Uncountable Sets and Cantor's Theorem

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# Uncountable Sets and Cantor's Theorem ## Cantor's Diagonal Argument The set of real numbers $\mathbb{R}$ is uncountable. **Proof:** Assume $\mathbb{R}$ is countable.

Uncountable Sets and Cantor's Theorem

Cantor's Diagonal Argument

The set of real numbers R\mathbb{R} is uncountable. Proof: Assume R\mathbb{R} is countable. List all reals in (0,1)(0,1) as a sequence r1,r2,r3,…r_1, r_2, r_3, \dots. Each rir_i has decimal expansion 0.ai1ai2ai3…0.a_{i1}a_{i2}a_{i3}\dots. Construct b=0.b1b2b3…b = 0.b_1b_2b_3\dots where bi=5b_i = 5 if aii≠5a_{ii} \neq 5, else bi=6b_i = 6. Then bb differs from every rir_i at the ii-th decimal place, so bb is not in the list. Contradiction. □\square

Cantor's Theorem

For any set AA, ∣A∣<∣P(A)∣|A| < |\mathcal{P}(A)|. Proof: The map f:A→P(A)f: A \to \mathcal{P}(A) defined by f(a)={a}f(a) = \{a\} is injective, so ∣A∣≤∣P(A)∣|A| \leq |\mathcal{P}(A)|. To show strict inequality, suppose a bijection g:A→P(A)g: A \to \mathcal{P}(A) exists. Define B={a∈A:a∉g(a)}B = \{a \in A : a \notin g(a)\}. Then B∈P(A)B \in \mathcal{P}(A), so B=g(b)B = g(b) for some b∈Ab \in A. But b∈B  ⟺  b∉g(b)=Bb \in B \iff b \notin g(b) = B, contradiction. □\square Join Discord PreviousCardinality and CountabilityNextCombinatorics & Probability
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