1. Introduction and Main Results
In many cases, the completeness properties of various objects of general topology or topological algebra can be characterized externally as closedness in ambient objects. For example, a metric space X is complete if and only if X is closed in any metric space containing X as a subspace. A uniform space X is complete if and only if X is closed in any uniform space containing X as a uniform subspace. A topological group G is Raĭkov complete if and only if it is closed in any topological group containing G as a subgroup.
On the other hand, for topological semigroups there are no reasonable notions of (inner) completeness. Nonetheless, one can define many completeness properties of semigroups via their closedness in ambient topological semigroups.
A topological semigroup is a topological space X endowed with a continuous associative binary operation , .
Definition 1. Let be a class of topological semigroups. A topological semigroup X is called
-closed if for any isomorphic topological embedding to a topological semigroup , the image is closed in Y;
injectively -closed if for any injective continuous homomorphism to a topological semigroup , the image is closed in Y;
absolutely -closed if for any continuous homomorphism to a topological semigroup , the image is closed in Y.
For any topological semigroup we have the implications:
Definition 2. A semigroup X is defined to be (injectively, absolutely)-closed if it is X endowed with the discrete topology.
In this paper, we are interested in the (absolute, injective) -closedness for the classes:
of topological semigroups satisfying the separation axiom ;
of Hausdorff topological semigroups;
of Hausdorff zero-dimensional topological semigroups.
A topological space satisfies the separation axiom if all its finite subsets are closed. A topological space is zero-dimensional if it has a base of the topology consisting of clopen (= closed-and-open) sets.
Since
, for every semigroup the following implications hold:
From now on, we assume that
is a class of topological semigroups such that
Semigroups having one of the above closedness properties are called categorically closed. Categorically closed topological groups and semilattices were investigated in [
1,
2,
3,
4,
5,
6,
7,
8,
9,
10,
11] and [
12,
13,
14,
15], respectively. This paper is a continuation of the papers [
3,
15,
16], which contain inner characterizations of semigroups possessing various categorically closed properties.
In this paper we shall characterize (absolutely and injectively) -closed unipotent semigroups.
A semigroup X is called
unipotent if X contains a unique idempotent;
chain-finite if any infinite set contains elements such that ;
singular if there exists an infinite set such that is a singleton;
periodic if for every there exists such that is an idempotent;
bounded if there exists such that for every the n-th power is an idempotent;
group-finite if every subgroup of X is finite;
group-bounded if every subgroup of X is bounded.
The following characterization of
-closed commutative semigroups was proved in the paper [
16].
Theorem 1. A commutative semigroup is -closed if and only if it is chain-finite, periodic, nonsingular and group-bounded.
For unipotent semigroups, this characterization can be simplified as follows:
Theorem 2. A unipotent semigroup X is -closed if and only if X is bounded and nonsingular.
Another principal result of this paper is the following characterization of injectively -closed unipotent semigroups.
Theorem 3. A unipotent commutative semigroup X is injectively -closed if and only if X is bounded, nonsingular and group-finite.
Example 1. For an infinite cardinal κ, the Taimanov semigroup
is the set κ endowed with the semigrop operation The semigroup was introduced by Taimanov in [17]. Its algebraic and topological properties were investigated by Gutik [18] who proved that the semigroup is injectively -closed. The same also follows for Theorem 3 because the semigroup is unipotent, bounded, nonsingular and group-finite. The Taimanov semigroups witness that there exist injectively -closed unipotent semigroups of arbitrarily high cardinality. For a semigorup
X, let
be the
center of
X. The center of an (injectively)
-closed semigroup has the following properties, proven in Lemmas 5.1, 5.3, 5.4 of [
16] (and Theorem 1.7 of [
19]).
Theorem 4. The center of any (injectively) -closed semigroup is chain-finite, periodic, nonsingular (and group-finite).
Corollary 1. The center of an injectively -closed unipotent semigroup X is injectively -closed.
Proof. By Theorem 4, the semigroup is chain-finite, periodic, nonsingular, and group-finite. By Theorem 1, the semigroup is -closed. By Theorem 2, is bouned. If is empty, then is injectively -closed. So, we assume that . Being bounded, the semigroup contains an idempotent. Being a subsemigroup of the unipotent semigroup X, the semigroup is unipotent. By Theorem 3, the unipotent bounded nonsingular group-finite semigroup is injectively -closed. □
Another corollary of Theorem 3 describes the center of an absolutely -closed unipotent semigroup.
Corollary 2. The center of an absolutely -closed unipotent semigroup X is finite and hence absolutely -closed.
Proof. By Theorem 4, the semigroup
is chain-finite, periodic, nonsingular, and group-finite. If
is empty, then
is finite and hence absolutely
-closed. So, we assume that
is not empty. Being periodic, the semigroup
contains an idempotent
e. Since
X is unipotent,
e is a unique idempotent of the semigroups
X and
. Let
be the maximal subgroup of the semigroup
. The group
is finite because
is group-finite. By Theorem 1.7 of [
19], the complement
is finite and hence the set
is finite, too. □
Corollaries 1 and 2 suggest the following open problems.
Problem 1. - 1.
Is the center of a -closed semigroup -closed?
- 2.
Is the center of an injectively -closed semigroup injectively -closed?
- 3.
Is the center of an absolutely -closed semigroup absolutely -closed?
3. Proof of Theorem 2
Theorem 2 will be derived from the following lemmas.
Lemma 2. Let X be a periodic commutative semigroup with a unique idempotent e and trivial maximal subgroup . If X is not bounded, then there exists an infinite subset such that .
Proof. To derive a contradiction, assume that X is not bounded but for every infinite set we have . Taking into account that X is periodic and unipotent, we conclude that . By Lemma 1, the maximal subgroup is an ideal in X.
Inductively we shall construct a sequence of points and a sequence of positive integer numbers such that for every the following conditions are satisfied:
- (i)
;
- (ii)
;
- (iii)
.
To start the inductive construction, take any
and let
be the smallest number such that
. Such number
exists as
X is periodic. Since
is an ideal in
X, it follows from
and
that
. Assume that for some
, we have chosen sequences
and
. For every
, consider the set
and observe that for every
the inductive condition (ii) implies
This means that and by our assumption, the set is finite. Since X is unbounded, there exists an element and a number such that but . Since the set consists of points, there exist a number such that . It follows from and that . This completes the inductive step.
After completing the inductive construction, consider the infinite set
. We claim that
for any
. For
this follows from the inductive condition (ii). So, assume that
. By the induction condition (iii) and the Pigeonhole Principle, there exist two positive numbers
such that
. Let
and observe that
. Then
Proceeding by induction, we can prove that for every . Since X is periodic and is an ideal in X, there exists such that and hence . Then, and hence , which contradicts our assumption. □
Lemma 3. Let X be a periodic commutative semigroup with a unique idempotent e and bounded maximal subgroup . If X is not bounded, then there exists an infinite subset such that .
Proof. Since is bounded, there exists a number such that for all . Assuming that X is not bounded, we conclude that the subsemigroup of X is not bounded. We claim that . Indeed, for every with , we have by Lemma 1 and hence . Since the maximal subgroup of P is trivial, one can apply Lemma 2 and find an infinite set such that . □
Our final lemma implies Theorem 2.
Lemma 4. For a unipotent commutative semigroup X, the following conditions are equivalent:
- 1.
X is -closed;
- 2.
X is periodic, nonsingular and group-bounded;
- 3.
X is bounded and not singular.
Proof. The equivalence follows from Theorem 1, and is trivial. The implication follows from Lemma 3. □
4. Proof of Theorem 3
In this section we prove Lemmas 5 and 6 implying the “only if” and “if” parts of the characterization Theorem 3, respectively.
Lemma 5. If a unipotent semigroup X is injectively -closed, then its center is bounded, nonsingular, and group-finite.
Proof. By Theorem 4, the semigroup is periodic, nonsingular and group-finite. If is empty, then is bounded. If is not empty, then by the periodicity, contains an idempotent and hence is unipotent, being a subsemigroup of the unipotent semigroup X. By Lemma 4, is bounded. □
Lemma 6. Every bounded nonsingular group-finite unipotent commutative subsemigroup X of a topological semigroup Y is closed and discrete in Y.
Proof. Replacing Y by the closure of X, we can assume that X is dense in Y.
Claim 1. For every and there exists a neighborhood of y such that the set is finite.
Proof. To derive a contradiction, assume that there exists and such that for every neighborhood of y the set is infinite. The periodicity of X ensures that where , see Lemma 1. This lemma also implies that the set
Let k be the largest number such that for every there exists a neighborhood of y such that the set is finite. Since , the number k is well-defined.
Subclaim 1. For every there exists a neighborhood of y such that the set is a singleton in X.
Proof. By the choice of k, for every there exists a neighborhood of y such that the set is finite and hence closed in the -space Y. Then
Since the space
Y is
there exists an open neighborhood
W of
such that
. By the continuity of the semigroup operation, the point
y has an open neighborhood
such that
. Then,
□
By the maximality of
k, there exists
such that for every neighborhood
of
y the set
is infinite. It follows from
that
and hence
and hence
. By Subclaim 1, there exists a neighborhood
of
y such that the set
is a singleton in
X. Choose any
. By Lemma 1,
and
. By Subclaim 1, there exists a neighborhood
such that
is a singleton in
X. Then, the set
is infinite but
is a singleton. However, this contradicts the nonsingularity of
X. □
Claim 2. For every and there exists a neighborhood of y such that is a singleton in X.
Proof. By Claim 1, there exists a neighborhood of y such that the set is finite and hence closed in the -space Y. Then,
Since the space
Y is
, there exists an open neighborhood
W of
such that
. By the continuity of the semigroup operation, the point
y has an open neighborhood
such that
. Then,
□
Claim 3. For every , and , there exists a neighborhood of y such that is a singleton in X.
Proof. For the statement follows from Claim 2. Assume that for some we know that for every and there exists a neighborhood of y such that is a singleton in X. By Claim 2, there exists a neighborhood of y such that is a singleton in X. Then is a singleton in X. □
Claim 4. For every the subspace of Y is discrete.
Proof. To derive a contradiction, assume that for some the subspace is not discrete and let k be the smallest number with this property. Since is finite, . Let y be a non-isolated point of . It follows that and and hence . By the minimality of k, the space is discrete. By the continuity of the semigroup operation, there exists a neighborhood of y such that . By Claim 2, we can additionally assume that .
By induction we shall construct a sequence of points in and a decreasing sequence of open sets in Y such that for every the following conditions are satisfied:
- (i)
;
- (ii)
and .
Assume that for some we have chosen a neigborhood of y and a sequence of points . Since y is a non-isolated point of , there exists a point satisfying the inductive condition (i). Observe that . By Claim 2, there exists a neighborhood of y such that . This completes the inductive step.
After completing the inductive construction, we obtain the infinite set such that . However, this contradicts the nonsingularity of X. □
Claim 5. For every the set is closed in Y.
Proof. To derive a contradiction, assume that for some k the set is not closed in Y. We can assume that k is the smallest number with this property. Since is finite, and hence . Fix any point and observe that
see Lemma 1. By Claim 4, the space
is discrete. Consequently, there exists a neighborhood
of
y such that
. Since
, the set
is infinite and
, which contradicts the nonsingularity of
X. □
The boundedness of X implies that for some . By Claims 4 and 5, the set is closed and discrete in Y. □