My textbook says that the set of permutations of the positive integers is uncountable. I want to prove this but my mind's completely blank. How do I write a proof for this?
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You could possibly assume there is an injection between the set of permutations of positive integers and the natural numbers and show a contradiction? I don't know if it will work but that would be my first attempt. – SeraPhim Oct 06 '20 at 16:35
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1Can you build an injection of the powerset of the positive integers into your set? – metamorphy Oct 06 '20 at 16:36
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Hint: Consider $\mathbb N$ in pairs: $\{1,2\},$ $\{3,4\},$ $\{5,6\},\dots$ Think of the permutations on $\mathbb N$ that are a permutaion on each such pair. On each pair $\{2n-1,2n\},$ such a permutation will either switch $2n-1$ with $2n,$ or will leave the pair alone. The set of all such permutations can naturally be identified with the set of binary sequences.
zhw.
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I wanted to write this as a comment but can’t do presently. But you can see this by making a diagonal argument akin to standard one used to show that the real numbers are uncountable.
Added Later:
The (accepted/first, see also Brian’s) answer of following post Is symmetric group on natural numbers countable? contains the diagonal argument I was trying to intimate in my comment/answer. (Apologies for not providing this detail earlier in my comment/answer).
Jack LeGrüß
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Yes, I can. I actually learnt/read a proof even here on math.stackexchange or math overflow a few years ago. I’d just recall or look it out. (I’d like to know if my not providing details is the reason for a downvote on this post/comment; if any advice, I’d appreciate it. I’m new here and may not be aware of some things). Thanks – Jack LeGrüß Oct 06 '20 at 16:58
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@Peter and SeraPhim: I have provided the details by referencing the original source where I had learnt this some time ago. – Jack LeGrüß Oct 06 '20 at 17:16