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we know that A Set $S$ together with a relation $\ge$ which is both $transitive$ and $reflexive$ such that for any two elements a,b in $S$, there exists another element c in $S$ with c $\ge$ a and c $\ge$ b In this case, the relation $\ge$ is said to "direct" the set.

But can we call A Set $S$ is A FULL directed Set : If Every subset From $S$ is a directed Set?

I need an example of a directed set but it is not A Full directed one.

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    $\begingroup$ It seems like a full directed set is necessarily a chain. $\endgroup$ – amrsa Dec 23 '17 at 12:18
  • $\begingroup$ after search yeb it is a chain can you help me with an example $\endgroup$ – Yaser Tarek Dec 23 '17 at 12:22
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    $\begingroup$ Well, in that case, any directed set that is not a chain will be an example $\endgroup$ – amrsa Dec 23 '17 at 12:23
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    $\begingroup$ Would this work? $S=P\bigl(\{0,1\}\bigr)=\bigl\{\emptyset,\{0\},\{1\},\{0,1\}\bigr\}$ with the inclusion as the transitive and reflexive relation. $S$ is clearly directed (the union plays the role of the join). Then the subset $\bigl\{\{0\},\{1\}\bigr\}$ of $S$ is not directed. $\endgroup$ – gniourf_gniourf Dec 23 '17 at 12:23
  • $\begingroup$ @gniourf_gniourf That would work. And without the empty set, it would work too (this would be the minimal example) $\endgroup$ – amrsa Dec 23 '17 at 12:26

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