How can I prove chain-completeness of a poset? In particular this is the problem:
$\langle D, \leq \rangle$ is a chain-complete poset (every chain has a least upper bound). Prove or disprove $\langle convex(D),\subseteq \rangle$ is a chain-complete poset.
Where $convex(D)=\{S \subseteq D |\forall x,y\in S, z\in D. (x\leq z \leq y \Rightarrow z\in S\}$
My naive opinion, probably wrong:
I think, it's quite obvious that is true, and given a chain $c_1,c_2,...,c_n \in convex(D)$, $c_1\subseteq c_2...\subseteq c_n$ we just take $c_n$ as LUB.
To show that $\langle \operatorname{convex}(D),\subseteq\rangle$ is chain-complete, it suffices to show that for any chain $C_i$ in $\operatorname{convex}(D)$ with $C_i\subseteq C_{i+1}$ the union $C=\bigcup_iC_i$ is also in $\operatorname{convex}(D)$, since this can then serve as the least upper bound. This is trivial if the chain is finite, since in that case $C$ is its last element.
To prove that $C\in\operatorname{convex}(D)$ also for infinite chains, let $x,y\in C$ be given. Then $x\in C_i$ and $y\in C_j$; without loss of generality assume $i\le j$. Then also $x\in C_j$, so $x,y\in C_j$ and thus for all $z\in D$ we have $x\le z\le y\Rightarrow z\in C_j$ and thus also $x\le z\le y\Rightarrow z\in C$. Thus $C\in \operatorname{convex}(D)$.