# Why is the completion of the rationals isometrically isomorphic to the reals constructed by Dedekind cuts? [duplicate]

Suppose that $\mathbb{R}$ has already been constructed by the use of Dedekind cuts.

If $(\mathbb{Q},d)$ is a metric space with $d(x,y)=|x-y|$, then its completion is also one way to view the real numbers.

If you already know the Dedekind cut construction of the reals, why is the completion of $(\mathbb{Q},d)$ isometrically isomorphic to $\mathbb{R}$, to see that both constructions give essentially the same structure? More explicitly, what exactly is the isomorphism here?

Added: I suppose what I want to ask is, if $\mathbb{Q}^*$ denotes the Cauchy completion of the rationals with the usual metric, then there should be some isomorphism $\varphi:\mathbb{R}\to\mathbb{Q}^*$, to conclude that $\mathbb{R}$ and $\mathbb{Q}^*$ are essentially the same.

In this case, if $r\in\mathbb{R}$, then what is $\varphi(r)$? I'm not sure what it goes to in $\mathbb{Q}^*$, since the elements of $\mathbb{Q}^*$ are equivalence classes of Cauchy sequences, where $\{p_n\}\sim\{q_n\}$ iff $\lim_{n\to\infty}|p_n-q_n|=0$. So really, what is $\varphi$, and how is it an isomorphism that preserves the metric? Thank you.

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## marked as duplicate by t.b., Alex Becker, Henning Makholm, Zhen Lin, Zev ChonolesJan 2 '12 at 6:23

You might find kahen's answer to a previous question concerning constructing the reals via Cauchy sequences: math.stackexchange.com/questions/11923/… – Alex Becker Jan 2 '12 at 4:49
Careful: it's a priori circular to construct the reals by completing the rationals, since metric completion is a notion that already requires that we know what the reals are! (I think there is a way around this problem, but I'm not sure.) – Qiaochu Yuan Jan 2 '12 at 4:52
Thanks, I saw those answers, but I didn't see it said anywhere what the isometric isomorphism actually is. – Gonzalez Jan 2 '12 at 4:52
It's $x \mapsto \sup \{r \in \mathbb Q \;\vert\; r < x\}$, obviously. – kahen Jan 2 '12 at 4:54
@Qiaochu: I don't think there's a cheap way: Bourbaki's workaround (in the topology book, if I remember correctly) is to construct the completion of the uniform space $(\mathbb Q,d)$ (the set of minimal Cauchy filters) and to work with this first. This gives a description of the completion without mentioning the reals. – t.b. Jan 2 '12 at 5:04

For each $r\in\mathbb{R}$ fix a sequence $\langle q_n(r):n\in\omega\rangle$ of rational numbers converging to $r$. If you want to be completely explicit, you can do this as follows: for each $p\in\mathbb{Q}$ let $q_n(p)=p$ for each $n\in\omega$, and if $r\in\mathbb{R}\setminus\mathbb{Q}$, let $\langle q_n(r):n\in\omega\rangle$ be the sequence of convergents of the continued fraction expansion of $r$. (There are other ways: you could use the sequence of truncated decimal or binary expansions of $r$, for instance.)
The desired map is then $$\varphi:\mathbb{R}\to\mathbb{Q}^*:r\mapsto \big[\langle q_n(r):n\in\omega\rangle\big]\;,$$ where $\big[\langle q_n(r):n\in\omega\rangle\big]$ is the equivalence class of $\langle q_n(r):n\in\omega\rangle$. Proving that $$d_{\mathbb{Q}^*}\big(\varphi(r),\varphi(s)\big)=\lim_{n\to\infty}|q_n(r)-q_n(s)|=|r-s|$$ for all $r,s\in\mathbb{R}$ is straightforward, assuming that you’ve already constructed $\mathbb{R}$ by Dedekind cuts and proved that it has the expected properties.
If you also want to prove directly that $\varphi$ preserves addition and multiplication, you’ll have to go back to the definitions of those operations in $\mathbb{R}$ (as constructed via Dedekind cuts) and in $\mathbb{Q}^*$ and verify that $\varphi$ does preserve them; this will be very tedious but not difficult in principle. Similarly, if you want to prove directly that $\varphi$ preserves the order, you’ll have to go back to the definitions of the order in $\mathbb{R}$ (as constructed via Dedekind cuts) and in $\mathbb{Q}^*$ and verify that $\varphi$ does preserve it.