By Joseph Breuer, Mathematics, Howard F. Fehr
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In the complementary set R let be a denumerable subset and let the complementary set of over R be , that is R = ∪ . It is possible that could be the empty set. , A, , and are disjoint. Hence Since A and are denumerable, by theorem (b) above we have A ∪ and also ~ . Hence: In outline form, the proof appears as follows: 13. A transcendental number is a real or imaginary number, that is, not an algebraic number. We can now prove the important theorem: The set of all real transcendental numbers is non-denumerable and has the cardinal number c.
This typical method of reasoning in carrying out set-theoretical proofs can easily be generalized to the case where M is an infinite set. Let M be an infinite set and U(M) the set of all subsets of M. Then |M| |U(M)|, that is, U(M) has either the same cardinal number as M—since the set of all subsets with only one element is clearly equivalent to the set M—or U(M) has a greater cardinal number than M. We shall prove | U(M)| > |M|. Let Um be a subset of U(M) which is equivalent to M. Hence Um ⊆ U(M) and Um ~ M.
V. Denumerable Sets 1. The simplest infinite set is the set of natural numbers, namely N = {1,2,3,…}. We assign to this set the transfinite number* a, and write |N| = a. All the sets having the cardinal number a are called denumerable sets. Denumerable sets are those sets which can be put into one-to-one correspondence with the set of natural numbers. In this respect, denumerable sets can have their elements ordered into a sequence, that is, they have a first element, a second, a third, and so on.