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Since $R$ could be a $\le$, or $\ge$, or all kinds of other possibilities, I kind of agree that "lower set" is not ideal (excuse the pun). "Leftward closed set" seems somewhat more reasonable, since n …

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 …

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 $.

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 …

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 …

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

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 …

Does $\stackrel{\mathrm L}{\preceq}$ well-quasi-order the regular languages over $A^*$? No, let $ L_n $ contain all strings of length $ n $, then the $ L_n $ form an infinite antichain. So $\stac …

Does $\stackrel{\mathrm L}{\preceq}$ well-quasi-order the prefix-closed regular languages over $A^*$? No, let $L_n$ be the set of all prefixes of $0^n10^\infty$. Then the $L_n$ are prefix-closed, …

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.

Try Rosenstein: Linear orderings http://books.google.com/books/about/Linear_orderings.html?id=y3YpdW-sbFsC

Many lattices do have this property (for example completely distributive, or 5-element nondistributive). Here is a small counterexample. Let $L$ be the 9-element lattice $$ L=\{s_1,t_1,s^*,t^*,s_1\we …

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 …