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What I do is for finite sets A, B, let $A={a_1, a_2, ...a_n}$ and $B={b_1, b_2, ...b_m}$

A function f assigns each element $a_i$ of $A$ to an element $b_j = f (a_i)$ of $B$; there are $m$ possibilities for each element of $A$, we have $m, m, ...m=m^n= \vert B \vert ^ {\vert A \vert}$ possible functions.

Since A and B are any sets, how can we extend above to an infinite countable or uncountable sets?

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    $\begingroup$ This depends on how you define $|B|^{|A|}$ for infinite cardinals. The definition is usually "the cardinality of all the set of all functions from $A$ to $B,$ which is really what you are trying to prove. $\endgroup$ – Thomas Andrews Jul 18 '19 at 14:33
  • $\begingroup$ Yes that is what I mean. It confused me when I tried to solve a problem mapping finite sets to infinite countable sets. Since $b_m$ cannot be shown in an infinite set, I am confused on how to extend this to A finite and B infinite countable. $\endgroup$ – WaterBro Jul 18 '19 at 14:50
  • $\begingroup$ It's not clear what you mean by "Yes that is what I mean." If this is your definition of $|B|^{|A|}$, then the problem is answered "by definition." The only thing that is really needed is the case when $A,B$ finite, then you'd want the definition to be the same as the natural number definition of exponentiation. $\endgroup$ – Thomas Andrews Jul 18 '19 at 15:04
  • $\begingroup$ Sorry I did not state it clearly, the question can be easily answered by definition if $A, B$ are finite, while for example, if $A$ is finite and $B$ is infinite countable, then the cardinality of the set of all functions mapping A to B should be $\aleph_0$, but I don't know how to generate a formal proof based on $A, B$ finite situation. $\endgroup$ – WaterBro Jul 18 '19 at 15:09
  • $\begingroup$ @WaterBro the point is that the definition is not limited to the finite case; that's also taken for the definition of cardinal exponentiation in the infinite case. If your question is 'how can I show that $\aleph_0^n=\aleph_0$ for all $n$?', that's a separate one, though also relatively straightforward. $\endgroup$ – Steven Stadnicki Jul 18 '19 at 16:18

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