Union of Initial Segments is Initial Segment or All of Woset

Theorem
Let $\left({X, \preccurlyeq}\right)$ be a well-ordered non-empty set.

Let $A \subseteq X$.

Let:


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

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

Then either:


 * $J = X$

or:


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

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.

By Subset of Well-Ordered Set is Well-Ordered, $X \setminus J$ is itself well-ordered.

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

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 $b \preccurlyeq y$.

By virtue of $J \ne X$, there must be some $a \in A$ with $J \subseteq S_a$.

Also, by the definition of set union:


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

By the definition of set equality, $J$ is itself an initial segment, as it is both a subset and superset of $S_a$ as defined above.