Saharon Shelah- Zero One Laws for Graphs with Edge Probabilties Decaying with Distance, Part II

8/3/2019 Saharon Shelah- Zero One Laws for Graphs with Edge Probabilties Decaying with Distance, Part II

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ZERO ONE LAWS FOR GRAPHS WITH EDGE PROBABILITIES

DECAYING WITH DISTANCE, PART II

SH517

SAHARON SHELAH

Abstract. Fixing an irrational (0, 1) we prove the 0-1 law for the randomgraph on [n] with the probability of {i, j} being an edge being essentially

1|ij|

.

0. Introduction

This continues [Sh:467] which is Part I, background and a description of theresults are given in [I,0]; as this is the second part, our sections are named 4 - 6and not 1 - 3.

In 5 we just state the necessary probabilistic inequalities quoting [Sh:E49]. In[Sh:E49] we prove the case with a successor function (this was originally 7).

Recall that we fix an irrational (0, 1)R and the random graph Mn = M0n isdrawn as follows:

(a) its set of elements is [n] = {1, . . . , n}

(b) for i < j in [n] the probability of {i, j} being an edge is p|ij| where

p = 1/ if > 1 as is 1/2 if = 1 or just 1 p = 1/ for > 1

(c) the drawing for the edges are independent

(d) Kn is the set of possible values ofMn,K is the class of graphs.

Our main interest is to prove the 0-1 laws (for first order logic) for this 0-1 context,but also to analyze the limit theory.

We can now explain our intentions.

Zero Step: We define relations


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2 SAHARON SHELAH

Remember from [I,1] that A


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ZERO ONE LAWS FOR GRAPHS WITH EDGE PROBABILITIES DECAYING WITH DISTANCE, PART I ISH5173

1. Applications

1(A). The Basics for the case of 1

.

We intend to apply the general theorems (Lemmas [I,2.17,2.19]), to our problem.That is, we try to answer: does the main context M0n with pi = 1/i

for i > 1satisfies the 0-1 law? So here our irrational number (0, 1)R is fixed. We workin Main Context (see 1.1 below, the other one, M1n , would work out as well, see7).

{1.1}Context 1.1. A particular case of [I,1.1]: pi = 1/i for i > 1, p1 = p2 (where (0, 1)R is a fix irrational) and the n-th random structure is Mn = M0n = ([n], R)(i.e. only the graph with the probability of {i, j} being p|ij|).

{1.2}Fact 1.2. 1) For any (finite) graph A, we have A K, that is

1 = limn

Prob(A is embeddable into Mn).

2) Moreover 2 for every > 0

1 = limn

Prob(A has n1 disjoint copies in Mn).

This is easy, still, before proving it, note that since by our definition of theclosure A cm,k(,Mn) implies that A has < n embeddings into Mn we get:

{1.3}Conclusion 1.3. cm,k

Mn() : n < satisfies the 01 law (being a sequence of

empty models).

Hence (see [I,Def.1.4,Conclusion 2.19]){1.4}

Conclusion 1.4. K = K, the class of finite graph, and for our main theorem it

suffices to prove simple almost niceness of K (see Def.[I,2.13]).(Now 1.3 explicate one part of what in fact we always meant by random enough

in previous discussions.)

Proof. Let the nodes of A be {a0, . . . , ak1}. Let the event Enr be:

a 2rk + 2 is an embedding of A into Mn.

The point of this is that for various values of r these tries are going to speak onpairwise disjoint sets of nodes, so we get independent events.

Now subfact Prob(Enr ) = q > 0 (i.e. > 0 but it does not depend on n, r). {1.5}(Note: this is not true,e.g. in the close context where the probability of{i, j} beingan edge when i = j is 1/n + 1/2|ij|, as in that case the probability depends onn. But still, we can have q > 0 which suffices); where:

q =


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p1. Anyway, in our case we multiplied by 2 to avoid this (in the definition of theevent)). For the second case, (the probability of edge being 1/n + 1/2(ij))

q


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e(A,B,) = e(A, B) = |e(A, B)| where

e(A, B) = {e : e an edge of B, e A, and e x/ for x B \ A}.

[This measures the number of expensive, long edges (e stands for edges).]

Story:v larger means that there are more candidates for copies of B,e larger means that the probability per candidate is smaller.

{1.8}Definition 1.7. 1) For (A,B,) T and our given irrational (0, 1)R we define(w stands for weight)

w(A,B,) = w(A, B) = v(A, B) e(A, B).

2) Let

(A, B) =: { : (A,B,) T, and if C B \ A is a non-empty-closed set then w(A, C A) > 0}.

3) If A B then we let (A, B) = Max{w(A, B) : (A, B)}.{1.9}

Observation 1.8. 1) (A,B,) TandA = B w(A, B) = 0.2) If A B C (hence A C) and (A,C,) T and B is -closed then

(a) (A,B, (B \ A)) T

(b) (B,C, (C \ B)) T

(c) w(A, C) = w(B\A)(A, B) + w(C\B)(B, C)(d) similarly for v and e.

3) Note that 1.8(2) legitimizes our writing instead of (C\A) or (B\(CA))when (A,B,) T and C is a -closed subset of B. Thus we may write, e.g.,w(A C, B) for w(A C,B, (B \ A \ C)).4) If (A,B,) T and D B \ A and D+ =

{x/ : x D} then wD+ (A, A

D+) wD(A, A D) and D+ is -closed.

Proof. 1) As is irrational and v(A, B) is not zero.2) Clauses (a),(b) are totally immediate, and clauses (c),(d) are easy or see theproof of 1.15 below.3) Left to the reader.

4) Clearly by the choice of D+

we have vD+ (A, A D+

) = |D+

/D+

| = |D/( D)| = vD(A, AD) and eD+ (A, AD

+) eD(A, AD) hence wD+ (A, AD+) wD(A, A D). 1.8

{1.10}Discussion 1.9. Note: w(A, B) measures in a sense the expected value of thenumber of copies of B over a given copy of A with saying when one node is nearto another. Of course, when is the identity this degenerates to the definition in[ShSp:304].We would like to characterize i and s (from Definition [I,1.4,(3)] and Definition[I,1.4,(4)]), using w and to prove that they are O.K. (meaning that they form a


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nice context). Looking at the expected behaviour, we attempt to give an effectivedefinition (depending on only).

All of this, of course, just says what the intention of these relations and functionsis (i.e.


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1(B). Investigating the xs.

{1.14}Claim 1.13. Assume A 0. If C = B \ A then w(A, A C) = w(A, B) > 0 and

w(A C, B) = w(B, B) = 0. Lastly if C / {, B \ A} we use 1.12()1[A, B , ]itself.]3) If (A,B,) T, A A, B B, A B and B \ A = B \ A then(A, B, ) T, w(A, B, ) w(A,B,) also e(A, B, ) e(A,B,), v(A, B, ) =v(A,B,).

4) In (3) if in addition AMA

B, i.e., no edge {x, y} with satisfying x A\A and

y B\A then the equalities hold.{1.16}

Claim 1.15. A s B if and only if either A = B or for some we have:(A,B,) T and for every non-empty -closed C B \ A, we have w(A, A

C, C) > 0, that is (A, B) = .

Proof. So we have A s B. If A = B we are done: the left side holds as its firstpossibility is A = B. So assume A 0

(exists as C = B is O.K. because then there is no such C).By 1.8(4):


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()2 for every non-empty C B\Cso not necessarily 0-closed we have w(C, C

C, 0 C) > 0 hence (C 0 and by ??(2)). Now(in the last line we change C to C twice), by ??(3) we will get

v(C, C D, ) = |D/| = |D1/1| + |D0/0|= v(C, C D1, D1) + v(C D1, C D1 D0, D0)= v(C, C D1, D1) + v(C, C D0, D0),

and (using ??(3)):

e(C, C D, ) = e(C, C D1, D1)+e(C D1, C D1 D0, D0) e(C, C D1, D1) + e(C, C D0, D0),

and hence

w(C, C D, ) = v(C, C D, ) e(C, C D, )= v(C, C D1, D1) + v(C, C D0, D0)e(C, C D1, D1)e(C D1, C D1 D0, D0) v(C, C D1, D1) + v(C, C D0, D0)

e(C

, C

D1, D1) e(C, C D0, D0)= w(C, C D1, D1) + w(C, C D0, D0) 0,

and the (strict) inequality holds by the irrationality of , i.e. by 1.8(1). So actu-ally (C, B , ) satisfies the requirements on C, 0 thus giving contradiction to theminimality of C.

The if direction:As the case A = B is obvious, we can assume that the second half of 1.15 holds.

So let be as required in the second half of 1.15.


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Suppose A 0 as requested. 1.15{1.17}

Claim 1.16. 1) i is transitive.

2) s is transitive.3) For any A C for some B we have A i B

s C.

4) If A


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By 1.8(2)(c) we have w(A, C) = w(A

, A0) + w(A0, C) and by the previous

sentence the latter is 0 + 0 = 0, so w(A, A) 0 as required.

2) We use the condition from 1.15. So assume A0 s A1

s A2 and witness

A s A+1 (i.e. (A, A+1, ) is as in 1.15). Let be the equivalence relation onA2 \ A0 such that for x A+1 \ A we have x/ = x/. Easily (A0, A2, ) T.Now, by 1.8(2)(c), ??(3) and 1.15 the triple (A0, A2, ) satisfies the second conditionin 1.15 so A0 s A2.3) Let B be maximal such that A i B

C, such B exists as C is finite 4 andfor B = A we get A i B

C. Now ifB s C we are done, otherwise by thedefinition of s in 1.10(4) there is B

such that B


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w(A, C) < 0. As 1 was any equivalence relation on C\ A

by Definition 1.10(2)we have shown that A w(A2, B2).

As the number of such is finite and as we have shown (A2, B2) (A1, B1) weget (A1, B1) > (A2, B2).10) This follows from 0 +1 +2 below and the finiteness of (A, B2) recallingDefinition 1.7(3).

0 (A, B2) = .

[Why? As we are assuming A


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2 (A, B2) w(A, B2) < w(A, B1).

[Why? Otherwise let be from (A, B2) and let C = {x/ : x B1\A} A soB1 C B2.]

Case 1: C = B2.So v(A, B1) = v(A, B1, (B1\A)) = |(B1\A)/| = |(B2\A)/| = v(A, B2, ).

By an assumption of part (10), for some x A, y B2\B1 the pair {x, y} is anedge so e(A, B1, (B1\A)) is a proper subset of e(A, B2, ) hence

e(A, B1) < e(A, B2)

hence

w(A, B1) > w(A, B2)

is as required.Case 2: C = B2.

As in case 1, w(A, B1) w(A, C). Now B1 C B2, (using the

case assumption) and B1 0, k N,Mn K and A


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{2.4}Theorem 2.4. Assume A 0, for some k N, for every random enoughMn we have:() if f : A [n] thenG,kA,B(f,Mn) =

2 A kgn/k

(b) wg(A+g , B

+g ) < w(A, B)

(c) moreover the difference is > .

Why? Clause (a) as g / G1 see the or just... in its definition, clause (c) follows

from clause (b), so we concentrate on clause (b).Let Bg = {b/g : b B\A} A

+ so Bg B. rm Without loss of generality

3.1 B+g = Bg .

[Why? Toward contradiction assume that B+g = Bg . Easily w(A, B) wg (A

+, Bg).

Also Bg 0. Choose R+

small enough such that

()1 C0 0 such that t1 > c/ and t1 > 2.

Choose t2 22t1 +k+

(overkill, we mainly need to apply twice the -systemlemma; but note that in the proof of 3.10 below we will use Ramsey Theorem).Lastly, choose t > k2t2 and let k

N be large enough which actually means thatk > kandk (k + 1) t2 so k

(k, ) =: (k + 1) t2 is O.K.Suppose we have m, M, a, b as in () but such that the set R has at least t

members. Let (ci, di) R for i < t be pairwise distinct5.

As di ck(ab,M), we can choose for each i < t a set Ci M such that:

(i) Ci M,

(ii) |Ci| k,

(iii) di Ci,(iv) Ci (Ci (ab)) 1

elements then d ck(ck

,m

+k(a, M))


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(v) Ci ck,m+k[i]+1(a, M) ck

,m+k[i](a, M).

So without loss of generality

(vi) i < t/k2 k[i] = k[0]and|Ci| = |C0| = k k,

remember t2 < t/k2; also

(vii) b Ci.

[Why? If not then by clause (iv) we have (Ci a)


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[Why? Otherwise, letting X = di/ for any i u, the triple (D, D X, X) Thas weight

w(D, D X, X) = v(D, D X, X)e(D, D X, X)= 1 |{e : e an edge of M with one end in

D and the other in X}|.

Now as ci ck,m(a, M) clearly ci D and the pairs {ci, di} edge(M) are

distinct for different i clearly the number above is 1 |{(ci, di) : i u}| =1 |u| = 1 t1 < 0; contradiction to (D

, D).]Let D0 = a

{di,s/ : s S2 and i u}, clearly D0 is -closed subset of D

though not necessarily Dom() = D\D because of a.

3 b = di,1 and 1 S1\S0 and 0 / S0 S1 S2 and S1\S0 S2 (hence b D0).

[Why? The first two clauses hold as b Ci, b {di,0, di,1} and 2 and (g) of 1.The last clause holds by 1(d),(f) and the hence b D0 by the definition ofD0, S1\Dom() S2 and the first clause. Also 0 / S0 S1 S2 should be clear.]

4 For each i u we have w(Ci D0, Ci) < 0.

[Why? As Ci (ab) Ci D0 by clauses (b) + (f) of 1 and by monotonicity of


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w(D, D1) = w(D,{x/ : x

iu

Ci \ D} D)

= w(D, D D0) + w(D

D0, D D0

{di,s/ : i u,s < k, s / S0 S2})[now by 3.9 below with Bi = {di,s : s < k, s / S2 S0} and

B+i = {di,s/ : s < k, s / S2}]

w(D, D D0) +iu

w(D D0, D

D0 {di,s : s < k, s / S2 S0})

[now as Ci = {di,s : s < k} and di,s D D0 if s < k, s S0 S2, i u]

w(D, D D0) +iu

w(D D0, D

D0 Ci)

[now as w(A1, B1) w(A, B)when A A1 B1, A B B1, B1 \ A1 = B \ A) by ??(3)]

w(D D0, D0) +iu

w(Ci D0, Ci)

[now by the choice of c, D0 i.e., ()2 and the choice of , u + 5 respectively] c + |u| () = c t1 < 0,

contradicting the choice of , i.e., 7. 3.8{6.7A}

Observation 3.9. Assume

(a) A A Bi A B+i

B for i u,

(b) B\A is the disjoint union of B+i : i u,

(c) is an equivalence relation on B\A,

(d) each B+i is -closed in fact,

(e) B+i = {x/ : x Bi\A}.

Then w(A, B) {w(A, Bi) : i u}.Proof. By clause (b) + (d)

v(A, B) = {v(A, A B+i ) : i u} = {v(A, A Bi) : i u}

and by clause (b) the set e(A, B) contains the disjoint union ofe(A, Bi) : i u.Together the result follows. 3.9

{3.10}Claim 3.10. For every k, m and from N for some m = m(k,,m), for anyk k(k, ) (the functionk(k, l) is the one from claim 3.8) satisfying k km

we have

() ifM K, a M andb M\ck,m(a, M) then for some m mm

we have

ck(ab,M) ck,m+m(a, M) ck

,m(a, M).

Proof. For k = 0 this is trivial so without loss of generality k > 0. Let t = t(k, ) beas in the previous claim 3.8. Choose m such that, e.g., m/(km) (t + 5)2

2k!+

in the usual notation in Ramsey theory. We could get more reasonable boundsbut no need as for now. Remember that k(k, ) is from 3.8 and k is any naturalnumber k(k, ) such that k km.

If the conclusion fails, then the set


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Z =: {j m k : ck(ab,M) ck,j+1(a, M) ck

,j(a, M)}

satisfies:

j m m k Z [j,j + m) = .

Hence |Z| (m m k)/m.For j Z there are Cj M and dj such that|Cj | k and (Cj (ab)) 0 be such that

(A, B, ) Tand|B| kandA = B w(A, B) / (, ).

Let i() > 1 . Define inductively ki for i i() as follows

k0 = max{k(k, ), m k, m t + 1}

ki+1 = 22k

i

and lastly let

k = k ki().

We shall prove that m, k, t are as required in 3.11. So let M, a, b be as inthe assumption of () of 3.11. So M K, a M and b M \ ck

,m(a, M),but this means that the assumption of () in 3.10 holds for k, m(1), , so we canapply it (i.e., as m = m(k, m(1), ), k k(k, ) where k(k, ) is from 3.8 andk k m as k ki() > k

0 m

k).

So for some r m m(1) we have

1 ck(ab,M) ck,r+m(1)(a, M) ck

,r(a, M).

Let us define

R= {(c, d) : d ck(ab,M) \ ck,r+m(1)(a, M) and

c ck,r+m(1)k(a, M) and

{c, d} is an edge of M}.

How many members does R have? By 3.8 (with r + m(1) k here standing for m

there are (as k k(k, )) at most t members. But by 1 above

R= {(c, d) : d ck(ab,M) \ ck,r(a, M) and

c ck,r+m(1)k(a, M) and

{c, d} is an edge of M}.

But t (m + 1 ) + 1 < m(1) k by the choice of m(1) (and, of course, ck,i(a, M)

increase with i) hence for some m {r + 1, . . . , r + m(1) k m} we have


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2 (c, d) R c / ck,m+m(a, M) \ ck

,m1(a, M).

So

3 r m 1 < m + m r + m(1) k.

Let

B =: {c ck,m1(a, M) : for some d we have (c, d) R} a.

So by the above B = {c ck,m+m(a, M) : (d)((c, d) R)} a.

Let us check the demands (i) (iv) of () of 3.11, remember that we are defining

B = (ck(ab,M) \ ck,m+m(a, M)) B, that is the submodel of M with this set

of elements.

Clause (i): |B| t.As said above, |R| t, hence clearly |B| t + g(a) t + t.

Clause (ii): a B ck(B, M) ck,m(a, M).

As by its definition B ck,m1(a, M), and k k clearly ck(B, M)

ck,m(a, M) and B ck(B, M) always and a B by its definition.

Clause (iii):Clearly

B = ck,m+m(a, M) ((ck(ab,M) \ ck

,m+m(a, M)) B)

= (ck,m+m(a, M)) B.

Now the no edges holds by the definitions of B and R.

Clause (iv): B s B.

Clearly B B by the definition of B before the proof of clause (i). Towardcontradiction assume (B s B

) then 1.16(2) for some D, B k (and B


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(c) |Di| ki(d) if is an equivalence relation on Ci \ B and for some d D \ Di one of

the clauses below holds then there is such d Di+1 where1,d for some x C

d \ C

i , there are no y C

d C

i , j

N and yj :j j such that yj Cd , yj = x, y0 = y, {yj, yj+1} an edge of M,(actually an empty case, i.e., never occurs see ()14 below)

2,d there are x Cd \ C

i , y (C

i \ C

d ) B and y

Cd Ci such

that {x, y} is an edge of M and y is connected by a path y0, . . . , y1inside Cd to x so x = yj , y = y0 and [yi C

i i = 0] and (y

y)

3,d there is an edge {x1, x2} of M such that we have:

(A) {x1, x2} Cd(B) {x1, x2} is disjoint to C

i

(C) for s {1, 2} there is a path ys,0, . . . , ys,js in C

d

, ys,js =xs, [ys,j Ci j = 0] and (y1,0y2,0)

(e) if is an equivalence relation on Ci \ B to which clause (d) does not applybut there are d1, d2 D satisfying one of the following then we can findsuch d1, d2 Di+1

4,d1,d2 for some x1 Cd1

\ Ci , x2 Cd2

\ Ci and y1 Cd1

Ci , y2 Cd2

Ci we have: for s = 1, 2 there is a path ys,0, . . . , ys,js in C

ds

, ys,js =x, ys,0 = ys, [ys,j Ci j = 0] and: x1 = x2and(y1y2)

5,d1,d2 for some x1, x2, y1, y2 as in 4,d1,d2

we have: (y1y2) and {x1, x2}an edge.

So |Di()| k/k (by the choice of k, i() and clause (c)), hence Ci() =:

dDi() C

d has k

members, ab B {b} D0 C

i() ck

(ab,M) and

Ci() ck,m+m(a, M) = B ck

,m1(a, M) hence necessarily B s Ci()hence there is (B, Ci()). Let i = (C

i \ B).

Now

(B, Ci , i) (B, Ci ).

[Why? Easy.]

Case 1: For some i and an equivalence relation i on Di\B, clauses (d) and (e) arevacuous for i.

Let i be the set of pairs (x, y) from C \ B where C =

dD

Cd which satisfies

() or () where

() x, y Ci \ B and xiy

() for some d D we have x C \ Ci , x Cd , y C

i C

d and there is a

sequence yj : j j, j 1 such that yj = x, yj Cd , y0 = y, {yj, yj+1}an edge of M and [j > 0 yj / Ci ].

This in general is not an equivalence relation.Let C = {x : for some (x1, x2)

i we have x {x1, x2}}


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+i = {(x1, x2) : for some y1, y2 Di we have

y1y2, (x1, y1) i , (x2, y2)

i }.

Now

()1 +i is a set of pairs from C

with +i Di = i

()2 x C (x, x) +i .

[Why? Read clauses (), () and the choice of +i .]

()3 for every x C for some y Ci we have xi y.

[Why? Read the choice of +i , i .]

()4 +i is a symmetric relation on C.

[Why? Read the definition of +i recalling is symmetric.]

()5 +i is transitive.

[Why? Looking at the choice of i this is reduced to the case excluded in ()6below]

()6 if (x, y1), (x, y2) i , {y1, y2} Di, x / Di, then y1y2.

[Why? Because clause (e) in the choice of Di+1 is vacuous. More fully, otherwisepossibility 4,d1,d2 holds for i.]

()7 for every x C \ Ci , clause () apply to x C

, i.e., C = C.

[Why? As x C there is d D such that x Cd , hence by 1,d of clause (d) of

the choice of Di+1 holds for x hence is not vacuous contradicting the assumptionon i in the present case.]

()8 +i is an equivalence relation on C

\B.

[Why? Its domain is C\B by ()7, it is an equivalence relation on its domain by()1 + ()2 + ()4 + ()5.]

Also

()9 +i C

i = i.

[Why? By the choice of +i that is by ()1.]

()10 every +i -equivalence class is represented in C

i .

[Why? By the choice of +i and i .]

()11 if x1, x2 C \ B and (x1

+i x2) but {x1, x2} is an edge then {x1, x2}

Ci .


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[Why ()11 holds?Assume {x1, x2} is a counterexample, so {x1, x2} Ci , so without loss of

generality x1 / Ci . Now for = 1, 2 if x / C

i then we can choose d Di and

y Cd Ci such that d witnesses that (x, y)

i that is, as in clause () there

is a path y,0, . . . , y,j such that y,0 = y, y,j = x and (j > 0 y,j / Ci ).

We separate to cases:

(A) x1, x2 / Ci , d1 = d2. This case cant happen as 3,d1

of clause (d) isvacuous.

(B) x1, x2 / Ci , d1 = d2. In this case by the vacuousness of 5i,d1,d2

of clause

(e) we get contradiction.

(C) x1 Cd1

and x2 Ci . By the vacuousness of

2i,d1

of clause (d) we geta contradiction.

Together we have proved ()11.]

As i (B, Ci ) by ()8 + ()9 + ()10 + ()11 and , easily +i (B, C),hence (see 1.15) B


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34 SAHARON SHELAH

()13 w(Ai,j , Cdi,j

) can be zero if Ai,j = Cdi,j

and is otherwise.

[Why? As Ai,j

i

Cdi,j

.]

()14 in clause (d), 1,d never occurs.

[Why? If x Cd is as there, let Y = {y Cd : y, x are connected in M C

d }.

So x Y Cd , Y Ci = and C

d = C

d (B {b}) = C

d C

0 C

i . Hence

(Cd \Y)

Saharon Shelah- Zero One Laws for Graphs with Edge Probabilties Decaying with Distance, Part II

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Transcript of Saharon Shelah- Zero One Laws for Graphs with Edge Probabilties Decaying with Distance, Part II