Ordering of R (via Dedekind Cuts)

Proof

We can order elements on $\mathbb{R}$ where $A\le B$ for our cuts $A,B$ means $A\subseteq B$. This is automatically transitive:

Now we can verify $\mathrm{\forall }A,B\in \mathbb{R}$ exactly one of $A<B,A=B,A>B$ holds. To show this, let $A,B\in \mathbb{R}$.

1. ($A<B$): Assume $A\ne B,A\ngtr B$. We'll show $A<B$. Since $A\ne B$ then it's sufficient to show that $A\subseteq B$ (automatically a proper subset). Pick an arbitrary $a\in A$. Since we know $B\nless A$ then that implies that $B⊈A$ so then $\mathrm{\exists }b\in B$ such that $b\notin A$. If we show that $a<b$ then we can finish. Using the properties from $\mathbb{Q}$ for order:

• We know that $a\ne b$ since $a\in A$ while $b\notin A$.
• If $b<a$ then by the definition of cuts then $b\in A$ since $A$ is a cut. Contradiction.
• Therefore $a<b$.

Therefore since $a<b$ and $B$ is a cut, then $a\in B$ consequently. Thus $A\subseteq B$. Thus $A<B$.

2. Is it possible that $A=B$ and $B<A$? No since if we assume both are true then $A=B$ while $B\subseteq A$ so then clearly $A\subseteq B$ which implies $A<B$ which is a contradiction as we assume that has to be false. Thus only one can be true at a time.
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