Total order on irrational numbers

Question

I am trying to prove that the set of irrational numbers $\mathbb{I}$ is totally ordered. I almost completed the proof but stuck at the very end...

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Some theory. The set of irrational numbers $\mathbb{I}$ consist of all possible $A)(A'$ cuts that can be constructed on rational numbers $\mathbb{Q}$. The $A)(A'$ means there is no biggest element in $A$ class and no lowest element in $A'$ class.

Now we define the equal and bigger relations between two arbitrary irrational numbers $\alpha = A)(A'$ and $\beta = B)(B'$.

$$ \alpha = \beta :\Leftrightarrow (A=B) \land (A' = B') \qquad \quad \alpha > \beta :\Leftrightarrow B\subset A $$

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So now I need to prove that the set of irrational numbers is totally ordered:

$$ \boxed{\forall \alpha, \beta \in \mathbb{I} : \ (\alpha = \beta) \lor \Big( (\alpha > \beta) \dot{\lor} (\beta > \alpha)\Big)} $$

Answer 1

I think it's easier to prove trichotomy by showing that if $\ \alpha\ne \beta\ $ and $\ \alpha\not>\beta\ $ then $\ \beta>\alpha\ $.

If $\ \alpha\not>\beta\ $ then, by definition $\ B\not\subset A\ $, and if $\ \alpha\ne\beta\ $ too, it follows that $\ B\not\subseteq A\ $. Therefore $\ B\cap A'\ne\emptyset\ $. Let $\ b\in B\cap A'\ $ and $\ a\in A\ $. Since $\ b\in A'\ $, then $\ b>a\ $ and therefore $\ a\in B\ $. Since this holds for any $\ a\in A\ $, it follows that $\ A\subseteq B\ $, and, since $\ A\ne B\ $, that $\ A\subset B\ $. Therefore $\ \beta>\alpha\ $.

For completeness, I note that I've here used a couple of properties of the Dedekind cut $\ A)(A'\ $ that you didn't mention in your description—namely $\ a'>a\ $ for all $\ a'\in A'\ $ and $\ a\in A\ $, and $\ (x\in A) \wedge (x>y)\implies y\in A\ $.

Answer 2

Pure logical proof.

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Lemma 1. "Simple ="

For arbitrary irrational numbers $\alpha = A)(A'$ and $\beta = B)(B'$ $$ (A=B)\Leftrightarrow (A' = B') \Leftrightarrow \alpha = \beta $$

Proof:

Lets prove that $(A=B)\Rightarrow (A'=B')$. We assume the negation is true: \begin{gather*} (A=B)\land(A'\neq B')\Leftrightarrow\Big( (A=B)\land(A'\nsubseteq B') \Big)\lor\Big((A=B)\land(B'\nsubseteq A')\Big) \end{gather*}

Dealing with the left bracket: $ (A=B)\land(A'\nsubseteq B') $. It means there is an element $a\in A'$ which is not in $B'$, thus $a\in B$ (by definition of $B)(B'$ being a cut). Since $A=B$ we have $a \in A$. Finally, $a\in A \land a\in A'$ which means $A)(A'$ can't be a cut. We have a contradiction. The same reasoning can be done for the expression in the right bracket.

Finally, both expressions in big brackets cause a contradiction, so our proposal that the negation is true was actually false. So we proved $\Rightarrow$. This means both bottom and top classes are equal. By definition of irrationals equality it means $\alpha = \beta$.

The same reasoning can be done when proving $ (A'=B')\Rightarrow(A = B) $. $\blacksquare$

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Lemma 2. "Simple $>$"

For arbitrary irrational numbers $\alpha = A)(A'$ and $\beta = B)(B'$ $$ (B\subset A)\Leftrightarrow (A'\subset B') \Leftrightarrow \alpha > \beta $$

Proof is mostly the same as for Lemma 1.

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Lemma 3

For arbitrary not equal irrational numbers $\alpha = A)(A'$ and $\beta = B)(B'$ the following expression is always true $$ A\subset B \Leftrightarrow B\nsubseteq A $$

Proof:

$\Rightarrow$) $$ A\subset B \Rightarrow B\nsubseteq A $$

We assume that the negation is true $$ (A\subset B) \land (B\subseteq A) $$

Expanding $A\subset B$: $$ \underbrace{(A\subseteq B) \land (B\subseteq A)}_{A=B} \land (A\neq B) $$

So we get a contradiction $(A=B)\land(A\neq B)$. Thus we proved $\Rightarrow$.

$\Leftarrow$)

$$ B\nsubseteq A \Rightarrow A\subset B $$

We assume again that the negation is true $$ (B\nsubseteq A)\land \overline{(A\subset B)} $$

Lets write up what $\overline{(A\subset B)}$ means:

$$ \overline{(A\subset B)} \Leftrightarrow (A\nsubseteq B)\lor(A=B) $$

Now we put this into our expression above:

$$ \Big((B\nsubseteq A)\land(A\nsubseteq B) \Big) \lor \Big( (B\nsubseteq A) \land (A=B) \Big) $$

In the right brackets we have $(A=B)$ which by Lemma 1 means $\alpha = \beta$ and we get a contradiction (since $\alpha \neq \beta$ by condition).

The expression in the left brackets means that $\exists b \in B \land b\notin A$ which means $b\in A'$ and $\exists a \in A \land a\notin B$ which means $a \in B'$. By definition of a cut every element in $A'$ is bigger than every element in $A$ thus $ b > a $. Again, every element in $B'$ is bigger than every element in $B$ thus $ a > b $. So we get $$ (a > b)\land (b>a) $$

This can't be true because $a$ and $b$ are both rational numbers and $\mathbb{Q}$ is totally ordered!

Finally, both expressions in the left and in the right brackets cause a contradiction. Thus we proved $\Leftarrow$. $\blacksquare$

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Proof that $\mathbb{I}$ is totally ordered

$$ \boxed{\forall \alpha, \beta \in \mathbb{I} : \ (\alpha = \beta) \lor \Big( (\alpha > \beta) \dot{\lor} (\beta > \alpha)\Big)} $$

Proof:

Lets assume we have two arbitrary irrational numbers $\alpha$ and $\beta$. If $\alpha = \beta$ then everything is OK.

Otherwise, $\alpha \neq \beta$.

$$ \alpha \neq \beta \Leftrightarrow (A\neq B) \lor (A' \neq B') $$

There are 3 possible cases here:

1. $(A\neq B)$ and $(A' = B')$
2. $(A=B)$ and $(A'\neq B')$
3. $(A\neq B)$ and $(A'\neq B')$

First two cases immediately lead $\alpha = \beta$ (by Lemma 1) which gives a contradiction. Thus (3.) is the only possible case of $\alpha$ and $\beta$ inequality.

$$ \alpha \neq \beta \Leftrightarrow (A\neq B) \land (A'\neq B') $$

Now we expand $(A\neq B)$ and $(A'\neq B')$:

$$ \alpha \neq \beta \Leftrightarrow \Big( (A\nsubseteq B)\lor(B\nsubseteq A) \Big) \land \Big( (A'\nsubseteq B')\lor(B'\nsubseteq A') \Big) $$

Applying Lemma 3 results to everty $\nsubseteq$:

$$ \alpha \neq \beta \Leftrightarrow \Big( (B\subset A)\lor(A\subset B) \Big) \land \Big( (B'\subset A')\lor(A'\subset B') \Big) $$

Now we apply Lemma 2 to both left and right expressions in brackets.

$$ \alpha \neq \beta \Leftrightarrow \Big( (\alpha > \beta) \lor (\beta > \alpha) \Big) \land \Big( (\beta > \alpha) \lor (\alpha > \beta) \Big) $$

$$ \alpha \neq \beta \Leftrightarrow (\alpha > \beta) \lor (\beta > \alpha) $$

The only thing left is to exclude the case when $(\alpha > \beta)\land(\beta > \alpha)$ simultaneously. If so then by definition of $>$ we have

$$ (B\subset A) \land (A\subset B) \Leftrightarrow \underbrace{(B\subseteq A)\land(A\subseteq B)}_{A=B} \land (A\neq B) \Leftrightarrow 0 $$

So, we proved that

$$\alpha \neq \beta \Leftrightarrow (\alpha>\beta) \dot{\lor} (\beta > \alpha) $$

$\blacksquare$

Answer 3

I think you are focusing far too much on symbolic manipulations of logical propositions, treating this like a simple algebra problem which can be "solved" by shuffling symbols around. Consequently you are getting lost in a thicket of symbols.

Let's use the definition of a Dedekind cut that's on Wikipedia. I'll use your $A)(A'$ notation.

Definition. The Dedekind cut $A)(A'$ is a pair of sets of rationals $(A, A')$ such that:

1. $A$ is non-empty and is not all of $\mathbb Q$.
2. If $y\in A$ and $x<y$, then $x\in A$.
3. $A$ contains no maximum element.

The only time in your entire question where you seem to use these assumptions is when you mention that if $A\neq B$ then $A'\neq B'$, where $A)(A'$ and $B)(B'$ are Dedekind cuts. Everything else in your reasoning seems to use no other facts about Dedekind cuts besides that a Dedekind cut is an ordered pair of two sets. This is what I mean when I say you seem to be focused too much on simple symbolic manipulation.

So, we have to show that if two Dedekind cuts $\alpha=A)(A'$ and $\beta=B)(B'$ are different, then either $A\subset B$ or $B\subset A$. Well, since they're different, we know $A\neq B$. We also know that both $A$ and $B$ satisfy the three properties listed above. How can we use that?

Suppose $A\neq B$. If every element of $A$ is in $B$ then $A\subset B$ and we're done. Therefore let $a\in A$ not be in $B$. It is visually obvious that $a$ must be "to the right" of $B$, in the interval "between" $\alpha$ and $\beta$. In particular we ought to have $a$ greater than every element of $B$, and from there, by point (2) above, we can conclude $B\subset A$. So how can we show this rigorously? Try the following:

1. Suppose $a$ is not greater than every element in $B$. Then show there is $b\in B$ such that $b\geq a$.
2. Deduce that $a\in B$, and therefore by contradiction, the assumption in step 1 was incorrect and $a$ is greater than every element in $B$.
3. Conclude that that $B\subset A$.