question about construction of real numbers

Question

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?

Answer 1

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.

Answer 2

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.