The Cartesian product of two sets X and Y, denoted X×Y, is the set of all ordered pairs (x,y) where x is in Xand y is in Y, that is X×Y={(x,y)∣x∈X and y∈Y}.
Since ∣A∣=∣B∣, there exists a bijection f:A→B. By analogy, the equality ∣C∣=∣D∣ implies that there exists a bijection g:C→D. Let us prove that the map h:A×C→B×D, h(a,c)=(f(a),g(c)), is a bijection as well.
First prove that h is an injection. Let h(a1,c1)=h(a2,c2). So (f(a1),g(c1))=(f(a2),g(c2)). It follows that f(a1)=f(a2) and g(c1)=g(c2). Since f and g are injections, a1=a2 and c1=c2. Therefore, (a1,c1)=(a2,c2), and h is an injection.
Finally, prove that h is a surjection. Let (b,d)∈B×D. Since f and g are surjections, there exist a∈A and c∈C such that f(a)=b and g(c)=d. Then h(a,c)=(f(a),g(c))=(b,d), and consequantly, h is a surjection.
It follows that h:A×C→B×D is bijection, and therefore, ∣A×C∣=∣B×D∣.
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