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- Jan 7th 2008, 10:58 AMyellow4321cardinals
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. - Jan 7th 2008, 11:07 AMJhevon
- Jan 7th 2008, 12:11 PMtopsquark
- Jan 7th 2008, 12:58 PMJhevon
- Jan 7th 2008, 02:08 PMyellow4321
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. - Jan 7th 2008, 02:16 PMJhevon
- Jan 7th 2008, 02:36 PMPlato
First, don’t be concerned about the statements on cardinality.

Once you have the basic idea down, all the rest is easy.

Secondly, the above is not the Schroeder-Bernstein. But nonetheless it is easy to prove anything about finite sets.

The basic idea of a finite set is:**Any finite set can be listed using positive integers as subscripts**.

Thus if $\displaystyle n = \left| X \right| \le \left| Y \right| = m$ then we can write $\displaystyle X = \left\{ {x_1 ,x_2 ,x_3 , \cdots ,x_n } \right\}\,\& \,Y = \left\{ {y_1 ,y_2 ,y_3 , \cdots y_m } \right\}$.

If we define $\displaystyle f:X \mapsto Y$ by $\displaystyle f\left( {x_j } \right) = y_j $ the result follows easily. - Jan 8th 2008, 03:22 AMyellow4321
.. for all x,y in X f(x)=f(y) then x=y ?

the less than/equal to sign is confusing me. - Jan 8th 2008, 03:42 AMPlato