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A poset or partially ordered set is a set endowed with a partial order, meaning a binary relation $\leq$ which is reflexive ($x \leq x$ for all $x$), antisymmetric ($x\leq y$ and $y\leq x$ implies $x=y$), and transitive ($x\leq y$ and $y\leq z$ implies $x \leq z$).

6 votes

Does the lattice of all topologies embed into the lattice of $T_1$-topologies?

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 …
Bjørn Kjos-Hanssen's user avatar
5 votes

Complete sets of incompatible totally ordered down-set in a partially ordered set

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 …
Bjørn Kjos-Hanssen's user avatar
5 votes

Quotients of $\text{Part}(X)$

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.
Bjørn Kjos-Hanssen's user avatar
4 votes

The set of complements equal to the complement of set

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.
Bjørn Kjos-Hanssen's user avatar
4 votes
Accepted

Hausdorff interval topology on distributive lattices

The countable atomless Boolean algebra is a counterexample. See E.S. Northam, The interval topology of a lattice, 1953 (Propositions 2 and 3).
Bjørn Kjos-Hanssen's user avatar
4 votes

Classification of countable posets?

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 …
Bjørn Kjos-Hanssen's user avatar
4 votes
Accepted

Order-embedding, but no lattice embedding between distributive lattices

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 …
Bjørn Kjos-Hanssen's user avatar
4 votes
Accepted

Is this ordering on the set of all covers of $\omega$ a (complete) lattice?

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 …
Bjørn Kjos-Hanssen's user avatar
3 votes

Antichain on $\mathcal{P}(\omega)/fin$ of cardinality $2^{\aleph_0}$?

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 …
Bjørn Kjos-Hanssen's user avatar
3 votes
Accepted

Is $({\cal P}(\omega), \leq_{\text{inj}})$ a distributive lattice?

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 $(\ …
Bjørn Kjos-Hanssen's user avatar
2 votes

Completion of a single totally ordered down-set

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 $.
Bjørn Kjos-Hanssen's user avatar
2 votes
Accepted

Count of lattices on finite set

1, 1, 1, 1, 2, 5, 15, 53, 222, 1078, 5994, 37622, 262776, 2018305, 16873364, 152233518, 1471613387, 15150569446, 165269824761, ... There is a lot of information in The On-Line Encyclopedia of Integ …
Bjørn Kjos-Hanssen's user avatar