First suppose the sets are finite, with cardinalities and . The power set and Cartesian product counting rules giveSince , the desired inequality follows.
The power-set product cardinality inequality also has an injection proof that does not assume finiteness. Choose distinct and distinct . For and , define the subset of These three encoding regions are disjoint. The row with second coordinate recovers ; the column with first coordinate and second coordinate different from recovers ; the extra position records whether . Thus is injective, proving for arbitrary setsEncoding by alone would fail: all pairs with either factor empty would have the same output.
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