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Possible Duplicate:
Completion of rational numbers via Cauchy sequences
Dedekind Cuts versus Cauchy Sequences

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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Gonzalez
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    You might find kahen's answer to a previous question concerning constructing the reals via Cauchy sequences: http://math.stackexchange.com/questions/11923/completion-of-rational-numbers-via-cauchy-sequences/12009#12009 – Alex Becker Jan 02 '12 at 04:49
  • 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 02 '12 at 04:52
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    Thanks, I saw those answers, but I didn't see it said anywhere what the isometric isomorphism actually is. – Gonzalez Jan 02 '12 at 04:52
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    It's $x \mapsto \sup {r \in \mathbb Q ;\vert; r < x}$, obviously. – kahen Jan 02 '12 at 04:54
  • Qiaochu has a good point. There certainly is a way around it, as you can see from the answers at Alex Becker's link. – Jonas Meyer Jan 02 '12 at 05:01
  • Thanks both, I'll try to fix the wording. – Gonzalez Jan 02 '12 at 05:02
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    @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 02 '12 at 05:04

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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.

Brian M. Scott
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