1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Homework Help: Cardinality of set

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

    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
    |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
    Looks good!
  6. Jul 22, 2011 #5
    I really appretiate for your help.
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook