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 Sep10 comment A lattice generated by two particular sublattices of the lattice of binary relations I'm not shure, how your all quantification works. I suppose you mean all partitons and a fixed mapping $Y:\mathfrak PU\to \mathfrak PU:X\mapsto Y_X$. Aug6 comment The preorder of countable order types Thanks for that link. You are right, I forgot to handle the case where two orders are equivalent, but both embeddings have to skip certain levels. On the other hand I also overlooked the case of levels that have finite cardinality. But what I don't understand is, how $Q^\omega$ can describe our situation with an injective mapping $h$. As we consider all suborders of $Q^\omega$ up to equivalence, I would think we must accept that $h$ is non-injective. But maybe there is something I'm missing. Aug6 revised The preorder of countable order types Add a warning about a possible wrong answer. Aug6 comment How many Homomorphisms are there from one bounded lattice to another? The general solution should be sufficiently complicated as the result does not depend only on the size of the lattice, but also on its internal structure. As finite lattices are always complete lattices, the question is: where can I embed the irreducibles. You may ask the question in the setting of Formal Concept analysis asking the question how many different ways are there to embed a reduced context of $L_1$ into a non-reduced context of $L_2$ that hits all non-reducible objects and attributes. Though such questions are discussed there, this one look very special. Aug6 comment How many Homomorphisms are there from one bounded lattice to another? If $L_1=(\{0,1\},≤)$ then $|Hom(L_1,L_2)| = |{≤}_2|$ where $≤_2$ is the order relation of $L_2$, $|Hom_{0,1}(L_1,L_2)|=1$. $Hom(L_2,L_1)$ provides the order ideals (downsets) or order filters (upsets) of all intervals of $L_2$, while $Hom_{0,1}(L_2,L_1)$ provides the nontrival order ideals and order filters of $L_2$ the missing trivial order filters are: $\emptyset, L_2$. It should be easy to proove this as the complement of an order ideal is an order filter and vice versa. Jul30 comment How many Homomorphisms are there from one bounded lattice to another? At first, you should define what is a homomorphism from one lattice from one such lattice into another one. Depending on that definition the answer for the trivial case you mention is something between trivial and simple. Jul2 comment Undistinguishable elements in posets It looks to me like a topological property. There is a bunch of topologies that are generated by ordered sets. You could have a look at them. Another way would be to consider undistinguishable elements as anti-lattice parts of the ordered set. May be a translation into the language of Formal Concept Analysis might bring you up with some suitable notion. Jun24 comment Defining principal elements of every poset. Is this a new idea? In fact a non-principal filter may be non-principal in a certain lattice. But that's true for every element of any lattice and out of the scope of this question. If you have a counterexample then you should revise your definitions. You don't provide a definition of a filter. There is a common definition of ultrafilters in topology, but not in order theory. So you are probably the one who confuses different notions. Regarding your last sentence, $0$ is not principal by your definition. I came to the conclusion that all other elements are principal. So where is the contradiction? Jun22 answered Defining principal elements of every poset. Is this a new idea? May25 comment All tree orders are lattice orders? With the restriction that both order relations are suborders of the same order you are right, but this restriction is necessary. Otherwise you end up with the general case of the intersection of linear orders. An example: $\bigl(\{a,b\},\{(a,a), (a,b), (b,b)\}\bigr)∩\bigl(\{a,b\},\{(a,a),(b,a),(b,b)\}\bigr)=\bigl(\{a,b\},\{(a,a),(b‌​,b)\}\bigr)$. There is another point that shows that this property depends mainly on the relation: You can extend each suborder to the whole set by the discrete order on the complementary set and get the same result. May24 comment All tree orders are lattice orders? Thanks for providing the proof. Just one remark: The intersection of linearly ordered sets is not necessarily linear (for a counterexample see the Szpilrajn's lemma). But the intersection of linearly ordered downsets is linearly ordered. May24 answered Number of join-irreducible elements of a lattice: is it monotonic? May24 comment All tree orders are lattice orders? Every binary relation can be considered as a directed graph, when we interpret all the pairs of the relation as edges. This allows us to use the definition of a connected graph (in the sense of undirected graphs), here. The Neigbourhood relation @seaturtles mentions is something different. Not every order relation has such a neighbourhood relation. For example $\mathbb Q$ has no Hasse diagram. May20 comment Addition on well ordered sets not-commutative by showing $[0,1) +_o \mathbb{N} =_o \mathbb{N} \neq_o \mathbb{N} +_o [0,1)$ @AndréNicolas Wouldn't it be better to provide answers in the answers area? This would show others that it is not necessary to deal with the question in detail. May17 awarded Yearling May17 comment Finite set with $\sup\{S\} \notin S$ This answer is not correct. An order relation needs not to be linear. Thus, the set may not contain any supremum, or if it exists it is not necessarily an element of $S$. May17 answered what kind of relationship is “is prefix of”? May16 comment Lattice from Preorder Don't you think that category theory is a little bit overkill here? You should keep in mind that we are not on Math Overflow, here. Many users in this forum are not familiar with category theory. I'd think it would be helpful to answer the question in the language of order relations and lattices first. Afterwards you may provide the category theoretical point of view. May16 comment Introductions to posets on algerbaic structures (Everything I need to know about them) A tree order is always a semilattice if it is connected. Even if it doesn't have a smallest element. On the other hand only tree orders whose order ideals (downsets: $\downarrow a:=\{x\mid x≤a\}$) are suborders of $\mathbb Z$ are graph theoretical trees. Otherwise the underlying undirected graph has circles. May16 comment Introductions to posets on algerbaic structures (Everything I need to know about them) A short introduction is available at algebra.uni-linz.ac.at/Students/UniversalAlgebra/s11/… while the fundamental book is „Funktionen- und Relationenalgebren” by Pöschel/Kalužnin” available from Springer: springer.com/new+%26+forthcoming+titles+%28default%29/book/… . This gives two other keywords “function(al) algebras“ and “relation(al) algebras“ as mathematical objects which have nearly nothing in common with the branches functional algbra and relational algebra from mathematics/computer science.