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Results tagged with posets
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user 4600
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 $x \leq z$).
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 …
- 24.8k
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 $.
- 24.8k
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 …
- 24.8k
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 …
- 24.8k
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.
- 24.8k
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.
- 24.8k
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 …
- 24.8k
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 $(\ …
- 24.8k
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 …
- 24.8k
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 …
- 24.8k
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).
- 24.8k
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 …
- 24.8k