Union of Initial Segments is Initial Segment or All of Woset/Proof 1

Theorem

Let $\struct {X, \preccurlyeq}$ be a well-ordered non-empty set.

Let $A \subseteq X$.

Let:

$\ds J = \bigcup_{x \mathop \in A} S_x$

be a union of initial segments defined by the elements of $A$.

Then either:

$J = X$

or:

$J$ is an initial segment of $X$.

Suppose the hypotheses of the theorem hold.

If $J = X$ then the theorem is satisfied.

Assume $J \ne X$.

Then $X \setminus J$ is non-empty.

Thus $X \setminus J$ has a smallest element; call it $b$.

We claim that $b \preccurlyeq y$ for every $y \in J$.

Aiming for a contradiction, suppose there exists a $y \in J$ with $b \preccurlyeq y$.

Then there exists some $x_0 \in A$ with $y$ in the initial segment $S_{x_0}$.

Thus $b \in S_{x_0}$ and so $b \in J$.

This contradicts the fact that $b \in X \setminus J$.

Thus there cannot exist a $y \in J$ with $b \preccurlyeq y$.

Any element strictly preceding $b$ is in $J$, by choice of $b$.

Also, $b \notin J$, because $b \in X \setminus J$.

From above, any element strictly succeeding $b$ is not in $J$.

Thus $J = S_b$ by definition of initial segment.

$\blacksquare$