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 com­mu­ta­tivegroup 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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