# Continuity of the reals in terms of decimal expansions

I was wondering about how we could prove the completeness of $$\mathbb R$$ when this set is defined to be the set of all decimal expressions of the form : $$\underbrace{-}_{\text{sign}}\underbrace{317}_{\text{finite sequence of digits}}. \underbrace{12554382...}_{\text{possibly infinite sequence of digits}}$$ By the completeness of $$\mathbb R$$, I mean any property equivalent to the completeness axiom in a theory of the real numbers : it informally states that $$\mathbb R$$ has "no gaps".

Let's restrict ourselves to the interval $$[0,1]$$. With that in mind, let's define the set $$D$$ of all possibly infinite sequences of digits. Of course, $$D$$ is the set of all possible decimal parts of a real number in decimal expansion, and $$D$$ is essentially in bijection with $$[0,1]$$ (once we identify expressions like $$5$$ and $$4999...$$, because $$0.5 = 0.4999...$$).

By Cantor's diagonal argument, we already know that such set of sequences of digits is uncountable. But uncountability isn't completeness yet.

So is there a way to prove the completeness of $$D$$ ?

• You say "By the completeness of R I mean any property equivalent to the completeness axiom in a theory of the real numbers : it informally states that R has "no gaps". In order to give a proof of a property you need to use the precise definition of that property. What is the exact statement of the "completeness axiom" you are given? Commented Feb 18, 2019 at 16:04

Sure. You want to prove that any non-empty subset $${\cal A}\subseteq D$$ has a least upper bound in $$D$$. But the least upper bound of $$\cal A$$ can be formed one digit at a time. Each element is represented as $$[0.a_0a_1a_2\ldots]$$. Take $$a_0^{*}$$ to be the largest $$a_0$$ of any element of $$\cal A$$; take $$a_1^{*}$$ to be the largest $$a_1$$ of any element of $${\cal A}$$ that has $$a_0=a_0^*$$; take $$a_2^*$$ to be the largest $$a_2$$ of any element of $${\cal A}$$ that has $$a_0=a_0^*$$ and $$a_1=a_1^*$$; and so on. Then $$[0.a_0^*a_1^*a_2^*\ldots]$$ is an upper bound of $${\cal A}$$ by construction, and any smaller element of $$D$$ (with $$a_k < a_k^*$$, say, and equal for all preceding digits) is smaller than some element of $${\cal A}$$ (an element that forced the value $$a_k^*$$ during the construction); so $$[0.a_0^*a_1^*a_2^*\ldots]$$ is in fact the least upper bound.

• Right, using the least upper bound property seems a good choice. Thanks ! Commented Feb 19, 2019 at 8:23