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Cardinality of set

  1. Jul 22, 2011 #1
    1. The problem statement, all variables and given/known data

    Hi!
    I want to show that lXl<lYl implies lXl[itex]\in[/itex]lYl where lXl and lYl are some cardinal numbers of two sets X and Y and the ordering < is defined on cardinal numbers .


    2. Relevant equations



    3. The attempt at a solution
    I tried to solve it by myself as follows:
    lXl < lYl [itex]\rightarrow[/itex] lXl[itex]\leq[/itex]lYl and not lXl=lYl( X is not equipotent to Y)
    [itex]\rightarrow[/itex] there is a function f on X into Y s.t. f is a 1-1 function, and
    not lXl=lYl( cardinal numbers lXl and lYl are not same)
    [itex]\rightarrow[/itex] there is a function f on X into Y s.t. f is a 1-1 function, and
    lXl[itex]\in[/itex]lYl or lYl[itex]\in[/itex]lXl since lXl and
    lYl are initial ordinals.

    But I can't determine why lXl must belong to lYl.

    Could you give me a hint??
     
  2. jcsd
  3. Jul 22, 2011 #2

    micromass

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    |X| and |Y| are ordinals, and |X|<|Y| as ordinals (prove this). So, what do you know about the order relation on the ordinals?
     
  4. Jul 22, 2011 #3
    I tried to prove it.
    I found that if i assume lXl>lYl as ordinals, then it leads to lYl is less than or equal to lXl as cardinals. Then cantor- bernstein's theorem makes a conclusion s.t. lXl=lYl(X is equipotent to Y) . But this is contradiction to the hypothesis lXl<lYl as cardinals. And if lXl=lYl as ordinals, then it is obviously contradiction to the hypothesis. So, lXl<lYl.
    Is my proof right??
     
    Last edited: Jul 22, 2011
  5. Jul 22, 2011 #4

    micromass

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    Looks good!
     
  6. Jul 22, 2011 #5
    I really appretiate for your help.
    Thanks!
     
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