# Show that the set of all down-closed subsets of a partially ordered set is chain-complete

How can I show that given a partially ordered set $(A\le)$ with bottom, the set $\operatorname{dc}(A)$ that contains all subsets of $A$ that are down-closed is chain-complete?

By "partially ordered set" I mean a set with a partial order relation $\le$ over which is reflexive, antisymmetric, and transitive, i.e., which satisfies for all $a, b,$ and $c$ in $P$: $a\le a$ (reflexivity: every element is related to itself); if $a\le b$ and $\le a$, then $a = b$ (antisymmetry: there exists at most one relation between two distinct elements); if $\le b$ and $b\le c$, then $a\le c$ (transitivity: if a first element is related to a second element, and, in turn, that element is related to a third element, then the first element is related to the third element).

By "bottom" I mean the least element of the set.

By "down-closed" I mean that for each element $x$ in a set $P\subseteq A$ if $y\in A$ and $y\le x$ then $y\in P$.

By "chain-complete" I mean a poset in which each chain has a least upper bound.

By "chain" I mean a set in which for all elements $x,y\in the chain we have either$x\le y$or$y\le x.\operatorname{dc}(P)$is ordered by set inclusion. • Please define what you mean by "partially ordered set", "bottom", "down-closed", "chain", and "complete", and the ordering relation on the set of down-closed sets. I thought I knew what all those terms meant, but according to my understanding the statement you want to prove is blatantly false, so I guess your terminology is somehow different from what I'm used to. Is$\operatorname{dc}(A)$ordered by set inclusion? Is a chain a totally ordered set? If$A$is not totally ordered, do you really think$\operatorname{dc}(A)$will be totally ordered? – bof Dec 15, 2016 at 10:11 • If$a$and$b$are incomparable elements of$A,$are not$\{x\in A:x\le a\}$and$\{x\in A:x\le b\}$incomparable elements of$\operatorname{dc}(A)?$So how is$\operatorname{dc}(A)$a chain? – bof Dec 15, 2016 at 10:15 • Thank you. (1) These definitions should be incorporated into the body of your question (using the edit button) rather than appended as comments. (2) They are what I expected, expect for the definition of "chain". (3) The definition of "chain" is circular: "For 'chain' I mean a poset in which each chain have least upper bound." (4) You still haven't been defined what you mean by "complete". – bof Dec 16, 2016 at 1:13 • Thank you I incorporeted the definitions in the question as you suggested. For me "complete" doesn't mean nothing alone. I mean "chain complete" for en.wikipedia.org/wiki/Chain_complete. I hope this help. – Mark Dec 16, 2016 at 7:44 • OK. So, in the title and the first sentence of your question, by "is a complete chain" you mean "is chain complete", right? – bof Dec 16, 2016 at 8:24 ## 1 Answer Consider a chain$\mathcal C$in$\operatorname{dc}(A);$we have to show that$\mathcal C$has a least upper bound in$\operatorname{dc}(A).$Let$U=\bigcup\mathcal C,$the union of all the members of$\mathcal C.$I claim that$U\in\operatorname{dc}(A),$i.e., that$U$is down-closed. Suppose$y\in A$and$y\le x\in U;$I have to show that$y\in U.$Since$x\in U=\bigcup U,$we have$x\in P$for some set$P\in U.$Since$P$is down-closed, we have$y\in P\subseteq U,$so$y\in U.$Clearly,$U=\bigcup\mathcal C$is the least upper bound of$\mathcal C$in$\operatorname{dc}(A).$• dc(A) and dc(P) are the same set? – Mark Dec 16, 2016 at 9:05 • Typo. I'm used to using$P\$ for a poset. I'll correct it.
– bof
Dec 16, 2016 at 9:34