Cardinality of set

  • Thread starter gotjrgkr
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  • #1
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Homework Statement



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 .


Homework Equations





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??
 

Answers and Replies

  • #2
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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?
 
  • #3
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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?
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:
  • #4
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Looks good!
 
  • #5
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Looks good!
I really appretiate for your help.
Thanks!
 

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