I learn domain theory and stack in definition of algebraic poset. Recall $P$ is algebraic if for every $x\in P$,the set of compact element $y$ below $x$ is directed and has $x$ as least upper bound. From this definition,I have a question: whether every elements of algebraic poset should be compact element? I think the answer yes. Since if $x \in P$ is not a compact element and $c_1,c_2,\cdots,c_n$ are compact elements with $c_i\neq x$ for all $i$ and below $x$ then $\sup\{u:u\ll u \leq x\}=c_i$ for some $i$.Hence $P$ is not algebraic. Is it true? Can someone improve my answer or give example?
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Another example is $D=\mathcal{P}(\mathbb{N})$, the power set of $\mathbb{N}$, ordered by set inclusion. Then $x\in D$ is compact iff $x$ is finite. Since any element of $D$ is the union of its finite subsets, $D$ is algebraic.
Your argument would work in a finite poset. However, in that case, every element is compact.
Bernard Hurley
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Your proof is not a proof, why should your sup equal some $c_i$? The answer is no in general. For instance, the collection of convex subsets of $\mathbb{R}^2$ ordered by inclusion is an algebraic poset, and the compact elements are the convex subsets generated by a finite set (i.e. convex polytopes). And obviously not all convex subsets are convex polytopes.
polmath
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