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I am currently reading Terence Tao's "Analysis I" and while progressing through the book, the reader is repeatedly asked to prove trichotomy properties of order for the natural numbers $\mathbb{N}$, the whole numbers $\mathbb{Z}$, the rational numbers $\mathbb{Q}$, the real numbers $\mathbb{R}$, and finally, the extended real numbers $\mathbb{R}^{*}$.

The construction of the real numbers.

In Tao's book, the construction of real numbers is defined by means of formal limits of rational Cauchy sequences (formal in the sense that limits are not properly introduced until later chapters, but these are later shown to be equivalent). The notion of order is defined by taking $x < y$ if the number $x - y$ is negative, which translates into the Cauchy sequence associated to $x - y$ being negatively bounded away from zero.

Searching yielded:

The questions I have found answered here on the same topic have been from a different perspective, mostly set theoretic, and I haven't found a way of applying what has been said.

Trichotomy of order:

Just to reiterate, the statement goes like this:

Let $x, y$ be real numbers. Then, exactly one of the statement $x < y$, $x = y$ or $x > y$ is true.

The proofs of these statements are typically divided into two steps:

  1. Show that at most one of the statements can be true at any given time. Do this by assuming that two of them hold simultaneously, and deduce a contradiction.
  2. Show that at least one statement must hold for any two real numbers $x, y$.

What I struggle with:

My struggles lies mainly with number 2, showing the fact that at least one statement must hold. I'm not quite sure where to approach this. I've tried considering both direct, contrapositive and proof by contradiction, however doing the logical manipulations to arrive at the proper implications have not yielded any success. It is probably herein my problem lies. In trying to convert the following statement to a contrapositive equivalent statement

$$ \begin{align*} \forall x, y \in \mathbb{R} \implies \left[ (x < y) \lor (x = y) \lor (x > y) \right] && (direct) \\ \end{align*} $$ i have to rely on the fact that the trichotomy of order is true, which I can't since that is what I am trying to show.

Any hints for where I should start is greatly appreciated! Thanks in advance.

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  • $\begingroup$ How you prove trichotomy will depend at least on what you are proving it for. However, depending on what you have developed, a nice simplification is to use that $x<y \Leftrightarrow x-a<y-a$, and so subtracting $x$ from both sides, we can reduce trichotomy to comparison with $0$. $\endgroup$ – Aaron Jul 27 '16 at 13:59
  • $\begingroup$ @Aaron Thanks, I'll have a look at once and see what I can come up with! :) $\endgroup$ – Ivar Stangeby Jul 27 '16 at 14:09
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You could try showing that $x \neq y$ implies $x < y$ or $x > y$. Consider Cauchy sequences $(x_n)$ and $(y_n)$ representing $x$ and $y$ respectively. Remember that "$x = y$" means: for any $\delta > 0$ there exists a threshold $N > 0$ past which $\lvert x_n - y_m \rvert < \delta$ (i.e. this holds for $n,m > N$). The negation of that statement, namely "$x \neq y$", thus reads: there exists some $\delta > 0$ such that for any threshold $N > 0$, we can find $n,m > N$ such that $\lvert x_n - y_m \rvert \geq \delta$.

We can use the fact that the threshold is arbitrary to our advantage: the idea is that we will let $N$ vary to construct subsequences of $(x_n)$ and $(y_m)$. We will prove the result for these subsequences. First set, say, $N = 1$. This yields indices $n_1,m_1 > 1$. Now pick a threshold $N$ larger than $\max(n_1,m_1)$. This yields indices $n_2,m_2 > N \geq n_1,m_1$. If we keep repeating this process, we obtain subsequences $(x_{n_k})$ and $(y_{m_k})$ (still Cauchy) of our original sequences satisfying $\lvert x_{n_k} - y_{m_k} \rvert \geq \delta$ for all $k$. It remains to show that $x_{n_k} < y_{m_k}$ for all large enough $k$ (or $x_{n_k} > y_{m_k}$ for all large enough $k$). I will reindex these sequences as $(x_n)$ and $(y_n)$ for simplicity of notation. By their Cauchy property, there exists some $N > 0$ such that $\lvert x_n - x_m \rvert < \delta/2$ and $\lvert y_n - y_m \rvert < \delta/2$ when $n,m > N$. Now suppose for the sake of contradiction that $x_n < y_n$ but $x_m \geq y_m$ for some $n,m > N$. The inequality $\lvert x_n - y_n \rvert \geq \delta$ yields $y_n \geq x_n + \delta$, which in turn yields $$x_m \geq y_m > y_n - \frac{\delta}{2} \geq x_n + \frac{\delta}{2},$$ which contradicts the hypothesis $\lvert x_n - x_m \rvert < \delta/2$.

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Theorem. Let $x := (x_i)$ and $y := (y_i)$ be Cauchy sequences of rational numbers. Then either

  1. $\forall \epsilon > 0 : \exists N : \forall n > N : |x_n - y_n| < \epsilon$,
  2. $\exists \epsilon > 0 : \exists N : \forall n > N : x_n + \epsilon \leq y_n$, or
  3. $\exists \epsilon > 0 : \exists N : \forall n > N : x_n \geq y_n + \epsilon$.

I'm not intimately familiar with Tao's definitions, but I hope this theorem satisfies your needs.

Proof. Assume property 1 fails. We will then show that either property 2 or property 3 holds. The failure of property 1 means we can find $\epsilon > 0$ such that $$\forall N : \exists n > N : |x_n - y_n| \geq \epsilon \text{.}$$ By the Cauchiness of $x$ and $y$, find $N$ large enough that \begin{align*} &\forall m,n > N : |x_m - x_n| < \frac{\epsilon}{3} \\ &\forall m,n > N : |y_m - y_n| < \frac{\epsilon}{3} \text{.} \end{align*}

By a statement above, take $n > N$ such that $|x_n - y_n| \geq \epsilon$. So, either $x_n + \epsilon \leq y_n$ or $x_n \geq y_n + \epsilon$. Without loss of generality, assume $x_n + \epsilon \leq y_n$; we will be showing that property 2 holds. (If $x_n \geq y_n + \epsilon$, we'd show that property 3 holds by the same argument.)

It remains to show that, for all $m > N$, $x_m + \frac{\epsilon}{3} \leq y_m$. (We will use $\frac{\epsilon}{3}$ in our demonstration of property 2.) Let $m > N$. Conclude: \begin{align*} x_m + \frac{\epsilon}{3} &< x_n + \frac{\epsilon}{3} + \frac{\epsilon}{3} \\ &\leq y_n - \epsilon + \frac{\epsilon}{3} + \frac{\epsilon}{3} \\ &= y_n - \frac{\epsilon}{3} \\ &< y_m \text{.} \end{align*}

We have therefore shown property 2, as required. $\square$

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