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Results tagged with posets
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user 1946
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 $x \leq z$).
3
votes
Accepted
ACC (DCC) implies upper (lower) sets are upper (lower) closure of antichains?
$\newcommand\P{\mathbb{P}}$Let $U$ be any upper set in a partial
order $\langle\P,\leq\rangle$, which satisfies the descending
chain condition, which asserts that every descending sequence of
points t …
- 236k
1
vote
Minimum set cardinality for a given partially ordered set
To get things started, there are a number of easy observations.
First, $k$ is bounded above by $|P|$, since you can map every $p$ in $P$ to the lower cone $\{q\in P\mid q\leq p\}$.
Second, $k$ is …
- 236k
3
votes
Proof of glb and lub of lexicographic product of poset
Surely this information is in Ralph Mackenzie's book on universal algebra.
But let me just sketch some quick proofs. First, note that you were right to consider complete lattices, since in general t …
- 236k
10
votes
Classification of countable posets?
$\newcommand\P{\mathbb{P}}
\newcommand\Q{\mathbb{Q}}
\newcommand\N{\mathbb{N}}$
Update. I claim that, in a precise sense I shall explain (using the ideas
of Borel equivalence relation theory), there …
- 236k
7
votes
Is the homomorphism poset directed if the codomain is directed?
If $D$ is countable, then the answer is yes. The reason is that
every countable directed order $D$ has a cofinal $\omega$ sequence
$$d_0<d_1<\cdots <d_n<\cdots,$$ which can be constructed by
iterative …
- 236k
1
vote
Minimal (semi)lattice containing a given poset
Every separative partial order $P$ has a unique completion as a
complete Boolean algebra, which is of course a complete
complemented lattice, and that construction shares certain
similarities with the …
- 236k
6
votes
Accepted
Terminology for posets.
A partial order with no infinite descending chains is said to be well-founded. Every well-founded partial order admits an ordinal ranking function, an assignment of nodes in the order to ordinals, suc …
- 236k
28
votes
How many orders of infinity are there?
François has given an excellent answer to this question.
What you call a cofinal collection, a family $\cal F$ such that every function is dominated by a function in $\cal F$, is known as a dominati …
- 236k
18
votes
Accepted
Are these two quotients of $\omega^\omega$ isomorphic?
Very nice question!
They are not isomorphic.
What I claim is that when we take the quotient with respect to density, there is a countably infinite antichain above $0$ having a minimal upper bound, b …
- 236k
5
votes
Accepted
Does the collection of coverings on a set $X$ form a lattice when ordered by refinement?
If there are only finitely many points, it is a lattice, and it is always an upper semi-lattice. But in the infinite case, it is not a lattice.
François's comment on Fedor's answer shows that is it a …
- 236k
7
votes
Accepted
Image of poset with Hausdorff interval topology
The answer is no, not necessarily.
For a counterexample, let $Q$ be any atomless complete Boolean algebra, and let us view it via Stone's theorem as a field of sets, so that $Q$ is a subalgebra for …
- 236k
9
votes
Accepted
Order dimension of $\omega^\omega/(fin)$
$\newcommand\Fin{\text{Fin}}$Theorem. The order dimension of
$\langle\omega^\omega/\Fin,\leq^*\rangle$ is precisely the
continuum.
Proof. It is easy to see that the dimension is at most the
continuum …
- 236k
3
votes
Accepted
Dedekind-MacNeille completion of the strictly increasing members of $\omega^\omega$
The answer is no.
To see this, consider the bottoms of $K$ and $\omega^\omega$ under
the pointwise $\leq$ order you have described. Both structures
have a least element:
The constant $0$ function i …
- 236k
3
votes
Accepted
Does the lattice of coverings embed in the lattice of partitions?
The answer is no, because in general there can be more proper coverings than partitions, which will prevent any injective mapping from coverings to partitions.
For example, if $X$ is countably infini …
- 236k
11
votes
Accepted
Are free ultrafilters as posets product-irreducible?
No. Every nonprincipal ultrafilter $U$, considered as a partial under $\subseteq$, is a nontrivial product order. To see this, suppose that $U$ is a nonprincipal ultrafilter on $\kappa$.
Partition $\k …
- 236k