All Questions
24 questions
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48
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Neighborhoods of idempotents in topological inverse semigroups
In a topological group, for any neighborhood $U$ of the origin, there is another such neighborhood with the property that $V.V\subseteq U.$ I conjecture a similar property for topological inverse ...
- 63.9k
3
votes
0
answers
161
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Making the powerset into a topological monoid
Every monoid $X$ induces a monoid structure $\circledast$ on $\mathcal{P}(X)$ via
$$U\circledast V := \{uv\ |\ u\in U,v\in V\}.$$
Moreover, a morphism of monoids $f\colon X\to Y$ induces a morphism of ...
- 1,074
3
votes
0
answers
157
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Closure of the inverse image under the projection map
Let $S$ be a subsemigroup of a semitopological semigroup $(T,+)$, let $e$ be an idempotent in $T\setminus S$ such that $e\in cl_T(S)$, let $\mathcal{E}$ be a subsemigroup of $S\times S$ such that $(e,...
- 85
4
votes
0
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74
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Is each TS-topologizable group TG-topologizable?
Definition 1. A topology $\tau$ on a group $X$ is called
$\bullet$ a semigroup topology if the multiplication $X\times X\to X$, $(x,y)\mapsto xy$, is continuous in the topology $\tau$;
$\bullet$ a ...
- 41.9k
12
votes
1
answer
624
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Stone–Čech compactification as a semigroup
Let $G$ be a topological group (we can assume that $G$ is countable and discrete) and let $\beta(G)$ be the Stone–Čech compactification of $G$. It is known that $\beta(G)$ can be turned into a left ...
- 38.6k
3
votes
0
answers
31
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Compactness of the minimal ideal of a compact Hausdorff polytopological semigroup
A semigroup $X$ endowed with a topology is called
$\bullet$ a topological semigroup if the semigroup operation $X\times X\to X$ is continuous;
$\bullet$ a semitopological semigroup if for every $a,b\...
- 41.9k
7
votes
0
answers
138
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The smallest cardinality of a cover of a group by algebraic sets
$\DeclareMathOperator\cov{cov}$A subset $A$ of a semigroup $X$ is called algebraic if $$A=\{x\in X: a_0xa_1x...xa_n=b\}$$ for some $b\in X$ and $a_0,a_1,...,a_n \in X^1=X\cup \{1\}$. The smallest ...
- 41.9k
29
votes
1
answer
1k
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Is the Golomb countable connected space topologically rigid?
The Golomb space $\mathbb G$ is the set of positive integers endowed with the topology generated by the base consisting of the arithmetic progressions $a+b\mathbb N_0$ with relatively prime $a,b$ and $...
- 63.9k
-8
votes
1
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351
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Are there overwhelmingly more finite monoids than finite spaces? [closed]
A function $f:\mathbb{Z}_{\geq 1}\to\mathbb{Z}_{\geq 1}$ overwhelms $g:\mathbb{Z}_{\geq 1}\to\mathbb{Z}_{\geq 1}$ if for any $k\in \mathbb{Z}_{\geq 1}$ the inequality $f(n)\leq g(n+k)$ holds only for ...
- 38.6k
4
votes
0
answers
72
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When is the submonoid preserving a subspace finitely generated?
Let $T$ be a topological space with at least one open set whose closure is not open.
Let $G$ be a finitely generated group acting by homeomorphisms on $T$. Let $S\subset T$ be a subspace.
Under what ...
- 51
6
votes
3
answers
472
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Spaces with unique endomorphism monoids
If $(X,\tau)$ is a topological space, let $\text{End}(X)$ denote the collection of all continuous maps $f: X\to X$. With composition, this becomes the endomorphism monoid $(\text{End}(X), \circ)$.
We ...
CommunityBot
- 1
25
votes
3
answers
1k
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What spaces $X$ do have $\text{End}(X) \cong \text{End}(\mathbb{R})$?
This is a follow-up on the following question. Let $\text{End}(X)$ denote the endomorphism monoid of a topological space $X$ (that is, the collection of all continuous maps $f:X\to X$ with composition)...
- 28.7k
10
votes
0
answers
314
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How much do idempotent ultrafilters generate in terms of semigroups?
It is known that the set of ultrafilters on, say, the natural numbers $\mathbb{N}$, can naturally be endowed with the structure of a compact topological left semigroup (which fails to be anything ...
- 4,713
6
votes
0
answers
132
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Generalization of pseudogroups
Pseudogroups are defined here: https://ncatlab.org/nlab/show/pseudogroup
One of the problems with defining manifolds in terms of pseudogroups is that it gives no notion of a morphism between manifolds,...
CommunityBot
- 1
6
votes
0
answers
117
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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 ...
- 4,535
3
votes
1
answer
244
views
Category of continuous self maps
Is there any way to reconstruct a topological space from the category of its continuous self maps (possibly under some assumptions)?
How can we tell whether a category is the category of continuous ...
- 2,160
3
votes
0
answers
80
views
On the compactification of partial semigroups
We begin by introducing some relevant definitions.
Definition: A $\textit{partial semigroup}$ is a pair $(S,.)$ where $.$ maps a subset of $S \times S$ to $S$ and for all $a,b,c \in S, (a.b).c=a.(b.c)...
- 745
9
votes
0
answers
373
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Embedding $\beta\mathbb{N}$ into a product of Cantor sets
Let us consider $\beta\mathbb{N}$, the Stone-Čech compactification of the natural numbers (where we do not take $0$ to be a natural number, so the only idempotent elements are nonprincipal ...
- 4,713
9
votes
4
answers
1k
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When $X \times Y \cong X \times Z$ implies $Y \cong Z$ (in the category of finite topological spaces)
The title has it all. I'm looking for a reference to the following:
Q. Let $X, Y, Z$ be finite, non-empty (topological) spaces. When does $X \times Y \cong X \times Z$ imply $Y \cong Z$ (in the ...
- 13k
8
votes
1
answer
229
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Embedding abelian cancellative Hausdorff topological semigroups into abelian Hausdorff topological groups
An abelian cancellative semigroup embeds (via a semigroup monomorphism) into an abelian group. What about an abelian cancellative Hausdorff topological semigroup that does not embed (via a ...
- 5,409
6
votes
1
answer
1k
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Who coined "mob" and "clan" and why these words?
A mob is a word used for a topological semigroup which is a Hausdorff space. A clan is a compact connected mob with a two-sided identity element.
Who used these words with these meanings first and ...
- 188k
2
votes
0
answers
122
views
First-countable topological monoids without local absorbing elements whose topology is induced by a semimetric
This is a follow up of Question 163246. For the reader's convenience, let me first copy&paste some basic definitions.
We let a semimetric on a set $X$ be a function $d: X \times X \to [0,\infty]$ ...
CommunityBot
- 1
2
votes
2
answers
241
views
If $(\mathbb M, \tau)$ is a topological monoid, is $\tau$ always induced by a [left] subinvariant semimetric?
Let me start by recalling some basic definitions (just for the sake of avoiding misunderstandings due to the vocabulary of the post).
Basically following some ideas of W. Lawvere (but not his ...
- 27.5k
7
votes
2
answers
417
views
Does every commutative monoid admit a translation-invariant measure?
Let $T$ be a commutative monoid, written additively. The set $T$ is equipped with a canonical pre-order, defined by $s \le t$ when there exists $s' \in T$ so that $s + s' = t$. Consequently, $T$ may ...
- 38.6k