# Real numbers are uncountable

We pre­sent a var­i­ant of Can­tor’s di­ag­o­nal­iza­tion ar­gu­ment to prove the real num­bers are un­countable. This con­struc­tively proves that there ex­ist un­countable sets note Since the real num­bers are an ex­am­ple of one..

We use the dec­i­mal rep­re­sen­ta­tion of the real num­bers. An over­line ( $$\bar{\phantom{9}}$$ ) is used to mean that the digit(s) un­der it are re­peated for­ever. Note that $$a.bcd\cdots z\overline{9} = a.bcd\cdots (z+1)\overline{0}$$ (if $$z < 9$$; oth­er­wise, we need to con­tinue car­ry­ing the one); $$\sum_{i=k}^\infty 10^{-k} \cdot 9 = 1 \cdot 10^{-k + 1} + \sum_{i=k}^\infty 10^{-k} \cdot 0$$. Fur­ther­more, these are the only equiv­alences be­tween dec­i­mal rep­re­sen­ta­tions; there are no other real num­bers with mul­ti­ple rep­re­sen­ta­tions, and these real num­bers have only these two dec­i­mal rep­re­sen­ta­tions.

The­o­rem The real num­bers are un­countable.

Proof Sup­pose, for con­tra­dic­tion, that the real num­bers are countable; sup­pose that $$f: \mathbb Z^+ \twoheadrightarrow \mathbb R$$ is a sur­jec­tion. Let $$r_n$$ de­note the $$n^\text{th}$$ dec­i­mal digit of $$r$$, so that the frac­tional part of $$r$$ is $$r_1r_2r_3r_4r_5\ldots$$ Then define a real num­ber $$r'$$ with $$0 \le r' < 1$$ so that $$r'_n$$ is 5 if $$(f(n))_n \ne 5$$, and 6 if $$(f(n))_n = 5$$. Then there can be no $$n$$ such that $$r' = f(n)$$ since $$r'_n \ne (f(n))_n$$. Thus $$f$$ is not sur­jec­tive, con­tra­dict­ing our as­sump­tion, and $$\mathbb R$$ is un­countable. $$\square$$

Note that choos­ing 5 and 6 as our al­low­able digits for $$r'$$ side-steps the is­sue that $$0.\overline{9} = 1.\overline{0}$$. %%

Parents:

• Uncountability

Some in­fini­ties are big­ger than oth­ers. Un­countable in­fini­ties are larger than countable in­fini­ties.

• Real number