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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 …

If the ambient order is a lattice, they your order is indeed a lattice order. To see this, suppose that we have intervals (a,b) and (a',b'), in your sense that $a\le b$ and $a'\le b'$. If $b\lt b'$, t …

Surely this information is in Ralph Mackenzie's book on universal algebra. But let me just sketch some quick proofs. First, note that you were right to consider complete lattices, since in general t …

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 …

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 …

There are two natural orders to put on the product of two lattices, the product order and the lexical order. Product order: (a,b) ≤ (a',b') if and only if a ≤ a' and b ≤ b' Lexical order: (a …

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 …

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 $B$ …

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 …

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 …

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 …

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 …

If the machine transition induces a partial order, then the machine cannot find itself in the same local configuration again after leaving it, since a partial order has no loops. It follows that the l …

You can use this to prove the Cantor-Schröder-Bernstein theorem, which asserts that whenever $A$ injects into $B$ and $B$ injects into $A$, then they are bijective. Namely, suppose that $f:A\to B$ and …

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 …