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These structures can be characterized as being exactly what you obtain from bounded lattices by removing their top and bottom elements.

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 …

$ Z (L) $ contains the subspaces that are orthogonal or comparable to all the other subspaces in $ L $. So if the Hilbert space has finite dimension $ d $ you can get $2^d $ many elements in $ Z (L) $ …

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 …

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.

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

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, …

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 $(\ …

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 $K=\{1,2,3,6,12,18,36\}$ 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 …

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 …