Cardinality of Set: Proving lXl<lYl Implies lXl\inlYl

[gotjrgkr]

Homework Statement

Hi!
I want to show that lXl<lYl implies lXl$\in$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 $\rightarrow$ lXl$\leq$lYl and not lXl=lYl( X is not equipotent to Y)
$\rightarrow$ 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)
$\rightarrow$ there is a function f on X into Y s.t. f is a 1-1 function, and
lXl$\in$lYl or lYl$\in$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??

 

[micromass]

|X| and |Y| are ordinals, and |X|<|Y| as ordinals (prove this). So, what do you know about the order relation on the ordinals?

 

[gotjrgkr]

|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??

 

[micromass]

Looks good!

 

[gotjrgkr]

Looks good!

I really appretiate for your help.
Thanks!