Let p be a permutation of N. We say that p is bounded if there exists k so that |p(i)−i|≤k for all i. If p is bounded, must there exist M>0 such that p({1,2,…,M})={1,2,…,M}? If so, can we bound M by a function of k?

misyjny76 2022-09-20 Answered
Let p be a permutation of N . We say that p is bounded if there exists k so that | p ( i ) i | k for all i. If p is bounded, must there exist M > 0 such that p ( { 1 , 2 , , M } ) = { 1 , 2 , , M }? If so, can we bound M by a function of k?
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Answers (2)

Dillon Levy
Answered 2022-09-21 Author has 6 answers
Nope. Define 𝑝 as follows:
p ( 2 i 1 ) = 2 i + 1 for i 1,
p ( 2 ) = 1,
p ( 2 i + 2 ) = 2 i for i 1.
For an interval [ 1 , M ] either M or M 1 is odd, so p preserves no interval. But | p ( i ) i | 2.
Note that when k = 1, either p ( 1 ) = 1 or p ( 1 ) = 2 , p ( 2 ) = 1. Note also that leaving an initial segment requires that there exist finite cycles, so the example above shows that boundedness is not a strong enough assumption to guarantee that finite cycles exist.
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Linda Peters
Answered 2022-09-22 Author has 1 answers
Here's an involution, that is, a product of disjoint transpositions:
( 1 , 5 ) ( 2 , 4 ) ( 3 , 7 ) k = 1 ( 8 k 2 , 8 k + 2 ) ( 8 k , 8 k + 4 ) ( 8 k + 1 , 8 k + 5 ) ( 8 k + 3 , 8 k + 7 )
Nothing gets moved by more than 4, and no initial segment is fixed.
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