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The answer is no. In the Boolean algebra $P(\omega)/\text{Fin}$ every strictly descending sequence $A_0>A_1>A_2>\cdots$ has a nonzero lower bound, by the famous construction of Hausdorff. Namely, one …

As Sam mentioned in the comments, the answer to question 1 is yes. One can map every condition $p$ in the partial order to the lower cone, the set $S_p=\{q\in P\mid q\leq p\}$ of conditions below $p$. …

One of the standard topologies to consider would be the lower-cone topology, whose basic open sets are the lower cones $i{\downarrow}=\{j\mid j\leq i\}$. In this topology, the open sets are exactly th …

It is equivalent to AC. Consider any collection $A$ of nonempty sets, and let $\newcommand\P{\mathbb{P}}\P$ be the set of partial choice functions, so that $p\in\P$ if and only if $p$ is a partial fun …

Here is a counterexample showing that the quotient of $\text{Fct}(X,Y)$ is not necessarily a lattice. Let $X=\{0,1,2\}$ and let $Y=\{0,1\}$. Let $g(0)=g(1)=0$ and $g(2)=1$, while $f(0)=0$ and $f(1)=f …

This is a fantastic question! I spent the whole morning thinking about it, and I finally have a solution. The answer is no, not necessarily. To build a counterexample, I claim first that there is a …

The answer is no, because in general there can be more proper coverings than partitions, which will prevent any injective mapping from coverings to partitions. For example, if $X$ is countably infini …

If there are only finitely many points, it is a lattice, and it is always an upper semi-lattice. But in the infinite case, it is not a lattice. François's comment on Fedor's answer shows that is it a …

Yes, and in fact, most familiar topologies do not have a pseudo-complement. To see this, notice that that it often happens with a topology $\tau$ on a set $X$ that there are non-open sets $A$ and $B$ …

Very nice question! They are not isomorphic. What I claim is that when we take the quotient with respect to density, there is a countably infinite antichain above $0$ having a minimal upper bound, b …

The answer is no. To see this, consider the bottoms of $K$ and $\omega^\omega$ under the pointwise $\leq$ order you have described. Both structures have a least element: The constant $0$ function i …

No, clearly not, because you could put junk on top. But even if you avoid this by insisting that the map is surjective, there are counterexamples. Consider the map of Eric Wofsey's recent answer, wh …

Here is a partial answer in the case of complete Boolean algebras, which I claim do all arise as Lindenbaum algebras. Let $\mathbb{B}$ be a complete Boolean algebra, and suppose that $M$ is any $\math …

Every separative partial order $P$ has a unique completion as a complete Boolean algebra, which is of course a complete complemented lattice, and that construction shares certain similarities with the …

I've got it! Theorem. The lattices of the form $D(U,V)$ admit a complete classification by the isomorphism classes of $U$ and $V$ and the question of whether $U\neq V$. The point is that the latti …