Let $A$ be a chain. $B,C$ subsets of $A$ s.t $A= B\cup C$. If $B$ and $C$ are well-ordered then A is well-ordered.

Question

My attempt:

Assume that $B$ and $C$ are well-ordered and A is not. Since A is not well-ordered it contains an infinite descending sequence $\{a_i\}_{i=0}^\infty$. If the sequence has a limit $l$ $\in$ $A$ then $l$ is in B or C. With no loss of generality assume $l \in B$, then we can find $N \in \mathbb{N}$ such that $\{a_i\}_{i=N}^\infty$ $\subseteq B$. Then $B$ contains an infinite descending sequence hence it is not well ordered and we have a contradiction.

Question: How can I treat the case $\{a_i\}_{i=0}^\infty$ is divergent?

Answer 1

If $(a_n)_{n\in\Bbb N}$ is a sequence, then either $B_a=\{x\in \{a_n\}_{n\in\Bbb N}\mid x\in B\}$ or $C_a=\{x\in \{a_n\}_{n\in\Bbb N}\mid x\in C\}$ are infinite, WOLG $B_a$ is infinite, then $n_k=\min(m\in\Bbb N\mid a_m\in B_a\land a_m <a_{n_{k-1}})=\min(m\in\Bbb N\mid a_m\in B_a\land m >n_{k-1})$, then $(a_{n_k})_{k\in\Bbb N}$ is infinite decreasing sequence in $B$

Answer 2

Alternative.

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Let $D$ be a non-empty subset of $A$.

Then $D\cap B$ is not empty or $D\cap C$ is not empty.

If both are not empty then let $b$ be smallest element of $D\cap B$ and let $c$ be smallest element of $D\cap C$.

$b$ and $c$ are comparable. If $b\leq c$ then let $d=b$ and if $c\leq b$ then let $d=c$.

Then it can be proved that $d$ is smallest element of $D$.

If only one of the sets $D\cap B$ and $D\cap C$ is not empty then let $d$ be the smallest element of the non-empty set.

Again it can be proved that $d$ is smallest element of $D$.

So $A$ is a totally ordered set such that every non-empty subset has a smallest element.

That means exactly that $A$ is well-ordered.