# Proposition 5.4.4. in Tao

I am trying to prove the following proposition in Tao's analysis textbook.

For ever real number $$x$$, exactly one of the following three statements is true: (a) $$x$$ is zero; (b) $$x$$ is positive; (c) $$x$$ is negative. A real number $$x$$ is negative if and only if $$-x$$ is positive. If $$x$$ and $$y$$ are positive, then so are $$x + y$$ and $$xy$$.

I am unsure on how to approach the first part. Tao defines real numbers as limits of Cauchy sequences of rationals, though without defining just yet what a limit is. He defines a positive real number as one that can be written as the limit of a Cauchy sequence of rationals positively bounded away from $$0$$ and a negative real number as one that can be written as the limit of a Cauchy sequence negatively bounded away from $$0$$. We have a law of trichotomy for the rationals, which could be extended to every element of the sequence, perhaps, to say that, upon throwing out a finite number of terms, the sequence is either identically zero, positively bounded away from zero, or negatively bounded away from $$0$$, and thus $$x$$ is either $$0$$, positive, or negative. I am still unsure on how to formalize this, though, or whether I am on the right track.

The second statement seems rather straightforward. If $$x$$ is negative it is negatively bounded away from $$0$$: we have $$x = \text{LIM}_{n \to \infty} a_n$$, and $$\exists - c < 0$$ (-$$c$$ rational) such that $$a_n \leq -c$$, Therefore, $$-x = \text{LIM}_{n \to \infty} -a_n$$, where we have, multiplying through by $$-1$$, that $$\exists c > 0$$ such that $$-a_n \geq c$$, meaning $$x$$ is positively bounded away from $$0$$ and is therefore positive. The opposite implication is similar.

As for the third part: let $$x = \text{LIM}_{n \to \infty} a_n$$ and $$y = \text{LIM}_{n \to \infty} b_n$$. $$x$$ and $$y$$ are positive, meaning $$a_n$$ and $$b_n$$ are positively bounded away from $$0$$, so $$\exists c > 0, a_n \geq c$$ and $$\exists d > 0, b_n \geq d$$. Thus, for any $$n$$, $$a_n + b_n \geq c + d$$, and since the positive reals are closed under addition, $$c + d > 0$$ and $$a_n + b_n$$ is also positively bounded away from zero, so $$x + y$$ is also positive. The fourth part is similar, but with a product, $$cd$$, in lieu of a sum.

Assuming I have not made a mistake or omission, I believe that I understand how to write the later parts of the problem, but the first part is still quite confusing to me. Any help or insights would be greatly appreciated.

Here's a [messy] partial outline:

Let $$y$$ be a real number.

We'll do this in two parts: 1) show that $$y$$ can be labeled positive, negative, or 0 2) show that $$y$$ cannot be simultaneously: 0 and positive; 0 and negative; or positive and negative.

1) Suppose first that $$y \not\equiv 0$$. Then by Lemma $$5.3.14$$ [in my copy of the book]

Let $$x$$ be a non-zero real number. Then $$x=LIM_{n\rightarrow \infty} a_n$$ for some Cauchy sequence $$(a_n)_{n = 1}^{\infty}$$ which is bounded away from zero.

So we can say $$y=LIM_{n\rightarrow \infty}a_n$$, where there exists some rational $$c>0$$ such that for all $$n$$, $$|a_n|\geq c$$.

a) Suppose that $$y$$ is not negative. We'll show that $$y$$ must be positive.

Lemma 1: Let $$(a_n)_{n=1}^{\infty}$$ be Cauchy. Fix $$m$$. Then the sequence $$(b_n)_{n=1}^{\infty}$$, where $$b_n=a_{m+n}$$, is Cauchy and equivalent to $$(a_n)_{n=1}^{\infty}$$.

[The proof of this lemma relies on the epsilon-N definition of 'Cauchy'-ness]

Claim 1: There exists $$N$$ such that for all $$n\geq N$$, $$a_n\geq c$$.

Proof of Claim 1: First, we know that for all $$N$$, there must exist some $$n\geq N$$ such that $$a_n>0$$, since otherwise we would have some $$N_0$$ such that for all $$n\geq N_0$$, we have $$a_n\leq -c<0$$, and so $$y=LIM (a_n)_{n=N_0}^{\infty}$$, where the sequence $$(a_n)_{n=N_0}^{\infty}$$ is Cauchy and negatively bounded away from zero (making $$y$$ negative).

Now suppose the claim is false.

Then, for all $$N$$, there must be some $$n1, n2\geq N$$ such that $$a_{n1}>c>0$$ and $$a_{n2}<-c<0$$. Fix the $$N_1$$ such that for all $$n, m\geq N_1$$, $$|a_n-a_m|. But then our aforementioned $$n1$$ and $$n2$$ give us a contradiction, since $$a_{n1}-a_{n2}>2c$$.

This proves Claim 1.

Hence, by Lemma 1, $$y=LIM (a_n)_{n=N}^{\infty}$$, where the sequence $$(a_n)_{n=N}^{\infty}$$ is positively bounded away from zero.

b) Now suppose $$y$$ is not positive. We'll show that $$y$$ must be negative. [the proof of this is similar to a]

So right now we know that any real $$y$$ must be at least one of: positive, negative, zero.

2) We now want to show $$y$$ cannot be more than one of our three options.

a) First, if $$y=0=LIM b_n$$, then for any $$c>0$$, there exists $$N$$ such that for all $$n \geq N$$, $$|b_n-0|=|b_n|. So if $$y=0$$, $$y$$ cannot be bounded away from zero, and so $$y$$ cannot be positive or negative.

b) Now, let $$y$$ be positive. Then there exists a rational $$c>0$$ so that $$y=LIM b_n$$ for some sequence $$b_n>c$$. Since $$0=LIM 0$$, and we always have $$|b_n-0|>c>0$$, the sequences $$(b_n)$$ and $$(0)$$ cannot be equivalent, and so $$y$$ is nonzero. Suppose for the sake of contradiction that $$y$$ is both negative and positive. Then there exists a rational $$d>0$$ such that $$y=LIM e_n$$ for some sequence such that $$e_n<-d$$ for all $$n$$. But then we have $$|b_n-e_n|>c+d>0$$ for all $$n$$, and so $$b_n$$ and $$e_n$$ cannot be equivalent sequences, a contradiction.

c) The case where $$y$$ is negative mimics 2b).