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1 The Proof

|\mathbb N| < |\mathbb R|

Before we get to the main proof, let’s do a lemma:

|\mathbb N| \leq |\mathbb R|

The lemma is easily proven by showing a total injection:

Consider the function f : \mathbb N \rightarrow \mathbb R defined as f(x) = x. The function f is total (it is defined for all x \in \mathbb N) and injective \big(f(x) = f(y)\big) \rightarrow (x = y). Hence, there exists a total injection from \mathbb N to \mathbb R, which means |\mathbb N| \leq |\mathbb R|

With this lemma, all we need to prove that |\mathbb N| < |\mathbb R| is to prove that |\mathbb N| \neq |\mathbb R|.

We proceed by contradiction.

Assume |\mathbb N| = |\mathbb R|. Then there exists a total invertible function f : \mathbb N \rightarrow \mathbb R. Let f' be one such function.

Let x \in \mathbb R be defined such that the nth digit of x is one more than the nth digit of f'(n). Formally, that means:

x = \sum_{n \in \mathbb N} \Big(\big\lfloor f'(n) 10^{n}\big\rfloor + 1 \mod 10\Big) 10^{-n}

Because x is a real number and f' is invertible, there must exist some n_x \in \mathbb N such that f'(n_x) = x. But by construction the n_xth digit of x differs from the n_xth digit of f'(n_x); formally, that is

\big\lfloor x 10^{n_x}\big\rfloor = \big\lfloor f'(n_x) 10^{n_x}\big\rfloor + 1 \mod 10

That means f'(n_x) \neq x, which contradicts our definition of f' and n_x.

Because assuming |\mathbb N| = |\mathbb R| led to a contradiction, it must be the case that |\mathbb N| \neq |\mathbb R|.

Because of our lemma, we know that |\mathbb N| \leq |\mathbb R|, which must mean |\mathbb N| < |\mathbb R|

2 Discussion

Why does the above proof technique not work for integers? Because if we try to apply it to integers, we end up with an infinite string of digits on the left side of the decimal point, which is not an integer.

Why does the above proof technique not work for rationals? Because the decimal expansion of any rational repeats, and the diagonal construction of x does not repeat, and thus is not rational.

There is no magic to the specific x we picked; it would just as well to do a different base, like binary

x_1 = \sum_{n \in \mathbb N} \Big( 1 - \big\lfloor f'(n) 2^{n}\big\rfloor\Big) 2^{-n}

or to use a different digit-changing rule like

x_2 = \sum_{n \in \mathbb N} \Big( 9 - \big\lfloor f'(n) 10^{n}\big\rfloor\Big) 10^{-n}

or to grab a different set of digits

x_3 = \sum_{n \in \mathbb N} \Big(\big\lfloor f'(n) 10^{n + 2(n \mod 2)}\big\rfloor + 1 \mod 10\Big) 10^{-n - 2(n \mod 2)}