Density and capacity of balleans generated by filters
UDC 519.51 We consider a ballean $\mathbb B=(X,P,B)$ with an infinite support $X$ and a free filter $\phi$ on $X$ and define $B_{P\times\phi}(x,(\alpha,F))$ for every $\alpha\in P$ and $F\in \phi.$ The ballean $(X,P\times\phi, B_{P\times\phi})$ will be called the ballean-filter mix of $\mathbb B$ an...
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| author | Brzeska, A. brzeska, anna Brzeska, A. |
| author_facet | Brzeska, A. brzeska, anna Brzeska, A. |
| author_sort | Brzeska, A. |
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We consider a ballean $\mathbb B=(X,P,B)$ with an infinite support $X$ and a free filter $\phi$ on $X$ and define $B_{P\times\phi}(x,(\alpha,F))$ for every $\alpha\in P$ and $F\in \phi.$ The ballean $(X,P\times\phi, B_{P\times\phi})$ will be called the ballean-filter mix of $\mathbb B$ and $\phi$ and denoted by $\mathbb B(B,\phi).$ It was introduced in [O. V. Petrenko, I. V. Protasov, Balleans and filters, Mat. Stud., 38, No. 1, 3–11 (2012)] and was used to construction of a non-metrizable Frechet group ballean. In this paper some cardinal invariants are compared. In particular, we give a partial answer to the question: if we mix an ordinal unbounded ballean with a free filter of the subsets of its support, will the mix-structure's density be equal to its capacity, as it holds in the original balleans? |
| doi_str_mv | 10.37863/umzh.v73i4.648 |
| first_indexed | 2026-03-24T02:03:31Z |
| format | Article |
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DOI: 10.37863/umzh.v73i4.648
UDC 519.51
A. Brzeska (Inst. Math., Univ. Silesia, Katowice, Poland)
DENSITY AND CAPACITY OF BALLEANS GENERATED BY FILTERS
ЩIЛЬНIСТЬ ТА ЄМНIСТЬ БОЛЕАНIВ, ЗГЕНЕРОВАНИХ ФIЛЬТРАМИ
We consider a ballean \BbbB = (X,P,B) with an infinite support X and a free filter \phi on X and define BP\times \phi (x, (\alpha , F )) for
every \alpha \in P and F \in \phi . The ballean (X,P \times \phi ,BP\times \phi ) will be called the ballean-filter mix of \BbbB and \phi and denoted by
\BbbB (B,\phi ). It was introduced in [O. V. Petrenko, I. V. Protasov, Balleans and filters, Mat. Stud., 38, № 1, 3 – 11 (2012)] and
was used to construction of a non-metrizable Fréchet group ballean. In this paper some cardinal invariants are compared.
In particular, we give a partial answer to the question: if we mix an ordinal unbounded ballean with a free filter of the
subsets of its support, will the mix-structure’s density be equal to its capacity, as it holds in the original balleans?
Розглядається болеан \BbbB = (X,P,B) з нескiнченним супортом X i вiльний фiльтр \phi на X та визначається
BP\times \phi (x, (\alpha , F )) для кожного \alpha \in P та F \in \phi . Болеан (X,P \times \phi ,BP\times \phi ) називають болеан-фiльтр мiксом
для \BbbB i \phi та позначають \BbbB (B,\phi ). Таку термiнологiю було введено у статтi [O. V. Petrenko, I. V. Protasov, Balleans
and filters, Mat. Stud., 38, № 1, 3 – 11 (2012)], де її застосовано для побудови болеана групи Фреше без метриза-
цiї. У цiй роботi порiвнюються деякi кардинальнi iнварiанти. Зокрема, наведено часткову вiдповiдь на питання:
якщо є мiкс ординально необмеженого болеана з вiльним фiльтром пiдмножин його супорту, то чи буде щiльнiсть
мiкс-структури рiвною її ємностi, як це має мiсце для оригiнальних болеанiв?
1. Introduction. Given sets X, P and a function B : X \times P \rightarrow \scrP (X), a triple \BbbB = (X,P,B) is
called a ball structure with a support X, a set of radiuses P and a ball function B. If (x, \alpha ) \in X\times P,
then B(x, \alpha ) is called a ball of radius \alpha around x. Consistently if A \subseteq X and \alpha \in P, then\bigcup
\{ B(x, \alpha ) : x \in A\} is called a ball of radius \alpha around the set A.
If B is a ball function, then a dual ball function B \star is defined as follows: for every x \in X and
\alpha \in P, we put
B \star (x, \alpha ) = \{ y \in X : x \in B(y, \alpha )\} .
The ball structure \BbbB \star = (X,P,B \star ) is called dual to the structure \BbbB = (X,P,B).
A ball structure \BbbB = (X,P,B) is called upper symmetric if for any \alpha , \beta \in P there exist \alpha \prime ,
\beta \prime \in P such that
B(x, \alpha ) \subseteq B \star (x, \alpha \prime ) and B \star (x, \beta ) \subseteq B(x, \beta \prime )
for every x \in X. A ball structure \BbbB = (X,P,B) is called upper multiplicative if, for any \alpha , \beta \in P,
there exists \gamma \in P such that
B(B(x, \alpha ), \beta ) \subseteq B(x, \gamma )
for every x \in X. A ball structure which is both upper symmetric and upper multiplicative is called a
ballean. Note that if \BbbB is upper symmetric/multiplicative, then \BbbB \star is upper symmetric/multiplicative.
Let \BbbB = (X,P,B) and \BbbB \prime = (X \prime , P \prime , B\prime ) be balleans. A mapping f : X \rightarrow X \prime is called:
\prec -mapping if, for any \alpha \in P, there exists \alpha \prime \in P \prime such that f(B(x, \alpha )) \subseteq B\prime (f(x), \alpha \prime ) for
every x \in X;
asymorphism (or isomorphism as in [3]) if f is a bijection and both f and f - 1 are \prec -mappings.
c\bigcirc A. BRZESKA, 2021
ISSN 1027-3190. Укр. мат. журн., 2021, т. 73, № 4 467
468 A. BRZESKA
If f : X \rightarrow X \prime is a \prec -mapping we will write \BbbB \prec \BbbB \prime . Balleans \BbbB = (X,P,B) and \BbbB \prime = (X \prime , P \prime , B\prime )
are called asymorphic if there exists an asymorphism f : X \rightarrow X \prime . We will write \BbbB = \BbbB \prime if X = X \prime
and \BbbB and \BbbB \prime are asymorphic. Note that \BbbB = \BbbB \star if and only if a ball structure \BbbB is upper symmetric.
Of course for every balleans we have \BbbB = \BbbB \star . A ballean \BbbB = (X,P,B) is called symmetric if
B(x, \alpha ) = B \star (x, \alpha ) for any two x \in X and \alpha \in P. Recall that every ballean is asymorphic to some
symmetric ballean [1].
Given \BbbB = (X,P,B), we define natural preordering \leq on P by the rule: \alpha \leq \beta iff B(x, \alpha ) \subseteq
B(x, \beta ) for every x \in X. A subset P \prime \subseteq P is called cofinal if, for each \alpha \in P there exists \beta \in P \prime
such that \alpha \leq \beta . A ballean \BbbB = (X,P,B) is called connected if, for every two points x and y
of X, there exists \alpha = \alpha (x, y) \in P such that y \in B(x, \alpha ). The connectedness is an equivalence
relation. A connected ballean \BbbB = (X,P,B) is called ordinal if P contains a cofinal subset P \prime
which is well-ordered by \leq . We put \mathrm{c}\mathrm{f}(\BbbB ) = min \{ \mathrm{c}\mathrm{a}\mathrm{r}\mathrm{d}(R) : R \subseteq P \wedge R is cofinal in \BbbB \} .
Observe that if we replace P by its minimal cofinal subset P \prime , we get an asymorphic bal-
lean [4, p. 175]. Hence, we can replace P by a regular cardinal \rho = \mathrm{c}\mathrm{f} \mathrm{c}\mathrm{a}\mathrm{r}\mathrm{d}(P ) and write \BbbB (X, \rho ,B)
in the place of \BbbB (X,P,B).
2. Examples.
Example 1. Let (X, d) be a metric space. Put Bd(x, \epsilon ) = \{ y \in X : d(x, y) \leq \epsilon \} for every
\epsilon > 0. The ballean \BbbB (X, d) = (X,\BbbR +, Bd) is called a metric ballean. Note that every metric ballean
is ordinal.
A filter \phi on infinite set X is called free if
\bigcap
\phi = \varnothing . A free filter \phi on infinite set X is called
uniform if \mathrm{c}\mathrm{a}\mathrm{r}\mathrm{d}(Y ) = \mathrm{c}\mathrm{a}\mathrm{r}\mathrm{d}(X) for every Y \in \phi . If \mathrm{c}\mathrm{a}\mathrm{r}\mathrm{d}(X) \geq \kappa \geq \omega , then the free filter of all
subsets F \subseteq X such that \mathrm{c}\mathrm{a}\mathrm{r}\mathrm{d}(X \setminus F ) < \kappa will be denoted by \scrF \kappa (X).
Example 2. Let X be a set and \phi be a free filter on X. For any x \in X and F \in \phi , we put
B\phi (x, F ) =
\Biggl\{
X \setminus F, if x \not \in F,
\{ x\} , if x \in F.
Then \BbbB (\phi ) = (X,\phi ,B\phi ) is a symmetric ballean.
Note that \chi (\phi ) = \mathrm{c}\mathrm{f}(\BbbB (\phi )) for every free filter on an infinite set X [4, p. 23].
Example 3. Let \BbbB = (X,P,B) be a ballean and \phi be a free filter on X. We put
B\phi (x, F ) =
\Biggl\{
B(x, \alpha ) \setminus F, if x \not \in F,
\{ x\} , if x \in F.
Then the ball structure \BbbB (B,\phi ) = (X,P \times \phi ,BP\times \phi ) is a ballean. We will call it a ballean-filter mix
of \BbbB and \phi [2].
Remark 1. Note that:
(a) if \phi and \psi are free filters on X such that \phi \subseteq \psi , then \BbbB (B,\phi ) \prec \BbbB (B,\psi );
(b) \BbbB (B,\phi ) \prec \BbbB for every free filter \phi on X;
(c) if \BbbB = (X,P,B), \BbbB \prime = (X,P \prime , B\prime ) and \BbbB = \BbbB \prime , then \BbbB (B,\phi ) = \BbbB (B\prime , \phi ) for every free
filter \phi on X.
ISSN 1027-3190. Укр. мат. журн., 2021, т. 73, № 4
DENSITY AND CAPACITY OF BALLEANS GENERATED BY FILTERS 469
Example 4. Given an uncountable set X and a regular cardinal \kappa < \mathrm{c}\mathrm{a}\mathrm{r}\mathrm{d}(X), we define \mathrm{S}\mathrm{e}\mathrm{q}(\kappa )
as a family of all subsets A of X such that \mathrm{c}\mathrm{a}\mathrm{r}\mathrm{d}(A) < \kappa . We put
PSeq(\kappa ) = \{ f : X \rightarrow \mathrm{S}\mathrm{e}\mathrm{q}(\kappa ) : \forall x\in X (x \in f(x)\wedge
\wedge \{ y \in X : x \in f(y)\} \in \mathrm{S}\mathrm{e}\mathrm{q}(\kappa ) ) \}
and BSeq(\kappa )(x, f) = f(x) for every (x, f) \in X \times PSeq(\kappa ). Then, thanks to regularity of \kappa ,
\BbbB (\mathrm{S}\mathrm{e}\mathrm{q}(\kappa )) = (X,PSeq(\kappa ), BSeq(\kappa )) is a ballean.
3. Subsets of balleans. Given a ballean \BbbB = (X,P,B), a subset V \subseteq X is called:
bounded if there exist \alpha \in P and x \in X such that V \subseteq B(x, \alpha );
unbounded if V is not bounded;
large if there exists \alpha \in P such that X = B(V, \alpha );
thick if X \setminus V is not large;
\alpha -discrete for \alpha \in P if \{ B(v, \alpha ) : v \in V \} is disjont family.
The set of all, respectively, bounded, unbounded, large, thick subsets of ballean \BbbB = (X,P,B)
will be denoted by, respectively, \mathrm{B}\mathrm{o}\mathrm{u}\mathrm{n}\mathrm{d}(B), \mathrm{U}\mathrm{n}\mathrm{b}\mathrm{o}\mathrm{u}\mathrm{n}\mathrm{d}(B), \mathrm{L}\mathrm{a}\mathrm{r}\mathrm{g}\mathrm{e}(B), \mathrm{T}\mathrm{h}\mathrm{i}\mathrm{c}\mathrm{k}(B). Note that every
dense subset of a metric space (X, d) is large in \BbbB (X, d). However it is not vice versa. For example,
the set of integers is large in \BbbB (\BbbR , d) but it is not dense in (\BbbR , d), where d is an Euclidean metric.
It is also known that for every \alpha \in P each maximal (with respect to inclusion) \alpha -discrete subset
of any ballean \BbbB = (X,P,B) is large in it [1, 4].
4. Density. We define a density of a ballean \BbbB as follows:
\mathrm{d}\mathrm{e}\mathrm{n}(\BbbB ) = min \{ \mathrm{c}\mathrm{a}\mathrm{r}\mathrm{d}(Y ) : Y \subseteq X \wedge Y is large in \BbbB \} .
Recall [1] that if X is infinite and filter \phi is free, then ballean \BbbB (\phi ) is unbounded and connected
(such a ballean is also called a conun) and \phi = \mathrm{L}\mathrm{a}\mathrm{r}\mathrm{g}\mathrm{e}(\BbbB (\phi )).
Remark 2. Note that \mathrm{L}\mathrm{a}\mathrm{r}\mathrm{g}\mathrm{e}(\BbbB (B,\phi )) \subseteq \phi \cap \mathrm{L}\mathrm{a}\mathrm{r}\mathrm{g}\mathrm{e}(\BbbB ). Indeed, if Y \subseteq X is large in \BbbB (B,\phi ),
then there exist F \in \phi and \alpha \in P such that B(Y, (\alpha , F )) = X. Hence F \subseteq Y and B(Y \setminus F, \alpha ) =
= X \setminus F. So, Y \in \phi \cap \mathrm{L}\mathrm{a}\mathrm{r}\mathrm{g}\mathrm{e}(\BbbB ).
Corollary 1. If \phi is a free filter on an infinite set X, then
\mathrm{d}\mathrm{e}\mathrm{n}(\BbbB (\phi )) \mathrm{d}\mathrm{e}\mathrm{n}(\BbbB ) \leq \mathrm{d}\mathrm{e}\mathrm{n}(\BbbB (B,\phi ))
for every ballean \BbbB = (X,P,B). So, if \phi is uniform, then
\mathrm{d}\mathrm{e}\mathrm{n}(\BbbB (B,\phi )) = \mathrm{d}\mathrm{e}\mathrm{n}(\BbbB (\phi )) = \mathrm{c}\mathrm{a}\mathrm{r}\mathrm{d}(X).
Proposition 1. Let \BbbB = (X,P,B) be a connected ballean with infinite support X and let \phi
be a free filter on X containing \scrF \omega (X) and such that X \setminus F is bounded for every F \in \phi . Then
\mathrm{L}\mathrm{a}\mathrm{r}\mathrm{g}\mathrm{e}(\BbbB (B,\phi )) = \phi .
Proof. Take F \in \phi and an arbitrary y \in F. Put G = F \setminus \{ y\} and choose x \in X and \alpha \in P such
that X \setminus G \subseteq B(x, \alpha ). By connectedness of \BbbB there exists \beta \in P such that x \in B(y, \beta ). By upper
multiplicative condition we can find \gamma \in P such that X \setminus G \subseteq B(y, \gamma ). So BP\times \phi (F, (\gamma ,G)) = X.
Hence, F \in \mathrm{L}\mathrm{a}\mathrm{r}\mathrm{g}\mathrm{e}(\BbbB (B,\phi )).
ISSN 1027-3190. Укр. мат. журн., 2021, т. 73, № 4
470 A. BRZESKA
Corollary 2. For every connected and bounded ballean \BbbB = (X,P,B) with an infinite support
X and every free filter \phi containing F\omega (X) the following equality holds:
\mathrm{d}\mathrm{e}\mathrm{n}(\BbbB (B,\phi )) = \mathrm{m}\mathrm{i}\mathrm{n} \{ \mathrm{c}\mathrm{a}\mathrm{r}\mathrm{d}(F ) : F \in \phi \} .
Proposition 2. Let \phi be a free filter on an infinite set X containing \scrF \omega (X) and let \BbbB =
= (X,P,B) be a bounded ballean. Then
\mathrm{d}\mathrm{e}\mathrm{n}(\BbbB (\phi )) = \mathrm{d}\mathrm{e}\mathrm{n}(\BbbB (B,\phi )) > \mathrm{d}\mathrm{e}\mathrm{n}(\BbbB ).
Proof. Let Y \in \phi and choose x0 \in X and \alpha \in P such that B(x0, \alpha ) = X. We consider two
cases:
Case 1: Assume that x0 \in Y. Then Y is large in \BbbB (B,\phi ). Indeed, if x0 \in Y, then
BP\times \phi (Y, (\alpha , Y \setminus \{ x0\} )) = X.
Case 2: If x0 \not \in Y, then Y \cup \{ x0\} appears large in \BbbB (B,\phi ), since
BP\times \phi (Y \cup \{ x0\} , (\alpha , Y )) = Y \cup (B(x0, (\alpha , Y )) \setminus Y ) = X.
Since \mathrm{c}\mathrm{a}\mathrm{r}\mathrm{d}(Y ) = \mathrm{c}\mathrm{a}\mathrm{r}\mathrm{d}(Y \cup \{ x0\} ), so \mathrm{d}\mathrm{e}\mathrm{n}(\BbbB (B,\phi )) \leq \mathrm{d}\mathrm{e}\mathrm{n}(\BbbB (\phi )). Of course, since \BbbB =
= (X,P,B) is bounded, we have \mathrm{d}\mathrm{e}\mathrm{n}(\BbbB ) = 1. By Corollary 1 the proof is complete.
Example 5. If d is the Euclidean metric and \phi = \scrF \omega (\BbbR ), then \aleph 0 = \mathrm{d}\mathrm{e}\mathrm{n}(\BbbB (\BbbR , d)) < \mathrm{d}\mathrm{e}\mathrm{n}(\BbbB (\phi )) =
= \mathrm{d}\mathrm{e}\mathrm{n}(\BbbB (Bd, \phi )) = \mathrm{c}\mathrm{a}\mathrm{r}\mathrm{d}(\BbbR ) although \BbbB (\BbbR , d) is not bounded.
Given a ballean \BbbB = (X,P,B), a set Y \subseteq X has asymptotically isolated \alpha -balls for some
\alpha \in P if, for every \beta > \alpha , there exists y \in Y satisfing B(y, \alpha ) = B(y, \beta ). If there exist \alpha \in P
and a set Y \subseteq X which has asymptotically isolated \alpha -balls, then we say that \BbbB has asymptotically
isolated balls. If \phi \alpha = \{ x \in X : B(x, \alpha ) = \{ x\} \} is nonempty for every \alpha \in P, we will say that \BbbB
has asymptotically isolated 0-balls.
Example 6. If \BbbB is unbounded and has asymptotically isolated 0-balls, then \scrB = \{ \phi \alpha : \alpha \in P\}
is a base for some filter \phi 0 on X. Then \BbbB \prec \BbbB (B,\phi 0) [2]. So, for every free filter on X such that
\phi 0 \subseteq \phi , we have
\BbbB = \BbbB (B,\phi 0) \prec \BbbB (B,\phi ) \prec \BbbB .
Hence
\mathrm{d}\mathrm{e}\mathrm{n}\BbbB = \mathrm{d}\mathrm{e}\mathrm{n}\BbbB (B,\phi 0) = \mathrm{d}\mathrm{e}\mathrm{n}\BbbB (B,\phi ).
5. Capacity. The capacity of a ballean \BbbB = (X,P,B) is determined by its thick subsets.
Namely
\mathrm{c}\mathrm{a}\mathrm{p}(\BbbB ) = sup \{ \mathrm{c}\mathrm{a}\mathrm{r}\mathrm{d}(\scrF ) : \scrF is a disjoint family of thick subsets of X\} .
Recall [4, p. 175], [1] (Theorem 3.1) that \mathrm{c}\mathrm{a}\mathrm{p}(\BbbB ) \leq \mathrm{d}\mathrm{e}\mathrm{n}(\BbbB ) for every ballean \BbbB and \mathrm{c}\mathrm{a}\mathrm{p}(\BbbB ) =
= \mathrm{d}\mathrm{e}\mathrm{n}(\BbbB ) for every ordinal ballean \BbbB .
Remark 3. Note that \mathrm{T}\mathrm{h}\mathrm{i}\mathrm{c}\mathrm{k}(\BbbB (B,\phi )) \supseteq \phi \cup \mathrm{T}\mathrm{h}\mathrm{i}\mathrm{c}\mathrm{k}(\BbbB ) and
\mathrm{c}\mathrm{a}\mathrm{p}(\BbbB ) \mathrm{c}\mathrm{a}\mathrm{p}(\BbbB (\phi )) \leq \mathrm{c}\mathrm{a}\mathrm{p}(\BbbB (B,\phi ))
for every free filter \phi and every ballean \BbbB = (X,P,B).
ISSN 1027-3190. Укр. мат. журн., 2021, т. 73, № 4
DENSITY AND CAPACITY OF BALLEANS GENERATED BY FILTERS 471
Remark 4. Note that if \phi is a free filter on an infinite X and \BbbB = (X, \rho ,B) is ordinal and
unbounded and such that \mathrm{d}\mathrm{e}\mathrm{n}(\BbbB ) = \mathrm{d}\mathrm{e}\mathrm{n}(\BbbB (B,\phi )), then
\mathrm{c}\mathrm{a}\mathrm{p}(\BbbB ) \leq \mathrm{c}\mathrm{a}\mathrm{p}(\BbbB (B,\phi )) \leq \mathrm{d}\mathrm{e}\mathrm{n}(\BbbB (B,\phi )) = \mathrm{d}\mathrm{e}\mathrm{n}(\BbbB ) = \mathrm{c}\mathrm{a}\mathrm{p}(\BbbB ).
Remark 5. If \phi is a free uniform ultrafilter on an infinite set X of the cardinality \kappa , then
cap(\BbbB (\phi )) = 1 and if also \mathrm{d}\mathrm{e}\mathrm{n}(\BbbB ) < \kappa , then \mathrm{c}\mathrm{a}\mathrm{p}(\BbbB (B,\phi )) < \mathrm{d}\mathrm{e}\mathrm{n}(\BbbB (B,\phi )) = \kappa .
Example 7. Note that if \phi = \scrF \omega (\BbbR ), then \phi \subseteq \mathrm{T}\mathrm{h}\mathrm{i}\mathrm{c}\mathrm{k}(\BbbB (\BbbR , d)). Indeed, if Y \in \phi (and, hence,
Y is large in \BbbB (\phi ) = (\BbbR , \phi ,B\phi )), then \BbbR \setminus Y is finite. So
\bigcup
\{ B(y, \epsilon ) : y \in R \setminus Y \} \not = \BbbR for every
\epsilon > 0. This means \BbbR \setminus Y cannot be large in \BbbB (\BbbR , d). Hence, Y \in \mathrm{T}\mathrm{h}\mathrm{i}\mathrm{c}\mathrm{k}(\BbbB (\BbbR , d)).
Recall [1] that if \mathrm{c}\mathrm{a}\mathrm{r}\mathrm{d}(X) = \kappa is regular, then, since \mathrm{L}\mathrm{a}\mathrm{r}\mathrm{g}\mathrm{e}(\BbbB (\mathrm{S}\mathrm{e}\mathrm{q}(\kappa ))) = [X]\kappa , we have
1 = \mathrm{c}\mathrm{a}\mathrm{p}(\BbbB (\mathrm{S}\mathrm{e}\mathrm{q}(\kappa ))) < \mathrm{d}\mathrm{e}\mathrm{n}(\BbbB (\mathrm{S}\mathrm{e}\mathrm{q}(\kappa ))) = \kappa .
Proposition 3. Let X be an uncountably infinite set and let \kappa < \mathrm{c}\mathrm{a}\mathrm{r}\mathrm{d}(X) be a regular cardinal.
Then
\mathrm{d}\mathrm{e}\mathrm{n}(\BbbB (\mathrm{S}\mathrm{e}\mathrm{q}(\kappa )) = \mathrm{d}\mathrm{e}\mathrm{n}(\BbbB (\scrF \kappa )) = \mathrm{d}\mathrm{e}\mathrm{n}(\BbbB (BSeq(\kappa ),\scrF \kappa )) = \mathrm{c}\mathrm{a}\mathrm{r}\mathrm{d}(X)
and
\mathrm{c}\mathrm{a}\mathrm{p}(\BbbB (\mathrm{S}\mathrm{e}\mathrm{q}(\kappa ))) = \mathrm{c}\mathrm{a}\mathrm{p}(\BbbB (\scrF \kappa )) = \mathrm{c}\mathrm{a}\mathrm{p}(\BbbB (BSeq(\kappa ),\scrF \kappa )) = 1.
Proof. To prove the statement we will note that \BbbB (\mathrm{S}\mathrm{e}\mathrm{q}(\kappa )) = \BbbB (\scrF \kappa ) = \BbbB (B,\scrF \kappa ). Indeed, put
\BbbB 1 = \BbbB (\mathrm{S}\mathrm{e}\mathrm{q}(\kappa )), \BbbB 2 = \BbbB (\scrF \kappa ) and \BbbB 3 = \BbbB (BSeq(\kappa ),\scrF \kappa ). We will show that the identity mapping is
an asymorphism in each of the following three cases.
Case 1: \BbbB 1 \prec \BbbB 2. Consider x \in X and f \in PSeq(\kappa ) and put
F = X \setminus f(x).
Then B1(x, f) = f(x) = B2(x, F ).
Case 2: \BbbB 2 \prec \BbbB 3. Take G \in \scrF \kappa and x \in X and put
fF (x) =
\Biggl\{
\{ x\} , x \in F,
X \setminus F, x \not \in F.
Then B2(x, F ) = B3(x, (fF , F )) for every x \in X.
Case 3: \BbbB 3 \prec \BbbB 1. If F \in \scrF \kappa , f \in PSeq(\kappa ) and x \in X, then we define g(x) as follows:
g(x) =
\Biggl\{
\{ x\} , x \in F,
f(x) \setminus F, x \not \in F.
Then B3(x, (f, F )) = B1(x, g). So \BbbB 1, \BbbB 2 and \BbbB 2 are pairwise asymorphic.
Observe that \mathrm{d}\mathrm{e}\mathrm{n}(\BbbB (B,\scrF \kappa )) = \mathrm{m}\mathrm{i}\mathrm{n} \{ F : F \in \scrF \kappa \} = \mathrm{c}\mathrm{a}\mathrm{r}\mathrm{d}(X).
To see that \mathrm{c}\mathrm{a}\mathrm{p}(\BbbB (\mathrm{S}\mathrm{e}\mathrm{q}(\kappa ))) = 1 suppose A \in \mathrm{T}\mathrm{h}\mathrm{i}\mathrm{c}\mathrm{k}(\BbbB (\mathrm{S}\mathrm{e}\mathrm{q}(\kappa ))). Then \mathrm{c}\mathrm{a}\mathrm{r}\mathrm{d}(A) = \mathrm{c}\mathrm{a}\mathrm{r}\mathrm{d}(X),
since otherwise we could find a bijection f : X \setminus A \rightarrow X and define g(x) = \{ f(x), x\} for every
x \in X. Then \BbbB Seq(\kappa )(X \setminus A, g) = X and hence X \setminus A would be large in \BbbB (\mathrm{S}\mathrm{e}\mathrm{q}(\kappa )), so we would
get a contradiction.
So there exists a bijection f : A\rightarrow X and g \in PSeq(\kappa ) (defined as above) such that \BbbB Seq(\kappa )(A, g) =
= X. Hence A \in \mathrm{L}\mathrm{a}\mathrm{r}\mathrm{g}\mathrm{e}(\BbbB (\mathrm{S}\mathrm{e}\mathrm{q}(\kappa ))). So since \mathrm{T}\mathrm{h}\mathrm{i}\mathrm{c}\mathrm{k}(\BbbB (\mathrm{S}\mathrm{e}\mathrm{q}(\kappa ))) \subseteq \mathrm{L}\mathrm{a}\mathrm{r}\mathrm{g}\mathrm{e}(\BbbB (\mathrm{S}\mathrm{e}\mathrm{q}(\kappa ))) we have
\mathrm{c}\mathrm{a}\mathrm{p}(\BbbB (\mathrm{S}\mathrm{e}\mathrm{q}(\kappa ))) = 1.
ISSN 1027-3190. Укр. мат. журн., 2021, т. 73, № 4
472 A. BRZESKA
Example 8. If \phi = \scrF \omega (\BbbR ), then density and capacity of \BbbB (\BbbB (\BbbR , d), \phi ) are equal and uncoun-
table. Indeed, if A \in \mathrm{L}\mathrm{a}\mathrm{r}\mathrm{g}\mathrm{e}(\BbbB (\BbbB (\BbbR , d), \phi )), then A \in \phi , so A is uncountable. For estimating
the capacity we put Ax = \{ x+ k : k \in \BbbZ \} for every x \in \BbbR . Then \{ Ax : x \in [0, 1)\} is a family of
pairwise disjoint thick subsets of \BbbB (\BbbB (\BbbR , d), \phi ). Hence
\mathrm{c}\mathrm{a}\mathrm{p} (\BbbB (\BbbB (\BbbR , d), \phi )) = \mathrm{d}\mathrm{e}\mathrm{n} (\BbbB (\BbbB (\BbbR , d), \phi )) > \mathrm{d}\mathrm{e}\mathrm{n}(\BbbB (\BbbR , d)) = \aleph 0.
Recall that \rho \leq \mathrm{d}\mathrm{e}\mathrm{n}(\BbbB ) for every unbounded ordinal ballean \BbbB = (X, \rho ,B) [1] (proof of the
Theorem 3.1).
Proposition 4. If \BbbB = (X, \rho ,B) is an unbounded ordinal ballean and \mathrm{c}\mathrm{a}\mathrm{r}\mathrm{d}(X) = \kappa > \aleph 0 is
such that \mathrm{d}\mathrm{e}\mathrm{n}(\BbbB ) < \kappa , then
\mathrm{c}\mathrm{a}\mathrm{p}(\BbbB (B,\scrF \rho (X))) = \mathrm{d}\mathrm{e}\mathrm{n}(\BbbB (B,\scrF \rho (X))) = \mathrm{c}\mathrm{a}\mathrm{r}\mathrm{d}(X)
and there exists a disjoint family \scrA \subseteq \mathrm{T}\mathrm{h}\mathrm{i}\mathrm{c}\mathrm{k}(\BbbB (B,\scrF \rho (X))) such that \mathrm{c}\mathrm{a}\mathrm{r}\mathrm{d}(\scrA ) = \kappa .
Proof. Let \kappa = \mathrm{c}\mathrm{a}\mathrm{r}\mathrm{d}(X) and \scrP = \{ X\tau : \tau < \rho \} be a disjoint family of sets, such that\bigcup
\scrP = X and \mathrm{c}\mathrm{a}\mathrm{r}\mathrm{d}(X\tau ) = \kappa for every \tau < \rho . Let \{ x\tau \lambda : \lambda < \kappa \} be an enumeration of X\tau for
every \tau < \rho . Then Z\lambda = \{ x\tau \lambda : \tau < \rho \} \in \mathrm{T}\mathrm{h}\mathrm{i}\mathrm{c}\mathrm{k}(\BbbB (B,\scrF \rho (X)) for every \lambda < \kappa . Indeed, suppose
there exist \lambda < \kappa , F \in \scrF \rho (X) and \alpha < \rho such that B\rho \times \scrF \rho (X)(X \setminus Z\lambda , (\alpha , F )) = X. Then
S = F \setminus (X \setminus Z\lambda ) \not = \varnothing . Hence B\rho \times \scrF \rho (X)(X \setminus Z\lambda , (\alpha , F )) \subseteq X \setminus S \not = X, a contradition.
So, \mathrm{c}\mathrm{a}\mathrm{p}(\BbbB (B,\scrF \rho (X))) = \kappa .
Theorem 1. Let \BbbB = (X, \rho ,B) be an ordinal unbounded ballean with an uncountable sup-
port X and let \phi be a free filter on X containing \scrF \omega (X) and such that X \setminus F \in \mathrm{B}\mathrm{o}\mathrm{u}\mathrm{n}\mathrm{d}(\BbbB )
for every F \in \phi . Then \mathrm{c}\mathrm{a}\mathrm{p}(\BbbB (B,\phi )) = \mathrm{d}\mathrm{e}\mathrm{n}(\BbbB (B,\phi )) and there exists a disjoint family \scrA \subseteq
\subseteq \mathrm{T}\mathrm{h}\mathrm{i}\mathrm{c}\mathrm{k}(\BbbB (B,\scrF \rho (X))) of the cardinality \mathrm{d}\mathrm{e}\mathrm{n}(\BbbB (B,\phi )).
Proof. Assume that \kappa = \mathrm{d}\mathrm{e}\mathrm{n}(\BbbB (B,\phi )). Let F \in \mathrm{L}\mathrm{a}\mathrm{r}\mathrm{g}\mathrm{e}(\BbbB (B,\phi )) be such that \mathrm{c}\mathrm{a}\mathrm{r}\mathrm{d}(F ) = \kappa .
We choose \alpha 1 < \rho and G \in \phi such that X = B\rho \times \phi (F, (\alpha 1, G)) and take an arbitrary x0 \in X \setminus G.
Since \rho \leq \mathrm{d}\mathrm{e}\mathrm{n}(\BbbB ), we shall consider two cases:
Case 1: \rho < \mathrm{c}\mathrm{f} \kappa . Inductively assume that, for some \beta < \rho , we have just defined a family
\{ Y\alpha \subseteq F : \alpha < \beta \} and a strictly increasing sequence \{ \gamma \alpha : \alpha < \beta \} such that, for each \alpha < \beta , the
following conditions hold:
(i) \mathrm{c}\mathrm{a}\mathrm{r}\mathrm{d}(Y\alpha ) = \kappa ;
(ii) \gamma \alpha > \alpha ;
(iii) Y\alpha \subseteq B(x0, \gamma \alpha ) \setminus B(x0, \alpha ).
Consider \beta = \alpha + 1 for some \alpha < \kappa . Let Z = F \setminus B(x0, \gamma \alpha ). Observe that \mathrm{c}\mathrm{a}\mathrm{r}\mathrm{d}(Z) = \kappa .
Indeed, on the contrary L = Z\cup \{ x0\} \in \mathrm{L}\mathrm{a}\mathrm{r}\mathrm{g}\mathrm{e}(\BbbB (B,\phi )). To see this we use the upper multiplicative
condition and find an \alpha 2 < \rho such that B(B(x, \alpha 0), \alpha 1) \subseteq B(x, \alpha 2) for every x \in X. So we obtain
the following:
X = B\rho \times \phi (F, (\alpha 1, G)) \subseteq B\rho \times \phi (Z, (\alpha 1, G)) \cup B\rho \times \phi (B(x0, \alpha 0), (\alpha 1, G)) \subseteq
\subseteq B\rho \times \phi (Z, (\alpha 1, G)) \cup B\rho \times \phi (x0, (\alpha 2, G)) \subseteq
\subseteq B\rho \times \phi (Z \cup \{ x0\} , (\mathrm{m}\mathrm{a}\mathrm{x}(\alpha 2, \alpha 1), G)).
So, Z \cup \{ x0\} \in \mathrm{L}\mathrm{a}\mathrm{r}\mathrm{g}\mathrm{e}(\BbbB (B,\phi )) and its cardinality is smaller then \kappa . A contradiction.
Since \rho < \mathrm{c}\mathrm{f} \kappa we can choose \gamma \beta > \gamma \alpha such that \mathrm{c}\mathrm{a}\mathrm{r}\mathrm{d}(Z \cap B(x0, \gamma \beta )) = \kappa . Put Y\beta =
= Z \cap B(x0, \gamma \beta ).
ISSN 1027-3190. Укр. мат. журн., 2021, т. 73, № 4
DENSITY AND CAPACITY OF BALLEANS GENERATED BY FILTERS 473
If \beta is a limit cardinal then, by regularity of \rho , there exists \gamma < \rho such that \tau > \gamma \alpha for each
\alpha < \beta . Then take Z = F \setminus B(x0, \tau ) and again choose \gamma \tau > \gamma such that \mathrm{c}\mathrm{a}\mathrm{r}\mathrm{d}(Z \cap B(x, \gamma \tau )) = \kappa
and put Y\beta = Z \cap B(x, \gamma \tau ) \subseteq F.
Let
\bigl\{
y\lambda \alpha : \lambda < \kappa
\bigr\}
be an enumeration of Y\alpha for every \alpha < \rho . Then T\lambda =
\bigl\{
y\lambda \alpha : \alpha < \rho
\bigr\}
\not \in
\not \in \mathrm{B}\mathrm{o}\mathrm{u}\mathrm{n}\mathrm{d}(\BbbB ) for every \lambda < \kappa . Indeed, if there exist x \in X and \alpha 0 < \rho such that T\lambda \subseteq B(x, \alpha 0),
then by connectedness and upper multiplicative condition we can find \alpha 1 < \rho satisfying T\lambda \subseteq
\subseteq B(x0, \alpha 1). But y\lambda \alpha \not \in B(x0, \alpha 1) for \alpha > \alpha 1. A contradiction.
So, T\lambda \in \mathrm{T}\mathrm{h}\mathrm{i}\mathrm{c}\mathrm{k}(\BbbB (B,\phi )) for every \lambda < \kappa .
Case 2: Now assume \mathrm{c}\mathrm{f} \kappa \leq \rho \leq \mathrm{d}\mathrm{e}\mathrm{n}(\BbbB ). Let g : \rho \rightarrow \kappa be an injection holding g(\rho ) cofinal in
\kappa . Inductively assume that for some \beta < \rho we have already defined a family \{ Y\alpha \subseteq F : \alpha < \beta \} and
a strictly increasing sequence \{ \gamma \alpha : \alpha < \beta \} such that following conditions hold:
(i) \gamma \alpha > \alpha ;
(ii) Y\alpha \subseteq B(x, \gamma \alpha ) \setminus B(x, \alpha );
(iii) \mathrm{c}\mathrm{a}\mathrm{r}\mathrm{d}(Y\alpha ) = g(\alpha ).
Consider \beta = \alpha + 1 for some \alpha < \rho . Put Z = F \setminus B(x0, \gamma \alpha ). Since \mathrm{c}\mathrm{a}\mathrm{r}\mathrm{d}(Z) = \kappa (compare with
Case 1) and \rho < \kappa there exists a \gamma \beta > \gamma \alpha such that \mathrm{c}\mathrm{a}\mathrm{r}\mathrm{d}(B(x0, \gamma \beta ) \cap Z) \geq g(\beta ). We choose
Y\beta \subseteq B(x0, \gamma \beta ) \cap Z of the cardinality g(\beta ).
If \beta is a limit cardinal then, by regularity of \rho , there exists \gamma < \rho such that \tau > \gamma \alpha for each
\alpha < \beta . Then take Z = F \setminus B(x0, \tau ) and choose \gamma \tau > \gamma such that \mathrm{c}\mathrm{a}\mathrm{r}\mathrm{d}(Z \cap B(x, \gamma \tau )) = g(\beta ). Put
Y\beta = Z \cap B(x0, \gamma \tau ).
Let
\bigl\{
y\lambda \alpha : \lambda < g(\alpha )
\bigr\}
be an enumeration of Y\alpha for every \alpha < \rho and let define
T\alpha
\lambda =
\Bigl\{
y\lambda \beta : \alpha + 1 \leq \beta < \rho
\Bigr\}
for every \alpha < \rho and \lambda < \kappa such that g(\alpha ) < \lambda < g(\alpha + 1). Then X \setminus T\alpha
\lambda \not \in \mathrm{B}\mathrm{o}\mathrm{u}\mathrm{n}\mathrm{d}(\BbbB ) for every \lambda
and \alpha (compare with Case 1), so T\alpha
\lambda \in \mathrm{T}\mathrm{h}\mathrm{i}\mathrm{c}\mathrm{k}(\BbbB (B,\phi )). Of course
\scrA = \{ T\alpha
\lambda : g(\alpha ) < \lambda < g(\alpha + 1) \wedge \alpha < \rho \}
is a disjoint family and thanks to choice of g, \mathrm{c}\mathrm{a}\mathrm{r}\mathrm{d}(\scrA ) = \kappa .
We conclude also that Theorem 3.3 of [1] implies the following statement.
Corollary 3. Let \BbbB = (X,P,B) be a ballean, \mathrm{c}\mathrm{a}\mathrm{r}\mathrm{d}(X) = \kappa , \mathrm{c}\mathrm{a}\mathrm{r}\mathrm{d}(P ) \leq \kappa . Then, for every
free filter \phi on X, there exists a disjoint family \scrF \subseteq \mathrm{T}\mathrm{h}\mathrm{i}\mathrm{c}\mathrm{k}(\BbbB (B,\phi )) such that \mathrm{c}\mathrm{a}\mathrm{r}\mathrm{d}(\scrF ) = \kappa and
provided that one of the following conditions is satisfied:
(i) there exists \kappa \prime < \kappa such that \mathrm{c}\mathrm{a}\mathrm{r}\mathrm{d}(B(x, \alpha )) \leq \kappa \prime for all x \in X and \alpha \in P ;
(ii) \mathrm{c}\mathrm{a}\mathrm{r}\mathrm{d}(B(x, \alpha )) < \kappa for all x \in X and \alpha \in P and \kappa is regular.
References
1. I. V. Protasov, Cellularity and density of balleans, Appl. General Topology, 8, № 2, 283 – 291 (2007).
2. O. V. Petrenko, I. V. Protasov, Balleans and filters, Mat. Stud., 38, № 1, 3 – 11 (2012).
3. I. V. Protasov, Coronas of balleans, Topology and Appl., 149, 149 – 160 (2005).
4. I. Protasov, M. Zarichnyi, General asymptology, Math. Stud. Monogr. Ser., 12, VNTL Publ., Lviv (2007).
Received 18.04.17,
after revision — 23.02.21
ISSN 1027-3190. Укр. мат. журн., 2021, т. 73, № 4
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| id | umjimathkievua-article-648 |
| institution | Ukrains’kyi Matematychnyi Zhurnal |
| keywords_txt_mv | keywords |
| language | English |
| last_indexed | 2026-03-24T02:03:31Z |
| publishDate | 2021 |
| publisher | Institute of Mathematics, NAS of Ukraine |
| record_format | ojs |
| resource_txt_mv | umjimathkievua/b5/9415527debb8c6e26ef00d4f25e623b5.pdf |
| spelling | umjimathkievua-article-6482025-03-31T08:48:15Z Density and capacity of balleans generated by filters DENSITY AND CAPACITY OF BALLEANSGENERATED BY FILTERS Density and capacity of balleans generated by filters Brzeska, A. brzeska, anna Brzeska, A. ballean ballean-lter mix ballean-ideal mix ballean ballean-lter mix ballean-ideal mix UDC 519.51 We consider a ballean $\mathbb B=(X,P,B)$ with an infinite support $X$ and a free filter $\phi$ on $X$ and define $B_{P\times\phi}(x,(\alpha,F))$ for every $\alpha\in P$ and $F\in \phi.$ The ballean $(X,P\times\phi, B_{P\times\phi})$ will be called the ballean-filter mix of $\mathbb B$ and $\phi$ and denoted by $\mathbb B(B,\phi).$ It was introduced in [O. V. Petrenko, I. V. Protasov, Balleans and filters, Mat. Stud., 38, No. 1, 3–11 (2012)] and was used to construction of a non-metrizable Frechet group ballean. In this paper some cardinal invariants are compared. In particular, we give a partial answer to the question: if we mix an ordinal unbounded ballean with a free filter of the subsets of its support, will the mix-structure's density be equal to its capacity, as it holds in the original balleans? UDC 519.51 Щiльнiсть та ємнiсть болеанiв, згенерованих фiльтрами Розглядається болеан $\mathbb B=(X,P,B)$ з нескінченним супортом $X$ і вільний фільтр $\phi$ на $X$ та визначається $B_{P\times\phi}(x,(\alpha,F))$ для кожного $\alpha\in P$ та $F\in \phi.$ Болеан $(X,P\times\phi, B_{P\times\phi})$ називають болеан-фільтр міксом для $\mathbb B$ і $\phi$ та позначають $\mathbb B(B,\phi).$ Таку термінологію було введено у статті [O. V. Petrenko, I. V. Protasov, Balleans and filters, Mat. Stud., 38, No. 1, 3–11 (2012)], де її застосовано для побудови болеана групи Фреше без метризації. У цій роботі порівнюються деякі кардинальні інваріанти. Зокрема, наведено часткову відповідь на питання: якщо є мікс ординально необмеженого болеана з вільним фільтром підмножин його супорту, то чи буде щільність мікс-структури рівною її ємності, як це має місце для оригінальних болеанів? Institute of Mathematics, NAS of Ukraine 2021-04-21 Article Article application/pdf https://umj.imath.kiev.ua/index.php/umj/article/view/648 10.37863/umzh.v73i4.648 Ukrains’kyi Matematychnyi Zhurnal; Vol. 73 No. 4 (2021); 467 - 473 Український математичний журнал; Том 73 № 4 (2021); 467 - 473 1027-3190 en https://umj.imath.kiev.ua/index.php/umj/article/view/648/9000 |
| spellingShingle | Brzeska, A. brzeska, anna Brzeska, A. Density and capacity of balleans generated by filters |
| title | Density and capacity of balleans generated by filters |
| title_alt | DENSITY AND CAPACITY OF BALLEANSGENERATED BY FILTERS Density and capacity of balleans generated by filters |
| title_full | Density and capacity of balleans generated by filters |
| title_fullStr | Density and capacity of balleans generated by filters |
| title_full_unstemmed | Density and capacity of balleans generated by filters |
| title_short | Density and capacity of balleans generated by filters |
| title_sort | density and capacity of balleans generated by filters |
| topic_facet | ballean ballean-lter mix ballean-ideal mix ballean ballean-lter mix ballean-ideal mix |
| url | https://umj.imath.kiev.ua/index.php/umj/article/view/648 |
| work_keys_str_mv | AT brzeskaa densityandcapacityofballeansgeneratedbyfilters AT brzeskaanna densityandcapacityofballeansgeneratedbyfilters AT brzeskaa densityandcapacityofballeansgeneratedbyfilters |