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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.

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  • $\begingroup$ 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? $\endgroup$
    – bof
    Dec 15, 2016 at 10:11
  • $\begingroup$ 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? $\endgroup$
    – bof
    Dec 15, 2016 at 10:15
  • $\begingroup$ 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". $\endgroup$
    – bof
    Dec 16, 2016 at 1:13
  • $\begingroup$ 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. $\endgroup$
    – Mark
    Dec 16, 2016 at 7:44
  • $\begingroup$ OK. So, in the title and the first sentence of your question, by "is a complete chain" you mean "is chain complete", right? $\endgroup$
    – bof
    Dec 16, 2016 at 8:24

1 Answer 1

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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 $\operatorname{dc}(A).$

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  • $\begingroup$ dc(A) and dc(P) are the same set? $\endgroup$
    – Mark
    Dec 16, 2016 at 9:05
  • $\begingroup$ Typo. I'm used to using $P$ for a poset. I'll correct it. $\endgroup$
    – bof
    Dec 16, 2016 at 9:34

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