Proof
(related to Proposition: Union of Countably Many Countable Sets)
- Since each \(D_n\) is countable by hypothesis, we can write \(D_n=\{x_{nm},~m\in\mathbb N\}\).
- We claim, that the union
\[M:=\bigcup_{n\in\mathbb N} D_n\] is again countable.
- In order to see it, we write the elements of the union in a table:
\[\begin{array}{ccccccc}
\left(x_{00}\right)_0,&\left(x_{01}\right)_1,&\left(x_{02}\right)_3,&\left(x_{03}\right)_6,&\left(x_{04}\right)_{10},&\cdots\\
\left(x_{10}\right)_2,&\left(x_{11}\right)_4,&\left(x_{12}\right)_7,&\left(x_{13}\right)_{11},&\cdots\\
\left(x_{20}\right)_5,&\left(x_{21}\right)_8,&\left(x_{22}\right)_{12},&\cdots\\
\left(x_{30}\right)_{9},&\left(x_{31}\right)_{13},&\cdots\\
\left(x_{40}\right)_{14},&\cdots\\
\vdots
\end{array}\]
- The outer index defines an injective function \(f:M\mapsto\mathbb N\). Therefore, \(M\) is countable.
∎
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References
Bibliography
- Forster Otto: "Analysis 1, Differential- und Integralrechnung einer Veränderlichen", Vieweg Studium, 1983
