Proof. Suppose that f and g are two diﬀerent bijective reloids from A to B. Then g
−1
◦ f is not
the identity reloid (otherwise g
−1
◦ f = I
dom f
RLD
and so f = g). But g
−1
◦ f is a bijective reloid (as a
composition of bijective reloids) from A to A wha t is impossible.
4 Rud in-Keisler equivalence and Rudin-Keisler o rder
Theorem 61. Atomic ﬁlter objects a and b (with possibly diﬀerent bases) are isomorphic iﬀ
a > b ∧ b > a .
Proof. Let a > b ∧ b > a. Then there are a monovalued reloids f and g such that dom f = a a nd
im f = b and dom g = b and im g = a. Thus g ◦ f is a monovalued morphism from a to a. By the
above we have g ◦ f = I
a
RLD
so g = f
−1
and f
−1
◦ f = I
a
RLD
so f is monovalued. Thus f is an injective
monovalued reloid from a to b and thus a and b are isomorphic.
The last the ore m cannot be generalize d from atomic f .o. to arbitrary f.o., as it’s shown by the
following two examples:
Example 62. A >
1
B ∧ B >
1
A but A is not isomorphic to B for some f.o. A and B.
Proof. Consider A = ↑
R
[0; 1] and B =
T
{↑
R
[0; 1 + ε) | ε > 0}. Then the function f = {(x;
x/2) | x ∈R} witnesses both inequalities A >
1
B and B >
1
A. But these ﬁlters cannot be isomor phic
because only one of them is principal.
Lemma 63. Let f
0
and f
1
are Set-morphisms. Let f(x; y) = (f
0
x; f
1
y) for a function f. Then
↑
FCD(Dst f
0
;Dst f
1
)
f
(A ×
RLD
B) = h↑f
0
iA ×
RLD
h↑f
1
iB.
Proof.
↑
FCD(Dst f
0
;Dst f
1
)
f
(A ×
RLD
B) =
↑
FCD(Dst f
0
;Dst f
1
)
f
T
{↑
Src f
0
×Src f
1
(A × B) | A ∈ up A,
B ∈ up B} =
T
{↑
Src f
0
×Src f
1
hf i(A × B) | A ∈ up A, B ∈ up B} =
T
{↑
Dst f
0
×Dst f
1
(hf
0
iA ×
hf
1
iB) | A ∈ up A, B ∈ up B} =
T
{↑
Dst f
0
hf
0
iA ×
RLD
↑
Dst f
1
hf
1
iB | A ∈ up A, B ∈ up B} =
(theorem 164?? in [3])=
T
{↑
Dst f
0
hf
0
iA | A ∈ up A} ×
RLD
T
{↑
Dst f
1
hf
1
iB | A ∈ up B} =
h↑f
0
iA ×
RLD
h↑f
1
iB.
Lemma 64. If an f.o. A is isomorphic to an f.o. B then if X is a set and ↑
Base(A)
X ∩ A is an
atomic f.o., then there exists a set Y such that ↑
Base(A)
X ∩ A is an atomic f.o. isomorphic to
↑
Base(A)
Y ∩ B.
[FIXME: See the book for a cor rected proof.]
Proof. Let A is isomorphic to B. Then there are sets A ∈up A, B ∈up B such that A ÷A is directly
isomorphic to B ÷ B. So there are a bijection f : PA ∩ up A → PB ∩ up B such that B = hf iA.
[FIXME: ?? equality is wrong.]
up
↑
Base(A)
X ∩ A
= up
↑
Base(A)
(X ∩ A) ∩ A
= ?? = hX ∩ A ∩ iup A = hX ∩ i(PA ∩ up A).
Thus hhf iiup
↑
Base(A)
X ∩ A
= hhf iihX ∩ i(PA ∩ up A) = hf (X) ∩ ihhf ii(PA ∩ up A) =
hf (X) ∩ i(PB ∩ up B) = hf(X) ∩ B ∩ iup B = hf (X) ∩ iup B = up
↑
Base(B )
(f (X)) ∩ B
.
So hf i
↑
Base(A)
X ∩ A
=
T
↑
Base(B )
hhf iiup
↑
Base(A)
X ∩ A
=
T
↑
Base(B)
up
↑
Base(B )
(f (X)) ∩ B
= ↑
Base(B )
(f (X)) ∩ B.
Finally we have ↑
Base(B )
(f (X)) ∩ B is isomorphic to ↑
Base(A)
X ∩ A from the last equality.
Theorem 65. Let f is a monovalued injective reloid. Then f is isomorphic to the f.o. dom f.
Proof. Let f is a monovalued injective reloid. There exists a bijection F ∈ up f . Consider the
bijective function p = {(x; Fx) | x ∈ dom F }.
hpidom F = F and consequently hpidom f =
T
↑
RLD(Dst f ;Src f )
hpidom K | K ∈ up f
=
T
↑
RLD(Dst f ;Src f )
hpidom(K ∩ F ) | K ∈ up f
=
T
↑
RLD(Dst f ;Src f )
(K ∩ F ) | K ∈ up f
=
T
↑
RLD(Dst f ;Src f )
K | K ∈ up f
= f . Thus p witnesses that f is isomorphic to the f.o. dom f .
Corollary 66. A monovalued injective reloid with atomic domain is atomic.
Rudin-Keisler equivalence and Rudin-Keisler order 11