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In the diagonal argument in which we prove $\mathbb{R}$ is uncountable, we take any function $f: \mathbb{N} \to \mathbb{R}$, suppose for the sake of contradiction that it is surjective, and consider any enumeration of $\mathbb{R}$, which will take the form: \begin{align*} f(0) & = a_{00} . a_{01} a_{02} a_{03} \ldots \\ f(2) & = a_{10} . a_{11} a_{12} a_{13} \ldots \\ f(3) & = a_{20} . a_{21} a_{22} a_{23} \ldots \\ \vdots \end{align*} The proof then constructs $x = b_0 b_1 b_2 \ldots$ where $b_i \neq a_{ii}$ for each $i$, so $x \not \in f(i)$ for any $i$, hence $x \not \in f(\mathbb{N})$.

Here is the question I have: what about ending in an infinite string of $9$'s? I can require when constructing $f(n)$ for each $n$ that I represent a real number in its decimal expansion not terminating in $9's$ (so I write $0.5\overline{0}$ instead of $0.4\overline{9}$, for example), but what is to stop the $x$ I construct from ending in $9$'s? If that's the case, I can't go digit-by-digit and rule out each line in this enumeration.

JeremyS
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    Never use $9$. If $a_{ii}=0$ use $b_i=1$, if $a_{ii} \ge 1$ use $b_i=0$. And I'm sure this question is already asked somewhere on this site... – Lee Mosher May 28 '21 at 21:31

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We can pick $b_i$ such that not only $b_i\ne a_{i,i}$, but additionally $b_i\ne 9$.

Hagen von Eitzen
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  • I hadn't even thought of that -- thank you. Are there other ways, though? I'd be interested in hearing if there are other tricks. – JeremyS May 28 '21 at 21:28