# Defining reals as infinite sequence of rationals

Considering the principle of dedekind cuts, would it be possible to define a real number $x$ out of an infinite sequence of rationals $r_1,r_2,\dots$ bounded upwards or downwards to the respective real.

Would it then be possible to define the real as an $\inf\{r_1,r_2,\dots\}$ or $\sup\{r_1,r_2,\dots\}$ respectively?

More precisely, my question is motivated from constructing the reals of $[0,1]$ from $[0,1]\cap\mathbb{Q}$.

• Not really an answer to your question regarding monoton sequences but the best approximation( in some sense)to a real number by rationals is given by the continued fraction expansion of a real number. Maybe you want to look into that. – Verdruss Jul 20 '17 at 19:35
• Well, the usual decimal expansion does that, no? e.g. $\pi= \{3,3.1, 3.14,3.141,3.1415,\cdots\}$. But of course such an expression is not unique so there's still work to be done. – lulu Jul 20 '17 at 19:35
• One of the standard constructions of the reals is via Cauchy sequences of rationals - is this what you had in mind? – Bungo Jul 20 '17 at 19:42
• From the text of the question, we see that the OP means "sequence" not "series" ... editing. – GEdgar Jul 20 '17 at 20:16
• Mmmm.....yeaahhhhh.... buutttttt.... The dedekin cuts are unique and distinct. Your rational sequences are not. So no. Because if we had {q_n} -> x, and {r_n} ->x but {q_n} != {r_n} he'd have to conclude they represent unequal values. – fleablood Jul 20 '17 at 20:27

First off, we can not talk of the $\inf$ and $\sup$ of rational sequences until after we have defined the real numbers.

So what you seem to be asking is can we define the real numbers by viewing them equivalent to bounded sequences.

And the answer to that is no, because different sequences with the same $\inf$s and $\sup$s would have to be considered to represent different values.

Example $\{2 - \frac 1n\}, \{2 - (\frac 34)^n\},$ and $\{2,2,2,2,2,......\}$ are three different sequences but they all (once we know what $\sup$s are) have the same $\sup$. We "want" them all to represent $2$ but seeing as they are all different, we have to believe there are different if this is our definition of the reals.

Unless somehow we can distinguish a concept of the limits solely in terms of the rationals.

Which is precisely what the dedekind cuts do.

Nowc we could have restrictions on the sequences. We could, for instance, say $q_n$ must equal $z_n*10^{-n}$ for some integer $z$ and that $\{q_n\}$ is strictly increasing and any $\{q_n\}$ where all $z_n = 10*z_{n-1}+ 9$ for all $n \ge M$ for some $M$ is the same representation of the sequence $\{q'_n=z'_n*10^{-n}\}$ where all $z'_M = z_M + 1$ and $z'_{n+1} = 10*z'_n$ for all $n > M$.

(Thus $\{2,2,2,2,2,2.....\}$ is the only acceptable sequence to represent $2$. And $\{3,3.1,3.14, 3.141, 3.1415, ....\}$ is the only acceptable sequence to represent $\pi$.)

But proving that is a good definition of the reals is a bit of a bear to analytically prove and misses a lot of the point.

First: I apologise for not writing this as a comment but I currently have under 50 reputation, so cannot.

Perhaps consider the completion of a metric space (https://en.wikipedia.org/wiki/Complete_metric_space#Completion), in this case the completion of $(\mathbb{Q}, | \cdot |)$?

• I don't think this result can be used directly in order to construct the reals without making a circular argument. Technically, you need the reals to even define a metric space. – Bungo Jul 20 '17 at 19:46
• On second thoughts I agree. But if we instead define $| \cdot | : \mathbb{Q} \to \mathbb{Q}$ and say that $x_n \to x$ in $\mathbb{Q}$ if for all rational $\epsilon > 0$ there exists $N$ such that $|x_n - x| < \epsilon$ whenever $n > N$, and then consider the construction, does that now work? – A. S. Jul 20 '17 at 21:34