Proving Non-Empty Finite Sets with Induction: A Challenge

AI Thread Summary
The discussion focuses on proving that for all non-empty finite sets A and B, there are |B|^|A| functions from A to B using induction. The initial approach involves assuming the statement holds for a proper subset of A and B, but the user struggles to finalize the proof. A key insight is that for each element in A, there are |B| choices for mapping, leading to the conclusion that the total number of functions is |B|^|A|. The user is encouraged to think of the problem in simpler terms, emphasizing the direct relationship between the elements of A and B. The thread highlights the importance of demonstrating understanding and effort in problem-solving.
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Hey guys. I need some help solving this proof using induction:

Prove that for all non-empty finite sets A and B, there are |B|^|A| functions from A to B

(where |B| and |A| obviously represents the cardinality of B and A respectively.)

Thanks in advance for the help, since I am pretty stuck. Thanks again!

Jeff

EDIT: Sorry, to reiterate...these are the practice problems we were told to look at before the test...so it's not HW.
 
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But you still have to demonstrate that you've tried solving this problem!
 
Oh haha! Sorry. Here is what I got and then got stuck:

b. Proof: For all non-empty finite sets A and B, there are |B||A| functions from A to B.
Assume for all non empty finite sets, for any proper subset Z C A and Y C B, we have |Y||Z| functions from Z to Y
Let z be an arbitrary element of A, let y be an arbitrary element of B, let Z=A\{z} and let Y=B\{y}
Therefore, we have functions from Z to {y}, Z to Y, {z} to {y} and {z} to Y


I thought I could use induction to say it = 1+|Y|^|Z|+1+|Y|...but I couldn't get that to boil down to |B|^|A|

Jeff
 
Look at it non-rigorously. For each element of A, there are |B| places to send it. That's |B|^|A| functions.
 
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