# Proof of the trichotomy of real numbers using Cauchy sequences.

I'm trying to prove the next proposition but I have stuck in one point of the proof. I would really appreciate some help, thanks in advance.

Definition: We shall say that a Cauchy sequences is bounded away from zero, if there is a positive rational number $$c$$ such that $$|a_n|\ge c$$ for every n.

We say that $$x$$ positive if there is a Cauchy sequence which is positively bounded away from zero and is the formal limit of the Cauchy sequences is $$x$$, i.e., if $$c$$ is a rational number such that $$c>0$$, then $$a_n \ge c$$ for each n. Similarly $$x$$ is said to be negative if there is a Cauchy sequence which is negatively bounded away from zero, .i.e., $$c$$ is a rational number such that $$c>0$$, then $$a_n \le -c$$ for each n.

Proposition: For every real number $$x$$ exactly one of the following three statements is true: (a) $$x$$ is zero, (b) $$x$$ is positive or (c) $$x$$ is negative.

Proof: First we need to show that at most one of (a), (b) and (c) hold.

Suppose that $$x$$ is zero. Then $$x$$ can be written as a Cauchy sequences which is equivalent to the zero sequence, i.e., $$\langle a_n \rangle \sim \langle 0\rangle$$ then $$\langle a_n \rangle$$ is not bounded away from zero, and so is neither positively or negatively bounded away from zero.

Now suppose that $$x$$ is positive by definition there is a Cauchy sequence such that is positively bounded away from zero and its formal limit is $$x$$. Let $$\langle \, a_n \, \rangle$$ be such sequence. So, $$a_n \ge c_P$$ for each n where $$c_P$$ is a positive rational number.

By the sake of the contradiction we may assume that also $$x$$ is negative in other words that there exists a Cauchy sequence such that is negatively bounded away from zero and its formal limit is $$x$$. So $$\,b_n \le -c_N$$ for each n where $$c_N$$ is a positive rational number. Let $$\,c = \text{min} (c_P ,c_N)$$. Since the formal limit of both sequences is $$x$$ then $$\langle a_n \rangle$$ is equivalent to $$\langle b_n \rangle$$, i.e., for every $$\varepsilon > 0$$ there exists a $$N_{\varepsilon}$$ such that $$|a_n-b_n| \le \varepsilon$$ for each $$n \ge N_{\varepsilon}$$. We choose that $$\varepsilon = \frac{c}{2}$$. But $$|a_n-b_n|$$ is always positive and so $$|a_n-b_n|= a_n-b_n > 2c$$. Then, both sequences are not equivalent which contradicts that the formal limit of each one is $$x$$.

To conclude the proof we need to show that at least one of (a), (b) and (c) always hold.

Either $$x$$ is zero or a nonzero real number. Suppose that $$x$$ is a nonzero real number. Let $$\langle b_n\rangle$$ be the Cauchy sequence for which $$x$$ is the formal limit. Since $$x$$ is a nonzero real, $$\langle b_n\rangle$$ cannot be equivalent to $$\langle 0\rangle$$, so is not eventually $$\varepsilon- \text{close}$$ to the zero sequence for each $$\varepsilon$$. At least exists one $$\varepsilon$$ such that $$|b_n -0| > \varepsilon$$. Let us fix this $$\varepsilon$$.

Moreover, it is a Cauchy sequence. So we have that there is an $$N$$ such that $$|b_k -b_j| \le \frac{\varepsilon}{2}$$ for every $$k,j \ge N$$. On the other hand we cannot have that $$|b_k|\le \varepsilon$$ if so, this would imply that is eventually equivalent to the zero sequences which contradicts our assumption. Thus, there must be some $$|b_k|> \varepsilon$$. It follows that $$|b_j| \ge \frac{\varepsilon}{2}$$ for every $$j \ge N$$.

Here is where I'm stuck I show that is bounded away from zero but I'm not completely sure of how to show that either is positively or negatively for every $$n \ge N$$. Could somebody give me a hint of how to do it, please?

Thanks

You actually need to say a little more before you choose $\epsilon$. You know that $\langle b_n\rangle\not\sim\langle 0\rangle$, so you know that there is an $\epsilon>0$ such that $|b_n|\ge\epsilon$ for infinitely many $n\in\Bbb N$; you don’t immediately know that $|b_n|>\epsilon$ for all sufficiently large $n$. To get that you must use the fact that $\langle b_n\rangle$ is Cauchy. Let $N_{\epsilon/2}$ be as in your argument, and fix $N\ge N_{\epsilon/2}$ such that $|b_N|\ge\epsilon$; then $|b_k-b_N|<\frac{\epsilon}2$ for $k\ge N$, so $|b_k|>\frac{\epsilon}2$ for $k\ge N$.

Now suppose that $k,\ell\ge N$, $b_k>0$, and $b_\ell<0$; what two contradictory conclusions can you draw about $|b_k-b_\ell|$?

• mmhh, since $b_k-b_\ell > 0$ clearly $|b_k-b_\ell| = b_k-b_\ell = |b_k| + |b_\ell| > \varepsilon$ which contradicts the fact that $|b_k-b_\ell|< \frac{\varepsilon}{2}$ is that right? – Jose Antonio Sep 13 '13 at 20:02
• @Jose: Yep, you’ve got it. – Brian M. Scott Sep 13 '13 at 20:04
• So, for each $k,\ell\ge N$, $b_k, b_\ell>0$ or $b_k, b_\ell<0$ if were not we have the above contradiction.Is that a correct conclusion? Brian as always thanks :) – Jose Antonio Sep 13 '13 at 20:09
• @Jose: Yes: once you get out as far as $N$, all terms must be on the same side of $0$. You’re welcome! – Brian M. Scott Sep 13 '13 at 20:13

Copying your derivation (this is part of proof of Lemma 5.3.14 in Tao's Analysis I book):

Let $ε>0$ be such that $\langle b_n \rangle$ is not eventually ε-close to zero. By $\langle b_n \rangle$ being Cauchy, there is an $N$ such that $|b_k−b_j|≤ε/2$ for every $k,j≥N$. On the other hand there exists $n_0 \geq N$ for which $|b_{n_0}| > ε$ (otherwise $b_n$ would be eventually ε-close to 0, a contradiction). So $|b_{n_0}−b_j|≤ε/2$ for all $j \geq N$.

Picking up where you left off:

Since $b_{n_0}$ is a rational number, and $|b_{n_0}| > ε >0$, we can only have exactly one of either $b_{n_0} > ε > 0$, or $b_{n_0} < -ε < 0$ (by trichotomy of rationals). Now since $\forall j \geq N$, $|b_{n_0}−b_j|≤ε/2$ (i.e., $b_j$ is only $ε/2$ away from $b_{n_0}$), it's not hard to see that if $b_{n_0} > ε > 0$, then $\forall j \geq N, b_j \geq ε/2 > 0$; similarly if $b_{n_0} < -ε < 0$, then $\forall j \geq N, b_j \leq -ε/2 < 0$. This shows that the non-zero sequence $b_n$ is eventually either positively bounded away from zero, or negatively bounded away from zero (exactly one of which is true). To finish the proof, in the first case, define $\langle a_n \rangle$ by setting $a_n:=ε/2$ for $n<N$, and $a_n:=b_n$ for $n\geq N$; in the second case, define $\langle a_n \rangle$ by setting $a_n:=-ε/2$ for $n<N$, and $a_n:=b_n$ for $n\geq N$; it's then easy to show that $\langle a_n \rangle=\langle b_n \rangle$, and $a_n$ is positively (negatively) bounded away from zero, and therefore $x$ is positive (negative).