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 $D$ ?
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.