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Here is a counterexample of size 23. Let $m=6$ and let $$P=\{0,a_1,\dots,a_m,1\}\cup\{b_{ij}: 1\le i<j\le m\}$$ where $0<a_i<b_{jk}<1$ whenever $i$ is distinct from $j$ and $k$. The cardinality of $P$ …

The first thing I thought of was order embedding and this is confirmed by an article on monotonicity in order theory.

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 …

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 $K=\{1,2,3,6,12,18,36\}$ 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 …

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 …

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.

You can build a perfect tree where the branching happens always and only at certain specified levels. There is an antichain of $2^n $ many finite strings $\sigma_{i, n} $ of length $2^n $. Consider …

Yes. We can take the collection of all tods $ s $ such that $ s \backslash t $ is a singleton and $ s $ is incompatible with $ t $. Any maximal chain extends exactly one of these, or $ t $.

Here's a simplified version of Dominic van der Zypen's counterexample: order finite binary strings by extension, with the empty string at the bottom. Consider the club $ D$ consisting of the tods gene …

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 …

This answer is to version 1 of the question. Yes, note that such a poset would have to be linear. Then, a countable dense linear order can have one of the following 6 types: Infinite and having no …