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.
 A: 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 $n, m\geq N_1$, $|a_n-a_m|<c/2$. 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 $n \geq N$, $|b_n-0|=|b_n|<c$. 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).
