Timeline for infinite permutations
Current License: CC BY-SA 2.5
8 events
| when toggle format | what | by | license | comment | |
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| Mar 10, 2010 at 18:24 | comment | added | Joel David Hamkins | Reiner, you correctly point out that not all infinite products of permutations give rise to infinite permutations. This is why I took pains in my answer to explain that my full product was well-defined and does define a permutation on the integers. (You have taken only part of my expression, and this part by itself does not define a permutation of the integers.) Nevertheless, that part of the expression is a bijection of the natural numbers with the positive integers, and the other factor is a bijection of the negative integers with the non-positive integers. Together, they make a permutation. | |
| Mar 10, 2010 at 14:11 | comment | added | Guntram | [...(0 3)(0 2)(0 1)] is not a bijection! | |
| Mar 10, 2010 at 8:38 | vote | accept | kakaz | ||
| Mar 10, 2010 at 1:29 | comment | added | Joel David Hamkins | Note that if one takes only the outermost n terms in each factor of the product, then the resulting finite permutation has the right answer on all integers in (-n,n). Thus, the infinite cycle is the limit of the finite approximations to it. It follows that every permutation is the pointwise limit of a sequence of finite permutations. | |
| Mar 9, 2010 at 22:33 | comment | added | Joel David Hamkins | This argument applies to permutations on any set X, not just on the natural numbers. You divide into disjoint cycles, and then represent each cycle as a product of transpositions as above. So altogether, you get an infinite well-defined product of transpositions which is equal to the given permutation. | |
| Mar 9, 2010 at 22:27 | history | edited | Joel David Hamkins | CC BY-SA 2.5 |
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| Mar 9, 2010 at 21:43 | history | edited | Joel David Hamkins | CC BY-SA 2.5 |
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| Mar 9, 2010 at 21:32 | history | answered | Joel David Hamkins | CC BY-SA 2.5 |