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This is an open problem. See Wikipedia article: http://en.m.wikipedia.org/wiki/Finite_lattice_representation_problem Palfy and Pudlak's result (see open-source description in Palfy's article Interva …

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 …

There is Burris and Sankappanavar's free book A Course in Universal Algebra.

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 …

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 …

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 …

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.

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.

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

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 …

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 …

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 $(\ …

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 …

Suppose $A$ and $B$ are (complete) ordered sets. Suppose $R\subseteq A\times B$, and $f(a)=\inf\{b : (a,b)\in R\}$ $g(b)=\inf\{a : (a,b)\in R\}$ then what can we call $f$ and $g$? Perhaps there is …