Answer to Question #136823 in Discrete Mathematics for Promise Omiponle

Question #136823
Show that if A, B, C, and D are sets with |A|=|B| and|C|=|D|, then |AxC|=|BxD|.
1
Expert's answer
2020-10-19T16:49:35-0400

The Cartesian product of two sets XX and YY, denoted X×YX\times Y, is the set of all ordered pairs (x,y)(x,y) where xx is in XXand yy is in YY, that is X×Y={(x,y)xX  and  yY}.{\displaystyle X\times Y=\{\,(x,y)\mid x\in X\ {\text{ and }}\ y\in Y\,\}.}

Since A=B|A|=|B|, there exists a bijection f:ABf: A\to B. By analogy, the equality C=D|C|=|D| implies that there exists a bijection g:CDg: C\to D. Let us prove that the map h:A×CB×Dh: A\times C\to B\times D, h(a,c)=(f(a),g(c))h(a,c)=(f(a), g(c)), is a bijection as well.

First prove that hh is an injection. Let h(a1,c1)=h(a2,c2)h(a_1, c_1)=h(a_2, c_2). So (f(a1),g(c1))=(f(a2),g(c2))(f(a_1), g(c_1))=(f(a_2), g(c_2)). It follows that f(a1)=f(a2)f(a_1)=f(a_2) and g(c1)=g(c2)g(c_1)=g(c_2). Since ff and gg are injections, a1=a2a_1=a_2 and c1=c2c_1=c_2. Therefore, (a1,c1)=(a2,c2)(a_1,c_1)=(a_2,c_2), and hh is an injection.

Finally, prove that hh is a surjection. Let (b,d)B×D(b,d)\in B\times D. Since ff and gg are surjections, there exist aAa\in A and cCc\in C such that f(a)=bf(a)=b and g(c)=dg(c)=d. Then h(a,c)=(f(a),g(c))=(b,d)h(a,c)=(f(a), g(c))=(b, d), and consequantly, hh is a surjection.

It follows that h:A×CB×Dh: A\times C\to B\times D is bijection, and therefore, A×C=B×D|A\times C|=|B\times D|.








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