Cantor’s Diagonalization Argument

Suppose that the infinity of decimal numbers between zero and one is the same as the infinity of counting numbers. Then all the decimal numbers can be denumerated in a list.

1 → d₁ = 0.d₁₁d₁₂d₁₃d₁₄ …….
2 → d₂ = 0.d₂₁d₂₂d₂₃d₂₄ …….

3 → d₃ = 0.d₃₁d₃₂d₃₃d₃₄ …….

4 → d₄ = 0.d₄₁d₄₂d₄₃d₄₄ …….

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n → d_n = 0.d_n1d_n2d_n3d_n4 …….
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Consider the decimal number x = 0.x₁x₂x₃x₄x₅ ……. , where x₁ is any digit other than d₁₁; x₂ is different from d₂₂; x₃ is not equal to d₃₃; x₄ is not d₄₄; and so on. Now, x is a decimal number, and x is less than one, so it must be in our list. But where? x can’t be first, since x’s first digit differs from d₁‘s first digit. x can’t be second in the list, because x and d₂ have different hundredths place digits. In general, x is not equal to d_n, since their nth digits are not the same.
x is nowhere to be found in the list. In other words, we have exhibited a decimal number that ought to be in the list but isn’t. No matter how we try to list the decimal numbers, at least one will be left out. Therefore, “listing” the decimal numbers is impossible, so the infinity of decimal numbers is greater than the infinity of counting numbers.

Last Updated February 7, 2022