Download equivalents of the compactness theorem for locally finite sets of

Adam Kolany
EQUIVALENTS OF THE COMPACTNESS THEOREM
FOR LOCALLY FINITE SETS OF SENTENCES
Abstract
We prove that the existence of consistent choices for dense relations, in short
Ff in , is tantamout in ZF to the existence of such choices for n–element sets
with n ≥ 3. Our interest in Ff in stems from [2] where it has been shown that
Ff in is equivalent to CTf in – the Compactness Theorem for locally finite sets of
propositional sentences. The condition CTf in , in turn, was suggested by Cowen
[1] in connection with the P = N P – problem.
The question whether Ff in is equivalent to ACf in , as well as the equivalence
f in
of F2 and AC2 are still open.
A binary relation R on a set X is said to be dense iff for every a ∈ X
{b ∈ X : ¬aRb} is finite;
S
let A be a family of sets and R a binary symmetric relation on A. A
choice set S on A is R–consistent iff aRb for every a 6= b, a, b ∈ S.
The existence of consistent choices, and its variants, are formulated as
follows:
F – for every nonempty family A of pairwise disjoint, finite and nonempty
S
sets and a binary symmetric relation R on A if every finite subfamily
of A has an R–consistent choice, then the whole family A has such a
choice set;
Ff in – as above with additional assumption that R is dense;
Fnf in – as above for families of exactly n–element sets;
f in
F≤n
– as in Ff in for families of at most n–element sets.
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Facts.
f in
Theorem 1. Fn+1
→ Fnf in for every n ≥ 2.
Proof. Let every finite subfamily A0 of A has an R–consistent choice.
With every A ∈ A we corelate two distinct new elements, say ∗1A and ∗2A .
Then we put A1∗ := A1 ∪ {∗1A }, A2∗ := A2 ∪ {∗2A }, where A1 , A2 are two
disjoint “copies” of A e.g. A1 : {(a, 1) : a ∈ A}, A2 := {(a, 2) : a ∈ A}.
Moreover let A∗ := {Ai∗ : A ∈ A, i = 1, 2}, A∗ : A1∗ ∪ A2∗ and π((a, i)) = a
for a ∈ A, i = 1, 2. It should be noticed that A∗ is a family of n+1–element
sets.
A
:
A1∗
:
A2∗
:
a
b
...
z
\|/
a1
|
a2
/|\
\|/
b1 . . .
|
b2 . . .
/|\
\|/
z1
|
z2
/|\
A binary symmetric relation R∗ on
diagram):
S
\|/
∗1A
∗2A
/|\
A∗ is defined as follows (see above
1.
If x ∈ A∗ , y ∈ B ∗ , A 6= B, A, B ∈ A, then
→ if x or y is an “asterisk”, then xR∗ y;
→ otherwise xR∗ y iff π(x)Rπ(y),
2.
If x, y ∈ A∗ , A ∈ A, then
→ if one of x or y is an “asterisk” then they are not in R;
→ otherwise xR∗ y iff π(x) = π(y).
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f in
Of course R∗ is dense, hence by Fn+1
, then exists an R∗ –consistent choice
S ∗ of A∗ . Let us notice that for no A ∈ A, and no j = i, 2 ∗jA ∈ S ∗ . Let
A ∈ A. If S ∗ ∩ A1 = {a1 } and S ∗ ∩ A2 = {a2 } then a1 R∗ a2 which shows,
by the definition of R∗ , that a1 and a2 are the copies of the same element
of A. Thus the set S := {π(x) : x ∈ S ∗ } is a choice on A. It is not difficult
to notice that S is R–consistent. 2
f in
Corollary 1. F f in → . . . → Fn+1
→ Fnf in → . . . → F3f in → F2f in .
Corollary 2. Fnf in →
V
k≤n
ACk .
Proof. One can easily show that Fkf in → ACk , for every k ≥ 2.
f in
Theorem 2. F3f in → F≤3
.
Proof. Let A be a family of at most three–element, nonempty and pairS
wise disjoint sets and R be a binary symmetric relation on A such that
every finite subfamily A0 of A has an R–consistent choice. For any A ∈ A
we define three–element set A∗ in a following way:
1.
2.
3.
If A is three–element, we just put A∗ := A;
If A is two–element, we choose aA ∈ A (by AC2 –comp. Cor. 2),
and take A∗ := A ∪ {a∗A }, where a∗A is a fresh “copy” of aA ;
If A has exactly one–element a, we choose its two “copies” a∗ and
a∗∗ , and take A∗ := A ∪ {a∗ , a∗∗ }.
Then we take A∗ := {A∗ : A ∈ A} and we define a binary relation R∗ on
S ∗
A in the following way:
xR∗ y iff π(x)Rπ(y), for x, y ∈ A∗ .
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where
(
π(x) :=
aA if x = a∗A , a∗∗
A
x otherwise
S
Now the number of those elements y ∈ A∗ such that ¬(xR∗ y), is not
S
greater than twice the number of those a ∈ A for which ¬(π(x)Ra).
Thus R is dense.
Since every R–consistent choice on A is also an R∗ –consistent choice
on A∗ , we get an R∗ –consistent choice S on the family A∗ . Then we easily
see that {π(x) : x ∈ S} is an R–consistent choice on A. 2
As it is known (see [2]) Ff in is equivalent to some statement about
propositional calculus. We consider the language {¬, ∧, ∨} and accept standard definitions of propositional formulae.
A set X of propositional formulas is said to be locally satisfiable iff
every finite subset X0 of X is satisfiable; X is locally finite iff every variable
p of X appears only in finite number of formulas of X . The Compactness
Theorem and its variants are the following statements:
CT – every locally satisfiable set of propositional formulas is satisfiable;
CTf in – every locally satisfiable and locally finite set of propositional formulas is satisfiable;
n-Sat – every locally satisfiable set of elementary disjunctions consisting of
at most n literals (a literal is a variable or its negation) is satisfiable;
n-Satf in – as above for locally finite sets of elementary disjunctions;
f in
Theorem 3. F≤3
→ 3-Satf in .
Proof. Let X be locally satisfiable family of at most three–literal disjunctions in which every variable appears only finitely often. Let now for
α ∈ X,
Aα := {(l, α) : l is a literal in α}
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and let A := {Aα : α ∈ X }. Of course A is a family of non–empty, at most
three–element, pairwise disjoint sets. Taking R defined as follows:
(l1 , α1 )R(l2 , α2 ) iff {l1 , l2 } is satisfiable,
f in
we easily see that A and R satisfy premises of F≤3
. Indeed, the density of
R follows from local finiteness of X . Let now A0 be a finite subfamily of
A. Then there is a finite X0 ⊆ X such that A0 ⊆ {Aa : α ∈ X0 }. Hence
X0 is satisfied by some valuation v0 . Since elements of X0 are elementary
disjunctions, the sets
{(l, α) ∈ Aα : v0 (l) = 1}, for α ∈ X0
are nonempty. We know that for finite families AC holds, so we can choose
exactly one (lα , α) from each of the above sets. Hence the set {(lα , α) : α ∈
X0 } is an R–consistent choice on A0 . Thus A has an R–consistent choice
S. Putting now for any variables p:
v(p) := 1 iff (p, α) ∈ S for some α ∈ X
we get a valuation satisfying X . 2
Corollary 3. Fnf in → 3-Satf in , for n ≥ 3.
R. Cowen proved in [1] that 3-Sat ↔ CT . The same argument can be
used to establish 3-Satf in → CTf in . Thus,
Corollary 4.
(i) Fnf in ↔ Ff in , n ≥ 3,
(ii) Fnf in ↔ CTf in , n ≥ 3.
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Proof. The proof follows clearly from the above considerations and from
[2].
The above could be seen as a variant of Fn ↔ F, n ≥ 3, proved by
Levy [3] and Mycielski [5]. Let us, however, note that the arguments are
quite different.
As an immediate corollary of the above considerations we also obtain
ACn →
| Fnf in , for n ≥ 3.
In case of
AC2 → F2f in and ACf in → Ff in
the problem is still open.
Remark 1. Levy proves in [3] that F2 →
| AC3 . Hence F2f in is not equivalent
to Ff in . Besides, using similar method as Levy, we can demonstrate that
| ACk for k 6= 2, 4.
F2 →
| ACk for every k 6= 2, 4. This shows F2f in →
Warning. Implication ACf in → Ff in cannot be falsified with the popular technic of “permutation models”. It is known that ACf in → ACcount
is true in any permutation model of ZF A (see [4], pp. 114, ex 11]) and
ACcount → Ff in is proved in [2].
References
[1] R.H. Cowen, Two hypergraph theorems equivalent to BPI, Notre
Dame Journal of Formal Logic 31 (1990), pp. 232–240.
[2] A. Kolany, P. Wojtylak, Some remarks on the Compactness Theorem, to appear.
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[3] A. Levy, Remarks on a paper by J.Mycielski, Acta Math. 14 (1963),
pp. 125–130.
[4] T.J. Jech, The axiom of choice, North Holland, Amsterdam–London
1973. Studies in Logic and Foundation of Mathematics 75.
[5] Mycielski, Some remarks and problems on the colouring of infinite
graphs and on the theorem of Kuratowski, Acta Math. Acad. Sci. Hung.,
12 (1961), pp. 125–129.
Department of Mathematics
Silesian University
40–007 Katowice, Bankowa 14
Poland
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