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Let A, B, C be sets. Show that the sets (A^B)^C and A^(BxC) have equal cardinality by constructing an explicit bijection between the two sets. Conclude that (a^b)^c=a^bc for any natural numbers a, b, c. Use a similar argument to also conclude a^b x a^c= a^b+c

#### Solution Preview

(i) Consider the mapping F:(A^B)^C-->A^(BxC) given as follows. It sends f:C-->(A^B) to F(f):BxC-->A defined by F(f)(b,c)=f(c)(b).

Let us show that F is injective. To this end, let us consider f,f':C-->(A^B) such that F(f)=F(f'). Let us show that f(c)=f'(c), for any c from C. For that it is sufficient to show that f(c)(b)=f'(c)(b). But f(c)(b)=F(f)(b,c), and similarly, f'(c)(b)=F(f')(b,c). Since F(f)=F(f'), F(f)(b,c)=F(f')(b,c). Therefore, f(c)(b)=f'(c)(b), and hence F is injective.

Let us now show that F is surjective. Consider any g:BxC-->A. Construct f:C-->(A^B) as follows. f(c):B-->A ...

#### Solution Summary

This solution helps with a proof about cardinality. It helps show that sets have equal cardinality by constructing an explicit bijection between two sets.

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