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Order types

  1. Dec 3, 2012 #1
    Hi all,

    Went over this today and I'm not grasping it: why is the order type of n + ω = ω, while ω + n ≠ ω? I'd really appreciate if someone could set up the requisite isomorphism in the former. Thanks!
  2. jcsd
  3. Dec 3, 2012 #2
    The set [itex]n+\omega[/itex] is essentially (order-isomorphic to) the following:


    Do you see that??

    The isomorphism between the above set and [itex]\omega[/itex] is given by the map T that does the following:

  4. Dec 3, 2012 #3
    Ahh yes, this helps a lot, thanks. So in case of $$\omega + n $$ we could try $$(0,0) < (1,0) < ... < (k,0) < ... < (0,1) < (1,1) < ... < (n-1,1)$$ but we wouldn't be able to set up an isomorphism between this and ##\omega##?
  5. Dec 3, 2012 #4
    Yeah exactly. Here we have the natural numbers and we paste n elements after it.
    So take (0,1) for example. That has an infinite number of predecessors. So it can't be [itex]\omega[/itex] since any element in [itex]\omega[/itex] has a finite number of predecessors.
  6. Dec 3, 2012 #5
    Excellent, thanks for your help!
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