Proof: Uncountable Set

Consider the following diagonal indexing (start counting in the left-upper corner, go to the right, then diagonally to the left, until you come back to the first column, and then start a new diagonal, beginning in the first row): \[\begin{array}{ccccccc} \left(\frac{1}{1}\right)_1,&\left(\frac{1}{2}\right)_2,&\left(\frac{1}{3}\right)_4,&\left(\frac{1}{4}\right)_7,&\left(\frac{1}{5}\right)_{11},&\cdots\\ \left(\frac{2}{1}\right)_3,&\left(\frac{2}{2}\right)_5,&\left(\frac{2}{3}\right)_8,&\left(\frac{2}{4}\right)_{12},&\cdots\\ \left(\frac{3}{1}\right)_6,&\left(\frac{3}{2}\right)_9,&\left(\frac{3}{3}\right)_{13},&\cdots\\ \left(\frac{4}{1}\right)_{10},&\left(\frac{4}{2}\right)_{14},&\cdots\\ \left(\frac{5}{1}\right)_{15},&\cdots\\ \vdots \end{array}\]
By the proposition about the union of countable many countable sets it follows that there is an injective function $f:\mathbb Q\to\mathbb N.$
By the definition of countable sets, this means that $\mathbb Q$ is countable.
∎

Corollaries: 1
Explanations: 2
Proofs: 3
Propositions: 4 5 6 7

Thank you to the contributors under CC BY-SA 4.0!

Forster Otto: "Analysis 1, Differential- und Integralrechnung einer Veränderlichen", Vieweg Studium, 1983