Let A and B be sets. Show that A x B and B x A have equal cardinality by constructing an explicit bijection between the two sets. Then use the following proposition to prove that multiplication is commutative. (Let n, m be natural numbers. Then nxm=mxn)

Proposition: Cardinal arithmetic

a) Let X be a finite set, and let x be an object which is not an element of X. Then X U (union) {x} is finite and #(X U {x})= #(X)+1

b) Let X and Y be finite sets. Then X U Y is finite and # (X U Y) is less than or equal to #(X) + #(Y). If in addition X and Y are disjoint (i.e., X intersection Y = the empty set), then #(X U Y)= #(X) + #(Y)

c) Let X be a finite set, and let Y be a subset of X. Then Y is finite, and #(Y) is less than or equal to #(X), If in addition Y does not equal X (i.e. Y is a proper subset of X), then we have #(Y) is less than #(X)

d) If X is a finite set, and f:X-->Y is a function, then f(X) is a finite set with #(f(X)) less than or equal to #(X). If in addition f is one to one, then #(f(X)) = #(X)

e) Let X and Y be finite sets. Then cartesian product X x Y is finite and # (X x Y) = #(X) x #(Y)

f) Let X and Y be finite sets. Then the set Y^X is finite and #(Y^X)= #(Y)^#(X)

Solution Preview

Consider the mapping f:AxB-->BxA which sends (a,b) to (b,a). This mapping is injective. Indeed, let f(a,b)=f(a',b'). We have f(a,b)=(b,a), and f(a',b')=(b',a'). Therefore (b,a)=(b',a'). This implies that b=b' and a=a'. Thus (a,b)=(a',b'). ...

Proof about union and cardinalities. ... A_i intersection A_ j = empty set for all i not equal to j, then their union is a finite set and the cardinality of their ...

... least ordinal of cardinality m. Thus U has a subcover of cardinality less than m, in contradiction to the definition of m. This completes the proof of Claim 2 ...

... End of Proof of Claim. ... The solution consists of step-by-step proofs of all the indicated ... A, as well as a step-by-step determination of the cardinality of the ...

Cardinality, countability. ... The third part, proof on how to get the integers and N into one to one, can I just show ordering hence; 1,2,3,4, ... and 0,1,-1,2,-2 ...

...Proof that f is onto: ... infinite sequences of 0's and 1's, so it follows from the Schroeder-Bernstein Theorem that P(A) and S have the same cardinality (ie, the ...

... f is an injective function between two sets of the same cardinality (the "two ... Definitions and proofs: In the above, I have freely assumed that you know all the ...

... number of elements in a set cardinality eg #∅ = 0 ... The proof rules involved are equational reasoning and structural ... rules that we used in our proofs using the ...