If A = {5,6,7} and B = {2,3}, {(2,6),(2,5),(3,7)} ⊆ B×A.
Consider the Cartesian product B×A={(2,5),(2,6),(2,7),(3,5),(3,6),(3,7)}B\times A=\{(2,5),(2,6),(2,7),(3,5),(3,6),(3,7)\}B×A={(2,5),(2,6),(2,7),(3,5),(3,6),(3,7)}. As we can see, the pairs (2,6),(2,5)(2,6), (2,5)(2,6),(2,5) and (3,7)(3,7)(3,7) belong to the set B×AB\times AB×A. Thus, {(2,6),(2,5),(3,7)}⊆B×A\{(2,6), (2,5),(3,7)\}\sube B\times A{(2,6),(2,5),(3,7)}⊆B×A.
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