Then the poset can be partitioned into at most $s$ chains. By pigeonhole principle, there is at least one chain of size $ge lceil frac{rs+1}{s}rceil= r+1. $ Thus either there is a chain of size $ge r+1$ or an antichain of size $ge s+1$.
answered Jan 19 at 8:55
SongSong
16.2k1739
answered Jan 19 at 8:55
SongSong
16.2k1739
answered Jan 19 at 8:55
SongSong
16.2k1739
answered Jan 19 at 8:55
SongSong
16.2k1739
16.2k1739
$begingroup$
Thank you I was stuck on finding the size of those chains.
$endgroup$
– NotAbelianGroup
Jan 19 at 9:02
add a comment |
$begingroup$
Thank you I was stuck on finding the size of those chains.
$endgroup$
– NotAbelianGroup
Jan 19 at 9:02
$begingroup$
Thank you I was stuck on finding the size of those chains.
$endgroup$
– NotAbelianGroup
Jan 19 at 9:02
$begingroup$
Thank you I was stuck on finding the size of those chains.
$endgroup$
– NotAbelianGroup
Jan 19 at 9:02
add a comment |
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$begingroup$
The proof of Dilworth's theorem is short and easy but a little bit tricky. With chains and antichains interchanged, it's completely obvious: if the maximum size of a chain is $r$, then the poset can be partitioned into $r$ antichains. This poor relation of Dilworth's theorem is just as good for proving that theorem about monotone subsequences.
$endgroup$
– bof
Jan 19 at 10:08