All Questions
Tagged with order-theory semigroups-and-monoids
18 questions
5
votes
0
answers
107
views
Structure of well-ordered commutative monoids
Let $(M,+)$ be a commutative monoid. Let $<$ be a well-ordering on $M$, where
$\forall a\in M,\ 0\leq a$
$\forall a,b,c\in M,\ a<b\Rightarrow a+c<b+c$
The first condition means $M$ will be ...
- 18.7k
1
vote
1
answer
142
views
Congruences that aren't "finite from above," take 2: semigroups
This is a hopefully less trivial version of this question. Briefly, say that a congruence is parafinite if it is the largest congruence contained in some equivalence relation with finitely many ...
- 21.2k
5
votes
2
answers
478
views
Generalization of the concept of a measure
Consider the following generalization of the concept of a measure:
Let $L = (X, \lor, \land, \bot)$ be a semi-bounded lattice.
Let $M = (Y, \bullet, e)$ be a commutative monoid.
An $(L, M)$-measure is ...
- 2,203
3
votes
1
answer
173
views
Well-foundedness of divisibility vs well-foundedness of right- and left-divisibility
Say that a preorder (i.e., a reflexive and transitive binary relation) $\preceq$ on a set $X$ is
artinian if there is no sequence $(x_n)_{n \ge 1}$ of elements of $X$ with $x_{n+1} \prec x_n$ for ...
- 10.5k
3
votes
1
answer
157
views
Is it true that the structure of a commutative ordered semiring is unique on a commutative ordered monoid?
Is it true that the structure of a commutative ordered semiring with identity is unique on a commutative ordered monoid (i.e., the structure of the monoid and the order are consistent)? I am not ...
- 6,962
11
votes
0
answers
286
views
Does every finite poset have a rigid endomorphism?
Crossposted on Mathematics.
In this post, an order-preserving self-map of a poset $X$ will be called an endomorphism of $X$, and such an endomorphism $f$ will be called rigid if the only automorphism ...
- 4,416
2
votes
0
answers
96
views
Non-commutative version of the order dimension of a poset
I view the order dimension of a poset $P$ as an inherently commutative notion. On the one hand, it can be defined via realizers, which I find fairly intuitive from an order-theoretic viewpoint. On the ...
- 51
2
votes
1
answer
176
views
Generating totally ordered free commutative monoids
Let’s say I have a set $A$. I build the free commutative monoid $M$ generated by $A$.
When can a well-order on $A$ be extended to $M$, in a way that is compatible with its monoid structure? I am ...
- 341
6
votes
0
answers
117
views
Closedness of the partial order in complete Hausdorff semitopological semilattices
First some definitions.
A semilattice is a commutative semigroup consisting of idempotents (i.e., elements such that $xx=x$). A typical example of a semilattice is the unit interval endowed with the ...
- 41.8k
2
votes
1
answer
302
views
Name for this algebraic structure?
I've found myself looking at a structure $\mathbb{M}$ whose important properties are:
$\mathbb{M}$ is a discretely ordered additive monoid.
$\mathbb{M}$ has a least element, and this least element is ...
- 10.1k
5
votes
0
answers
64
views
Characters on monotone functions
Characters on the semigroup $(C_{+}^{b}(\mathbb{R}^{d}),+)$, i.e. on bounded positive continuous functions with the ususal pointwise addition, are known to be of the form $C_{+}^{b}(\mathbb{R}^{d})\ni ...
- 289
8
votes
2
answers
483
views
Posets obtained from a semigroup by the definition $x \leq y \iff x \cdot y = x$
A po-groupoid is a groupoid $\langle A,\cdot\rangle $ such that the relation
defined by
$$
x \leq y \text{ if and only if } x \cdot y = x
$$
is a partial order on $A$, the order related to $\langle ...
- 1,124
2
votes
1
answer
294
views
Finitely generated ordered monoids and noetherian subsets
(This question was asked a long time ago on MSE but got no answer so far...)
Let $E$ be an additively written cancellable commutative monoid with no non-trivial units. We furnish $E$ with the order ...
- 6,700
4
votes
2
answers
393
views
Embedding a linearly ordered free monoid into a linearly ordered group
A linearly ordered (shortly, l.o.) monoid is a triple $\mathbb M = (M, \cdot, \le)$ for which $(M, \cdot)$ is a (multiplicatively written) monoid and $\le$ is a total order on $M$ such that $xy < ...
- 10.5k
4
votes
4
answers
1k
views
Why do we choose the standard total order on the integers?
I understand why the set of natural numbers $\mathbb N = \{ 0, 1, 2, \cdots \}$ is equipped with a total order. Indeed, every monoid has a pre-order, where $$n' \succeq n \quad \mathrm{if~and~only~if} ...
- 8,512
7
votes
2
answers
544
views
A linearly orderable monoid which does not embed into a linearly orderable group
It is known (after an example of A.I. Mal'cev) that there exist cancellative semigroups which do not embed into a group. On the other hand, it is not difficult to see that every linearly orderable ...
- 10.5k
4
votes
1
answer
518
views
Strictly totally ordered semigroups - Looking for references
Let $\mathfrak A = (A, \cdot)$ be a semigroup (written multiplicatively). We say that $\mathfrak A$ is linearly orderable if there exists a total order $\le$ on $A$ such that $ac < bc$ and $ca < ...
- 10.5k
1
vote
2
answers
378
views
Is this a pre-ordered commutative semigroup?
Motivation
I'm studying an approach to axiomatic thermodynamics based on the notion of commutative semigroup $(S,+)$ with a preorder relation $\to$ on $S$. In other words, $S$ is non-empty set, the ...
- 33.3k