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Suppose A is a (possibly infinite, possbily even uncountable) set, and let F be the set of functions from A to {0, 1}. Show that |F| = |P(A)|.

User Jovon
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Final Answer:

The cardinality of the set of functions from A to {0, 1} (denoted as |F|) is equal to the cardinality of the power set of A (denoted as |P(A)|).

Step-by-step explanation:

The key insight lies in establishing a one-to-one correspondence between the set of functions F and the power set P(A). Consider an element x in A. For each function f in F, there are two possibilities: f(x) = 0 or f(x) = 1. This choice for each element x defines a unique subset of A. Consequently, we can associate each function in F with a subset of A.

To formalize this correspondence, let's define a function g: F → P(A) such that for any function f in F, g(f) is the subset of A consisting of all elements x for which f(x) = 1. The function g is injective, as distinct functions lead to distinct subsets. Additionally, it is surjective, as for any subset S in P(A), we can define a function f such that f(x) = 1 if x is in S, and f(x) = 0 otherwise. This establishes a bijection between F and P(A), demonstrating that |F| = |P(A)|.

In summary, the cardinality of the set of functions from A to {0, 1} is equal to the cardinality of the power set of A, and the establishment of a bijective function solidifies this mathematical equivalence.

User Konrad Neuwirth
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