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Uncountable Sets and Cantor's Theorem
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Uncountable Sets and Cantor's Theorem
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Now · Cantor's Diagonal Argument
Uncountable Sets and Cantor's Theorem
Cantor's Diagonal Argument
The set of real numbers R is uncountable.
Proof: Assume R is countable. List all reals in (0,1) as a sequence r1,r2,r3,…. Each ri has decimal expansion 0.ai1ai2ai3….
Construct b=0.b1b2b3… where bi=5 if aii=5, else bi=6.
Then b differs from every ri at the i-th decimal place, so b is not in the list. Contradiction. □
Cantor's Theorem
For any set A, ∣A∣<∣P(A)∣.
Proof: The map f:A→P(A) defined by f(a)={a} is injective, so ∣A∣≤∣P(A)∣. To show strict inequality, suppose a bijection g:A→P(A) exists. Define B={a∈A:a∈/g(a)}. Then B∈P(A), so B=g(b) for some b∈A. But b∈B⟺b∈/g(b)=B, contradiction. □
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