One Succeeds Zero in Well-Ordered Integral Domain

Theorem

Let $\struct {D, +, \times, \le}$ be a well-ordered integral domain.

Let $0$ and $1$ be the zero and unity respectively of $D$.

Then $0$ is the immediate predecessor of $1$:

$0 < 1$

$\neg \exists a \in D: 0 < a < 1$

Aiming for a contradiction, suppose there exists an element $a \in D$ such that $0 < a < 1$.

Let us create the set $S$ of all such elements:

$S = \set {x \in D: 0 < x < 1}$

We know $S$ is not empty because it has already been asserted that $a$ is in it.

And all the elements in $S$ are strictly positive by definition.

Because $D$ is well-ordered it follows that $S$ has a minimal element, which we will call $m$.

Thus we have $0 < m < 1$.

$0 < m \times m < m$

Thus $m \times m \in S$ is an element of $S$ which is smaller than $m$.

But $m$ was supposed to be the minimal element of $S$.

From this contradiction we deduce that there can be no such element $a$ such that $0 < a < 1$.

$\blacksquare$

1969: C.R.J. Clapham: Introduction to Abstract Algebra ... (previous) ... (next): Chapter $2$: Ordered and Well-Ordered Integral Domains: $\S 8$. Well-Order: Theorem $12$