Bijection between power set and binary n-tuples

[jmjlt88]

Let A={1,...,n}. Show that there is a bijection of P(A) with the cartesian product Xⁿ, where X is the two element set X={0,1} and P(A) is the power set of A.


Below is the start of my proof. I just want to make sure that my function "makes sense." Proof: Let A={1,...n}, and X={0,1}. Define f: P(A) -> Xⁿ by f(A₀)=(x₁,...,x_n), where A₀ is a subset of A (and therefore an element of P(A)) and (x₁,...,x_n) is the element of Xⁿ such that x_i=1 if i ε A₀ and x_i=0 is i is not an element of A₀...In the next step, I let A₀=A₁, and show that their image in Xⁿ is the same. I suppose my question is really "how do I ensure f is well-defined?"

 

[clamtrox]

I'd just take an arbitrary element of P(A) and show that it maps to a single element in Xⁿ, then take an arbitrary element in Xⁿ and show exactly one element of P(A) is mapped there. That's enough right?

 

[HallsofIvy]

Given any subset, A, of A, in P(A), assign to every member, x, of A the value "1" if x is in S, "0" if not. That assigns a member of Xⁿ to every member of P(A).