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Total boundness of reloids
by Victor Porton
June 6, 2013
Abstract
I generalize total boundness of uniform spaces for arbitrary reloids (lters on a cartesian pro-
duct of sets). For reloids total boundness splits into two different concepts: α-total boundness
and β-total boundness.
Notation
I call reloid a triple (A; B; F) where A, B are sets and F is a filter on the cartesian product A × B
of these sets. I will denote GR(A; B; F) = F for a reloid (A; B; F). Relo ids are a generalization
of uniform spaces. Source of a reloid is Src(A; B; F) = A and destination Dst(A; B; F) = B. I also
denote xyGR(A; B; F ) = {(A; B; f ) | f F }.
I will r efer to a pair f = (U; E) of a (small) set U and a binary relation E E × E as Rel-
endomorph ism. Furthermore Ob f = U and GR f = E.
I denote hE iX = {y | x X: x E y} for a binary relation E and set X and hE iX = hGR Ei X for
a Rel-endomorphism E.
I define composition of reloids (B; C; G) (A; B; F ) = (A; C; H) where H is the filter induced by
the filter base {g f | f F , g G}.
The reverse reloid is defined by the formula (A; B; F )
1
= (B; A; F
1
).
I define partial order on the set of reloids as (A; B; F ) (A; B; G) F G. The set of reloids with
given source and destination is a complete lattice with join denoted and meet denotes .
See http://www.mathematics21.org/algebraic-general-topology.html for mor e de tails.
Thi ck binary relations
Definition 1. I will call α-thick and denote thick
α
(E) a Rel-endomorphism E when there exists
a finite cover S of Ob E such that A S: A × A GR E.
Definition 2. CS(S) =
S
{A × A | A S} for a collection S of sets.
Remark 3. CS means “cartesian squares”.
Obvious 4. A Rel-endomo rphism is α-thick iff there exists a finite cove r S of Ob E such that
CS(S) GR E.
Definition 5. I will ca ll β-thick and de note thick
β
(E) a Rel -endomorphism E when iff there
exists a finite set B such that hE iB = Ob E.
Proposition 6. thick
α
(E) thick
β
(E).
1
Proof. Let thick
α
(E). Then there exists a nite cover S of the set Ob E such that A S:
A × A GR E. Without loss of generality assume A
for every A S. So A hE i{x
A
} for some
x
A
for every A S. So hE i{ x
A
| A S} =
S
{hE i{x
A
} | A S} = Ob E and thus E is β-thick.
Obvious 7. Let X be a set, A an d B are Rel-endorelations on X and B A. Then:
thick
α
(A) thick
α
(B);
thick
β
(A) thick
β
(B).
Example 8. There is a β-thick Rel-morphism which is not α-thick.
Proof. Consider the Rel-morphism on [0; 1] with the below graph:
Γ = {(x; x) | x [0; 1]} {(x; 0) | x [0; 1]} {(0; x) | x [0; 1]}.
Γ is β-chick because hΓi{0} = [0; 1].
To prove that Γ is not α-thick it’s enough to prove that every set A such that A × A Γ is finite.
Suppose for the contrary th at A is infinite. Then A contains more than one non-zero points y, z
(y
z). Without loss of generality y < z. So we have that (y, z) is not of the form (y; y) nor (0;
y) nor (y; 0). Theref ore A × A isn’t a subset of Γ.
Totally bounded en do-reloids
The below is a straightforward generalization of the customary definition of to tally bounded sets
on uniform spaces (it’s proved be low that for uniform spaces the below definitions are equivalent).
Definition 9. An endoreloid f is α-totally bounded (totBound
α
(f )) if every E xyGR f is α-thick.
Definition 10. An endo reloid f is β -totally bounded (totBound
β
(f )) if every E xyGR f is β-
thick.
Remark 11. We could rewrite the above definitions in a more algebraic way like xyGR f thick
α
(with thick
α
would be defined as a se t rather than as a predicate), but we don’t really need thi s
simplification.
Proposition 12. If an endoreloid is α-totally bounded then it is β-totally bounded.
2
Proof. Because thick
α
(E) thick
β
(E).
Proposition 13. If an endoreloid f is reflexive and Ob f is finite then f is both α-totally bounded
and β-totally bounded.
Proof. It enough to prove that f is α-totally bounded. Really, every E xyGR f is reflexive.
Thus {x} × {x} E for x Ob f and thus {{x} | x Ob f } is a sought for finite cover of Ob f.
Obvious 14.
A principal endo-reloid induced by a Rel-morphism E is α-totally bounded iff E is α-thick.
A principal end o-reloid induced by a Rel-morphism E is β-totally bounded iff E is β-thick.
Example 15. There is a β-totally bounded endoreloid which is not α-totally bounded.
Proof. It follows from the example above and properties of principal endoreloids.
Special case of uniform spaces
Definition 16. Uniform space is essentially the same as symmetric, reflexive and transitive endo-
reloid.
Exercise 1. Prove that it is essentially the same as the standard definition of a uniform space (see Wikipedia
or PlanetMath).
Theorem 17. Let f is such a endoreloid that f f
1
f . Then f is α-totally bounded iff it is
β–totally bounded.
Proof.
. Proved above.
. For every ε GR f we have that hεi{c
0
},
, hεi{c
n
} covers the space. hεi{c
i
} × hεi{c
i
}
ε ε
1
because for x hεi{c
i
} (the same as c
i
hε
1
i{x}) we have hhεi{c
i
} × hεi{c
i
}i{x} =
hεi{c
i
} hεihε
1
i{x} = hε ε
1
i{x}. There exists ε
GR f such that ε ε
1
ε
because
f f
1
f. Thus for every ε
we have hεi{c
i
} × hεi{c
i
} ε
and so
hεi{c
0
},
, hεi{c
n
}.
is a sought for finite cover.
Corollary 18. A uniform space is α-totally bounded iff it is β-totally bounded .
Proof. From the theorem and the defi nition of uniform spaces.
Relationshi ps with other properties
Theorem 19. Let µ and ν are endoreloids. Let f is a principal C
(µ; ν) continuous, monovalued,
surjective re loid. Then if µ is β-totally bounded then ν is also β-totally bounded.
3
Proof. Let ϕ is the monovalued, surjective f unction, which induce s the relo id f .
We have µ f
1
ν f .
Let F GR ν. Then there exists E GR µ such that E ϕ
1
F ϕ.
Since µ is β-totally bounded, ther e exists a finite subset A of Ob µ such that hEiA = Ob µ.
We claim hF ihϕiA = Ob ν.
Indeed let y Ob ν be an arbitrary point. Since ϕ is surjective, there exists x Ob µ such that
ϕx = y. Since hE iA = Ob µ there exists a A such that a E x and thus a (ϕ
1
F ϕ) x. So
(ϕa; y) = (ϕa; ϕx) F . Therefore y hF ihϕiA.
Theorem 20. Let µ and ν are endoreloids. Let f is a principal C
′′
(µ; ν) continuous, surjective
reloid. Then if µ is α-totally bounded then ν is also α-totally bounded.
Proof. Let ϕ is the s urjective bina ry re lat ion which induces the re loid f.
We have f µ f
1
ν.
Let F GR ν. Then there exists E GR µ such that ϕ E ϕ
1
F .
There exists a finite cover S of Ob µ such that
[
{A × A | A S} E.
Thus ϕ (
S
{A × A | A S }) ϕ
1
F that is
S
{hϕiA × hϕiA | A S} F .
It remains to prove that {hϕiA | A S } is a cover of Ob ν. It is true because ϕ is a surjection and
S is a cover of Ob µ.
A stronger statement (principality req uirement removed):
Conjecture 21. The image of a uniformly continuous entirely defined monovalued s urjective reloid
from a (α-, β-)totally bounded endoreloid is also (α-, β-)totally bounded.
Can we remove the requirement to be entirely defined from the above conjecture?
Question 22. Under which conditions it’s true that join of (α-, β-) totally bo unded reloids is also
totally bounded?
Additiona l predicates
We may consider also the following predicates expressing different kinds of what is intuiti vely is
understood as boundness. Their usefulne ss is unclear, but I present them for completeness.
E GR f n N: thick
α
(E
n
)
E GR f n N: thick
β
(E
n
)
E GR f n N: thick
α
(E
0
E
n
)
E GR f n N: thick
β
(E
0
E
n
)
n N: totBound
α
(f
n
)
4
n N: totBound
β
(f
n
)
n N: totBound
α
(f
0
f
n
)
n N: totBound
β
(f
0
f
n
)
totBound
α
(f
0
f
1
f
2
)
totBound
β
(f
0
f
1
f
2
)
Some of the above defined predicates are equivalent:
Proposition 23.
E GR f n N: thick
α
(E
n
) n N: totBound
α
(f
n
).
E GR f n N: thick
β
(E
n
) n N: totBound
β
(f
n
).
Proof. Because every F GR f
n
is a superset of E
n
for some E GR f.
Proposition 24.
E GR f n N: thick
α
(E
0
E
n
) n N: totBound
α
(f
0
f
n
).
E GR f n N: thick
β
(E
0
E
n
) n N: totBound
β
(f
0
f
n
).
Proof. f
0
f
n
= f
0
f
n
. Thus every F GR(f
0
f
n
) we have F f
k
, thus F E
k
k
for all k for some E
k
GR f and so F E
0
E
n
where E = E
0
E
k
GR f .
Proposition 25. All predicates in th e above list are pairwise equivalent in the case if f is a
uniform space .
Proof. Because f f = f.
5