Union of Initial Segments is Initial Segment or All of Woset

Theorem
Let $X$ be a well-ordered non-empty set.

Let $A \subseteq X$.

Let:


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

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

Then either:


 * $J = X$

or:


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

Proof
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.

Furthermore, $X \setminus J$ is itself well-ordered.

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

There cannot exist a $y \in J$ with $y \succ b$.

To see this, observe that $y \succ b$ would imply the existence of some $x_0 \in A$ with $y$ in the initial segment $S_{x_0}$.

This would imply $b \in S_{x_0}$, contradicting the minimality of $b$.

Hence $J \subseteq S_{x_0}$ for some $x_0 \in A$.

By the definition of union:


 * $\displaystyle S_{x_0} \subseteq \bigcup_{x \mathop \in A} S_x = J$

By the definition of set equality, $J$ is itself an initial segment.