The reals (constructed as classes of Cauchy sequences of rationals) form a field

The real num­bers, when con­structed as equiv­alence classes of Cauchy se­quences of ra­tio­nals, form a to­tally or­dered field, with the in­her­ited field struc­ture given by

  • \([a_n] + [b_n] = [a_n+b_n]\)

  • \([a_n] \times [b_n] = [a_n \times b_n]\)

  • \([a_n] \leq [b_n]\) if and only if ei­ther \([a_n] = [b_n]\) or for suffi­ciently large \(n\), \(a_n \leq b_n\).

Proof

Firstly, we need to show that those op­er­a­tions are even well-defined: that is, if we pick two differ­ent rep­re­sen­ta­tives \((x_n)_{n=1}^{\infty}\) and \((y_n)_{n=1}^{\infty}\) of the same equiv­alence class \([x_n] = [y_n]\), we don’t some­how get differ­ent an­swers.

Well-defined­ness of \(+\)

We wish to show that \([x_n]+[a_n] = [y_n] + [b_n]\) when­ever \([x_n] = [y_n]\) and \([a_n] = [b_n]\); this is an ex­er­cise.

Since \([x_n] = [y_n]\), it must be the case that both \((x_n)\) and \((y_n)\) are Cauchy se­quences such that \(x_n - y_n \to 0\) as \(n \to \infty\). Similarly, \(a_n - b_n \to 0\) as \(n \to \infty\).

We re­quire \([x_n+a_n] = [y_n+b_n]\); that is, we re­quire \(x_n+a_n - y_n-b_n \to 0\) as \(n \to \infty\).

But this is true: if we fix ra­tio­nal \(\epsilon > 0\), we can find \(N_1\) such that for all \(n > N_1\), we have \(|x_n - y_n| < \frac{\epsilon}{2}\); and we can find \(N_2\) such that for all \(n > N_2\), we have \(|a_n - b_n| < \frac{\epsilon}{2}\). Let­ting \(N\) be the max­i­mum of the two \(N_1, N_2\), we have that for all \(n > N\), \(|x_n + a_n - y_n - b_n| \leq |x_n - y_n| + |a_n - b_n|\) by the tri­an­gle in­equal­ity, and hence \(\leq \epsilon\). <div><div>

Well-defined­ness of \(\times\)

We wish to show that \([x_n] \times [a_n] = [y_n] \times [b_n]\) when­ever \([x_n] = [y_n]\) and \([a_n] = [b_n]\); this is also an ex­er­cise.

We re­quire \([x_n a_n] = [y_n b_n]\); that is, \(x_n a_n - y_n b_n \to 0\) as \(n \to \infty\).

Let \(\epsilon > 0\) be ra­tio­nal. Then

$$|x_n a_n - y_n b_n| = |x_n (a_n - b_n) + b_n (x_n - y_n)|$$
us­ing the very handy trick of adding the ex­pres­sion \(x_n b_n - x_n b_n\).

By the tri­an­gle in­equal­ity, this is \(\leq |x_n| |a_n - b_n| + |b_n| |x_n - y_n|\).

We now use the fact that cauchy se­quences are bounded, to ex­tract some \(B\) such that \(|x_n| < B\) and \(|b_n| < B\) for all \(n\); then our ex­pres­sion is less than \(B (|a_n - b_n| + |x_n - y_n|)\).

Fi­nally, for \(n\) suffi­ciently large we have \(|a_n - b_n| < \frac{\epsilon}{2 B}\), and similarly for \(x_n\) and \(y_n\), so the re­sult fol­lows that \(|x_n a_n - y_n b_n| < \epsilon\). <div><div>

Well-defined­ness of \(\leq\)

We wish to show that if \([a_n] = [c_n]\) and \([b_n] = [d_n]\), then \([a_n] \leq [b_n]\) im­plies \([c_n] \leq [d_n]\).

Sup­pose \([a_n] \leq [b_n]\), but sup­pose for con­tra­dic­tion that \([c_n]\) is not \(\leq [d_n]\): that is, \([c_n] \not = [d_n]\) and there are ar­bi­trar­ily large \(n\) such that \(c_n > d_n\). Then there are two cases.

  • If \([a_n] = [b_n]\) then \([d_n] = [b_n] = [a_n] = [c_n]\), so the re­sult fol­lows im­me­di­ately. noteWe didn’t need the ex­tra as­sump­tion that \([c_n] \not \leq [d_n]\) here.

  • If for all suffi­ciently large \(n\) we have \(a_n \leq b_n\), then this part

Ad­di­tive com­mu­ta­tive group struc­ture on \(\mathbb{R}\)

The ad­di­tive iden­tity is \([0]\) (for­mally, the equiv­alence class of the se­quence \((0, 0, \dots)\)). In­deed, \([a_n] + [0] = [a_n+0] = [a_n]\).

The ad­di­tive in­verse of the el­e­ment \([a_n]\) is \([-a_n]\), be­cause \([a_n] + [-a_n] = [a_n-a_n] = [0]\).

The op­er­a­tion is com­mu­ta­tive: \([a_n] + [b_n] = [a_n+b_n] = [b_n+a_n] = [b_n] + [a_n]\).

The op­er­a­tion is closed, be­cause the sum of two Cauchy se­quences is a Cauchy se­quence (ex­er­cise).

If \((a_n)\) and \((b_n)\) are Cauchy se­quences, then let \(\epsilon > 0\). We wish to show that there is \(N\) such that for all \(n, m > N\), we have \(|a_n+b_n - a_m - b_m| < \epsilon\).

But \(|a_n+b_n - a_m - b_m| \leq |a_n - a_m| + |b_n - b_m|\) by the tri­an­gle in­equal­ity; so pick­ing \(N\) so that \(|a_n - a_m| < \frac{\epsilon}{2}\) and \(|b_n - b_m| < \frac{\epsilon}{2}\) for all \(n, m > N\), the re­sult fol­lows. <div><div>

The op­er­a­tion is as­so­ci­a­tive:

$$[a_n] + ([b_n] + [c_n]) = [a_n] + [b_n+c_n] = [a_n+b_n+c_n] = [a_n+b_n] + [c_n] = ([a_n]+[b_n])+[c_n]$$

Ring structure

The mul­ti­plica­tive iden­tity is \([1]\) (for­mally, the equiv­alence class of the se­quence \((1,1, \dots)\)). In­deed, \([a_n] \times [1] = [a_n \times 1] = [a_n]\).

\(\times\) is closed, be­cause the product of two Cauchy se­quences is a Cauchy se­quence (ex­er­cise).

If \((a_n)\) and \((b_n)\) are Cauchy se­quences, then let \(\epsilon > 0\). We wish to show that there is \(N\) such that for all \(n, m > N\), we have \(|a_n b_n - a_m b_m| < \epsilon\).

But

$$|a_n b_n - a_m b_m| = |a_n (b_n - b_m) + b_m (a_n - a_m)| \leq |b_m| |a_n - a_m| + |a_n| |b_n - b_m|$$
by the tri­an­gle in­equal­ity.

Cauchy se­quences are bounded, so there is \(B\) such that \(|a_n|\) and \(|b_m|\) are both less than \(B\) for all \(n\) and \(m\).

So pick­ing \(N\) so that \(|a_n - a_m| < \frac{\epsilon}{2B}\) and \(|b_n - b_m| < \frac{\epsilon}{2B}\) for all \(n, m > N\), the re­sult fol­lows. <div><div>

\(\times\) is clearly com­mu­ta­tive: \([a_n] \times [b_n] = [a_n \times b_n] = [b_n \times a_n] = [b_n] \times [a_n]\).

\(\times\) is as­so­ci­a­tive:

$$[a_n] \times ([b_n] \times [c_n]) = [a_n] \times [b_n \times c_n] = [a_n \times b_n \times c_n] = [a_n \times b_n] \times [c_n] = ([a_n] \times [b_n]) \times [c_n]$$

\(\times\) dis­tributes over \(+\): we need to show that \([x_n] \times ([a_n]+[b_n]) = [x_n] \times [a_n] + [x_n] \times [b_n]\). But this is true:

$$[x_n] \times ([a_n]+[b_n]) = [x_n] \times [a_n+b_n] = [x_n \times (a_n+b_n)] = [x_n \times a_n + x_n \times b_n] = [x_n \times a_n] + [x_n \times b_n] = [x_n] \times [a_n] + [x_n] \times [b_n]$$

Field structure

To get from a ring to a field, it is nec­es­sary and suffi­cient to find a mul­ti­plica­tive in­verse for any \([a_n]\) not equal to \([0]\).

Since \([a_n] \not = 0\), there is some \(N\) such that for all \(n > N\), \(a_n \not = 0\). Then defin­ing the se­quence \(b_i = 1\) for \(i \leq N\), and \(b_i = \frac{1}{a_i}\) for \(i > N\), we ob­tain a se­quence which in­duces an el­e­ment \([b_n]\) of \(\mathbb{R}\); and it is easy to check that \([a_n] [b_n] = [1]\).

\([a_n] [b_n] = [a_n b_n]\); but the se­quence \((a_n b_n)\) is \(1\) for all \(n > N\), and so it lies in the same equiv­alence class as the se­quence \((1, 1, \dots)\).

Order­ing on the field

We need to show that:

  • if \([a_n] \leq [b_n]\), then for ev­ery \([c_n]\) we have \([a_n] + [c_n] \leq [b_n] + [c_n]\);

  • if \([0] \leq [a_n]\) and \([0] \leq [b_n]\), then \([0] \leq [a_n] \times[b_n]\).

We may as­sume that the in­equal­ities are strict, be­cause if equal­ity holds in the as­sump­tion then ev­ery­thing is ob­vi­ous.

If \([a_n] = [b_n]\), then for ev­ery \([c_n]\) we have \([a_n] + [c_n] = [b_n] + [c_n]\) by well-defined­ness of ad­di­tion. There­fore \([a_n] + [c_n] \leq [b_n] + [c_n]\).

If \([0] = [a_n]\) and \([0] \leq [b_n]\), then \([0] = [0] \times [b_n] = [a_n] \times [b_n]\), so it is cer­tainly true that \([0] \leq [a_n] \times [b_n]\). <div><div>

For the former: sup­pose \([a_n] < [b_n]\), and let \([c_n]\) be an ar­bi­trary equiv­alence class. Then \([a_n] + [c_n] = [a_n+c_n]\); \([b_n] + [c_n] = [b_n+c_n]\); but we have \(a_n + c_n \leq b_n + c_n\) for all suffi­ciently large \(n\), be­cause \(a_n \leq b_n\) for suffi­ciently large \(n\). There­fore \([a_n] + [c_n] \leq [b_n] + [c_n]\), as re­quired.

For the lat­ter: sup­pose \([0] < [a_n]\) and \([0] < [b_n]\). Then for suffi­ciently large \(n\), we have both \(a_n\) and \(b_n\) are pos­i­tive; so for suffi­ciently large \(n\), we have \(a_n b_n \geq 0\). But that is just say­ing that \([0] \leq [a_n] \times [b_n]\), as re­quired.

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