# question about construction of real numbers

So I've gotten myself confused over a seeming tautology one uses when constructing the real numbers as equivalence classes of Cauchy sequences of rational numbers. Having constructed the real numbers in this way, once one makes sense of the term "positive real number" it is easy to extend the absolute value and in particular get a notion of convergence.

If one has a Cauchy sequence of rational numbers which represents the class of some real number, surely this sequence converges to this number, but how does one actually see this without begging the question?

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An important point. After constructing the reals after all the trouble, you should prove that a Cauchy sequence of reals does in fact converge to a real number. If this is not satisfied, you will have to repeat the process of inventing more numbers for this purpose. – Srivatsan Sep 10 '11 at 17:50
@Srivatsan: right, and I'm asking about an even more basic fact, which is that any Cauchy sequence of rational numbers converges to a real number. This, together with the density of the rationals in the reals, implies the completeness of the real numbers. – Justin Campbell Sep 10 '11 at 18:48
@Srivatsan: Slightly off at a tangent, but I think that it's an interesting fact that in constructive logic, the reals constructed in this way are not Cauchy complete. – George Lowther Sep 10 '11 at 23:03
@George I didn't know that. Thanks :) – Srivatsan Sep 10 '11 at 23:06
@Srivatsan: I should back up that statement with a link. Although it's not where I originally learned this from, a google search came up with the following. On the Cauchy Completeness of the Constructive Cauchy Reals math.fau.edu/lubarsky/Cauchyreals.pdf – George Lowther Sep 10 '11 at 23:24

## 2 Answers

More precisely, it’s the sequence of embedded copies of those rational numbers that converge to the given real number.

I’ll represent the set of equivalence classes of Cauchy sequences of rationals by $\hat{\mathbb{R}}$. To try to keep things as clear as possible, I’ll use $\|\cdot\|$ for the absolute value in $\hat{\mathbb{R}}$ and $\vert\cdot\vert$ for the absolute value in $\mathbb{Q}$. I’m assuming that for $[\sigma]\in \hat{\mathbb{R}}$ with $\sigma = \langle p_n:n\in\omega\rangle$ you’ve defined $\| \sigma\|$ to be $\langle \vert p_n \vert:n\in \omega \rangle$ and shown that this has the expected properties.

Now suppose that $[\sigma]$ is as above. For each $n\in\omega$ let $\sigma_n$ be the constant sequence whose terms are all $p_n$; $[\sigma_n]$ is of course the copy of $p_n$ in $\hat{\mathbb{R}}$. You want to show that the sequence $\langle [\sigma_n]:n \in \omega\rangle$ converges to $[\sigma]$ in $\hat{\mathbb{R}}$. Assuming that you’ve done the necessary ground work, it suffices to show that for each positive $r \in \mathbb{Q}$ there is an $n_r\in\omega$ such that $\|[\sigma]-[\sigma_n]\| = \langle \vert p_k-p_n \vert:k\in\omega\rangle \le [\rho]$ whenever $n\ge n_r$, where $\rho$ is the constant sequence whose terms are all $r$. (The definition is usually stated in terms of $<$ rather than $\le$, but the two are easily seen to be equivalent.)

By the definition of $\le$ in $\hat{\mathbb{R}}$, $\langle \vert p_k-p_n \vert:k\in\omega\rangle \le [\rho]$ provided that there is some $k_{r,n}\in\omega$ such that $\vert p_k-p_n \vert \le r$ whenever $k \ge k_{r,n}$. The sequence $\sigma = \langle p_n:n\in\omega\rangle$ is Cauchy in $\mathbb{Q}$, so there is an $m_r \in \omega$ such that $\vert p_k-p_n \vert < r$ whenever $k,n \ge m_r$. Take $n_r = m_r$, and for each $n \ge n_r$ let $k_{r,n} = m_r$ as well. Then for each $n \ge n_r$ we have $\vert p_k-p_n \vert < r$ whenever $k \ge k_{r,n}$, and hence $\|[\sigma]-[\sigma_n]\| \le [\rho]$ for each $n \ge n_r$, as desired.

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It seems we both made the same mistake independently? As @Srivatsan pointed out in a comment to my answer, that's not quite the definition of $\lt$ in $\hat{\mathbb{R}}$. – joriki Sep 10 '11 at 19:51
@joriki: Yep; it doesn’t behave like an ultrapower! Thanks; I think that I’ve made all of the necessary changes. – Brian M. Scott Sep 10 '11 at 20:15
I don't think it works like that. As long as you have "if and only if" there, it doesn't matter whether you use $\le$ or $\lt$, since $\le$ is just the negation of $\gt$. I think you only need the implication in one direction; at least that's how I resolved this in my answer. – joriki Sep 10 '11 at 20:37
@joriki: Argh. You’re right. It might almost have been easier to define $[\langle p_n\rangle] <[\langle q_n\rangle]$ iff there is $r\in\mathbb{Q}^+$ such that $q_n-p_n$ is eventually greater than $r$, but I’m simply deleting and only if. – Brian M. Scott Sep 10 '11 at 20:55
This is the definition that @TonyK suggested in a comment to my answer. – joriki Sep 10 '11 at 21:02

Yes, I find it difficult to keep these things straight in my mind, too – in fact I thought I'd gotten them straight but found while writing this that they were less straight than I thought. :-)

I think a good starting point is to convince yourself that some of the operations involved commute. That is, if $a$ and $b$ are sequences of rational numbers and $[a]$ and $[b]$ denote their equivalence classes, then

$$|[a]|=[|a|]$$

and

$$[a]-[b]=[a-b]\;.$$

This doesn't work for $\le$, which instead obeys

$$[a]\le [b] \Leftarrow\exists n_0\in\mathbb N:\forall n\gt n_0: a_n\le b_n\;.$$

Now we can write the definition of the convergence of $[\{a_n\}]$ to $[a]$ (where $\{a_n\}$ is the constant sequence with elements $a_n$),

$$\forall[\epsilon]\gt0:\exists n_0\in\mathbb N:\forall n\gt n_0:|[\{a_n\}]-[a]|\le[\epsilon]$$

(where I've written $[\epsilon]$ to make it easier to visually distinguish rationals and reals). Commuting the operations on the left-hand side of the inequality yields

$$\forall[\epsilon]\gt0:\exists n_0\in\mathbb N:\forall n\gt n_0:[|\{a_n\}-a|]\le[\epsilon]\;,$$

and this is implied by

$$\forall[\epsilon]\gt0:\exists n_0\in\mathbb N:\forall n\gt n_0:\exists m_0\in\mathbb N:\forall m\gt m_0: (\{a_n\}-a)_m\le\epsilon_m\;.$$

But $(\{a_n\}-a)_m=\{a_n\}_m-a_m=a_n-a_m$, so this becomes

$$\forall[\epsilon]\gt0:\exists n_0\in\mathbb N:\forall n\gt n_0:\exists m_0\in\mathbb N:\forall m\gt m_0: a_n-a_m\le\epsilon_m\;,$$

which looks a lot more helpful, since the left-hand side of the inequality is now the difference of two rationals in the sequence, which converges since $a$ is Cauchy. We can simplify this to

$$\forall[\epsilon]\gt0:\exists n_0\in\mathbb N:\forall m,n\gt n_0:a_n-a_m\le\epsilon_m\;,$$

since moving the existential quantifier in front of the universal one and requiring $m_0$ and $n_0$ to be the same only makes the condition stronger. Now it's recognizable that this condition is implied by $a$ being Cauchy, since $\epsilon_m$ must eventually be greater than some rational $\epsilon'\gt0$ and then we obtain $n_0$ from the Cauchy condition.

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How is the equivalence for $<$ correct? For instance, taking $a_n = 0$ and $b_n = 1/n$ obviously satisfies the RHS, but both sequences converge to $0$ (so the left hand side is violated). Am I missing something? (I am all confused, so this perhaps doesn't make sense to you. :)) – Srivatsan Sep 10 '11 at 19:27
@mixedmath: We all seem to be confused about different things here, so who knows what might confuse or unconfuse Justin :-). – joriki Sep 10 '11 at 19:43
@joriki: :) ${}{}{}$ – mixedmath Sep 10 '11 at 19:46
$[a] < [b] \Leftrightarrow \exists q \in \mathbb Q_{>0} : \exists n_0 \in \mathbb N : \forall n\gt n_0: a_n + q < b_n$ – TonyK Sep 10 '11 at 20:33
@Srivatsan: Wikipedia also talks about the completion as a metric space, despite defining a metric as a real-valued function. But one could first define a metric as a rational-valued function and then transfer it to the reals, much like the arithmetic operations are first defined on the rationals and then transferred to the reals -- so I think this perspective still has some value, even though one needs to be careful about what one means by a "metric". – joriki Sep 10 '11 at 20:59