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No. If $p_1\sim p_2$ and similarly for $q$, then $p_1a=p_2a$ and $$ p_1\vee q_1 \sim p_2\vee q_2 $$ iff $$ (p_1\vee q_1)a=( p_2\vee q_2)a $$ Let $a$ be projection on the $x$-axis, $p_1$ the diagonal, …

Yes. The l.u.b. of $\mathcal A$ and $\mathcal B$ is $\mathcal A \cup \mathcal B$. The g.l.b. of $\mathcal A$ and $\mathcal B$ is $\{A\cap B: A\in\mathcal A , B\in\mathcal B\}$. We can even generalize …

If $A\cap B$ contains a greatest element $y$ and another element $x$ then $$\neg x := y\setminus x\quad \in A\cap B$$ is a "complement" of $x$ within $A\cap B$. So in that sense, $A\cap B$ will alway …

Let $N$ be a countably infinite set and let $\mathcal P$ denote power set. I get that the automorphisms of $(\mathcal P(N),\subseteq)$ are all induced by permutations of $N$. But what can be said abo …

A year after you posted the question, Fritz showed the common theory of all such lattices is undecidable: https://arxiv.org/abs/1607.05870 In reponse to @MattF's query I'll post an example of how i …

Let $K=\{1,2,3,6,12,18,36\}$ ordered by divisibility. Let $L=\{1,2,3,6,12,24,36,72\}$ ordered by divisibility. Then $6=2\vee 3=12\wedge 18$ would have to be sent to both $6$ and $12$, but it can onl …

Claim: Any such $\varphi$ would have to map into a set on which all homomorphisms of $\text{Top}^{T_1}(\kappa)$ are constant. This follows from Theorems 1 and 2 of Hartmanis, Juris, On the latti …

Let $$A=\{a_0<a_0+a_1<a_0+a_1+a_2<\dots\},\qquad B=\{b_0<b_0+b_1<\dots\},$$ so that the $a_i$ and $b_i$ are the gaps in $A$ and $B$. Then $A\le_{\mathrm{inj}}B$ iff $a_i\le b_i$ for each $i$. Thus $(\ …

The countable atomless Boolean algebra is a counterexample. See E.S. Northam, The interval topology of a lattice, 1953 (Propositions 2 and 3).

I think you can remove the assumption of 0-preservation in @KeithKearnes' answer by replacing $\mathbf 3$, $\mathbf 4$, and the 0 element of each lattice by three mutually non-embeddable bounded latt …

This sounds like an ultrafilter without the intersection condition. So while I don't know if it already has a name, you could call it an ultra-upset (as opposed to downset) or ultra-final segment.

Yes. Since $\text{Part}(X)$ has a least element and a greatest element, just let $L$ not have that, e.g., let $L$ be the ordering of the integers.

Does property (A) have a name in the literature? Is it a studied notion? Don't know, but we could call it the property of being a strong antichain (since by taking $m=n=1$, it implies being …

No, any finite distributive lattice is isomorphic to (hence can be identified with) a collection of finite sets (closed under $\cap$ and $\cup$) ordered by inclusion. Let $1_A$ be the characteristic f …

By Exercise V.4.7 of Lattice theory: foundation by George Grätzer, we can take $M=N+1$. As for possible sharpness of this result, note that $P(N)$ has sizes 1,2,4,8 for $N=0,1,2,3$, whereas PART$(N+1 …