Saharon Shelah- Cardinalities of topologies with small base

8/3/2019 Saharon Shelah- Cardinalities of topologies with small base

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Cardinalities of topologies with small base

Saharon Shelah

Department of Mathematics

The Hebrew University, Jerusalem, Israel

and

Department of MathematicsRutgers University , New Brunswick N.J. USA

August 1991

revised March 1993

last revision April 20, 1993May 31, 1993

Abstract

Let T be the family of open subsets of a topological space (notnecessarily Hausdorff or even T0). We prove that if T has a base ofcardinality , < 2, strong limit of cofinality 0, then Thas cardinality or 2. This is our main conclusion (21). InTheorem 2 we prove it under some set theoretic assumption, whichis clear when = ; then we eliminate the assumption by a theoremon pcf from [Sh 460] motivated originally by this. Next we prove thatthe simplest examples are the basic ones; they occur in every example(for = 0 this fulfill a promise from [Sh 454]). The main result for

the case = 0 was proved in [Sh 454].

Partially supported by The Basic research Fund, Israeli Academy of Sciences. Publi-cation no. 454A done 8/1991, 3-4/1993. I thank Andrzej Roslanowski for proofreading,pointing out gaps and rewriting a part more clearly.

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* * *

Why does we deal with strong limit of cofinality 0? Essentially asother cases are closed.

Example 1 If I is a linear order of cardinality with Dedekind cuts thenthere is a topology T of cardinality > with a base B of cardinality .

CONSTRUCTION: Let B be {[, x)I : x I} where [, x)I = {y I : I |= y < x} 21

Remarks: as it is well known, if = , this limits possible generalizations of our mainTheorem. Also there are results guaranteeing the existence of such treesand linear orders, e.g. if is strong limit singular of uncountable cofinality, < 2 (see [Sh 262], [Sh 355, 3.5 +5]) and more (see [Sh 430]).

So we naturally concentrate on strong limit cardinals of countable cofi-nality. We do not try to save in the natural numbers like n() + 6 usedduring the proof.

Theorem 2 (Main) Assume

(a) n for n < are regular or finite cardinals, 2n < n+1 and =

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2. In fact, if |T| > then for some set X0 of points, {U X0 : U T} has cardinality 2. Moreover, for some B T of cardinality ,{X0 U : U is the union of a subfamily of B } has cardinality 2.

Definition 3 ([Sh 355, 5.1]) cov(, +, +, ) = min{|P| : P a family ofsubsets of each of cardinality , such that if a , |a| then forsome < and ai P (for i < ) we have a

i for < n

such that:

(a) the sets Z for < n are pairwise distinct

(b) for < n and V T we have: V T iff V Z = Z U V.

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PROOF: We shall use (*) freely. Define an equivalence relation E on Z

:

xEy iff |{U T : x U y U}|

(check that E is indeed an equivalence relation).Let Z Z be a set of representatives. Now for V T we have:

() {U T : U Z = V Z} xEz,{x,z}Z{U T

: z U x U but U Z = V Z} {V}where V = {y Z: for the x Z such that yE x we have x V}.

[Why? assume U is in the left side i.e. U T and UZ = V Z; now weshall prove that U is in the right side; ifU = V this is straight, otherwise forsome x Z, x U x V; as Z is a set of representities for E for somez Z, we have zEx so by the definition of V, x V z V. But asU Z = V Z we have z V z U. Together x U z / Uand we are done.]

Now the right side of () is the union of |Z|2 sets, each of cardinality (by the definition of xEz). Hence the left side in () has cardinality |Z|2 . Let {Vi : i < i

} T be maximal such that: Vi Z are

pairwise distinct and Vi T. So clearly |T| = |

i .

Hence (as is strong limit) necessarily |Z| , so we can let z Z for < n be distinct. For < < n we know that zEz hence for sometruth value t, we have |{U T

: z U z U t,}| > . But B isa base of T of cardinality , hence for some V, B the set

S, = {U T : z U z U t,, and {z, z} U V, U}

has cardinality > .Choose U1, S, such that < |S

1a,| where

S1,

def

= {U S, : U {z : < n} = U1, {z : < n}},

note that U1, exists as 2n < < |S,|.

By observation 4 we can find a family R of cardinality , members ofR has the form u = u : r , where r n, u B such that for everysequence u = u : < n of members ofB, there is a partition r : < n

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of n (so r = r[u] n for < n) such that ur R (remember B).Wlog if u = u : r R for = 1, 2 then u = u : r R where

r = r1 r2 and u =

u1 r

1

u2 r2 \ r1

.

For each V T we can find u[V] = u[V] : < n, such that (remem-ber B):

u[V] B,

z V z u[V] V,

z V u[V] = .

Clearly there is U2, S

1, such that:

() for any finite subset w n and < < n, the following family hascardinality > :

S2,,wdef= {U S1, : ( < n)(r[u[U]] = r[u[U

2,]]) and

( w)(u[U]r = u[U2,]r)}.

By the Erdos Rado theorem for some set M [n]+n :

(a) for every < from M, t, are the same

(b) for every < M , , M the truth values of z V,, z U2,, z u[U

2,] and the value of Min{ < n : r[u[U

2,]]}

depend just on the order and equalities between , , and .

Let M = {(i) : i < +n} where [i < j (i) < (j)], let t be 0 ifi < j t(i),(j) =truth and 1 if i < j t(i),(j) =false.

Case 1 If i < j < +n and < i > j then z() U2(i),(j).

So for some 1 < n

for every i < +n

, 1 = min{ : (i + t) r[u[U2(i),(i+1)

]]}.

We let Z = {z(i) : i < n}, Zi = {z(2i+t)}, Ui = u(2i+t)[U2(2i),(2i+1)].

Clearly Ui Z U2(2i),(2i+1) and Ui Z = Zi, lastly let Ti = S

2(2i),(2i+1),{1}

;now Z, Zi, Ui, Ti are as required.

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Case 2: If i < j < +

n then

< i z() U2(i),(j)

> j z() U2(i),(j)

So for some 1 < n, 2 < n

(a) for < i < j < +n

1 = min{ : () r[u[U2(i),(j)]]}

(b) for i < +n

2 = min{ : (i + t) r[u[U2(i),(i+1)]]}

Let Z = {z(i) : i < n}, Zi = {z() : < 2i} {z(2i+t)}, Ui ={u()[U

2(2i),(2i+1)] : 2i + 1} and Ti = S

2(2i),(2i+1),{1 ,2}

.

Case 3: If i < j < +n then

< i z() U2(i),(j)

+n > > j z() U

2(i),(j).

So for some 1 < n, 2 < n

(a) for i < +n , 1 = min{ : (i + t) r[u[U2(i),(i+1)]]}

(b) for i < j < < +n , 2 = min{ : () r[u[U2(i),(j)]]}

Let Z = {z(i) : i < n}, Zi = {z() : = 2i + t or 2i + 1 < < n}Ui = {u()[U2(2i),(2i+1)] : = 2i + t or 2i + 1 < < n} and Ti =

S2(2i),(2i+1),{1 ,2}.

Case 4: If i < j < +n then

< i z() U2(i),(j)

> j z() U2(i),(j)

So for some 1, 2, 3 < n

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(a) for < i < j < +

n , 1 = min{ : () r[u[U2

(i),(j)]]}(b) for i < +n , 2 = min{ : (i + t) r[u[U

2(i),(i+1)]]}

(c) for i < j < < +n , 3 = min{ : () r[u[U2(i),(j)]]}

Let Z = {z(i) : i < n}. Zi = {z() : < n and = 2i + 1 t}, Ui ={u()[U

2(2i),(2i+1)]: < n, = 2i + 1 t} and Ti = S

2(2i),(2i+1),{1,2,3}

.Now in all cases we have chosen Z, T, U, Z( < n) as required thusfinishing the proof of the claim. 25

Claim 6 If Z , |Z| , then {U Z : U T} has cardinality .

PROOF: Assume not. We can find T T such that:

() for U1, U2 T we have U1 = U2 U1 Z

= U2 Z.

() |T| > .

By induction on n we define T, Z1 , Z

2 , U :

(b) if then T T

(c) if = then T = T, Z1 = Z

2 = , U =

(d) Z1 Z2 Z

and |Z2 | lg , Z2 disjoint to {U : < lg }

(e) U T

(f) if V T then U V and V Z2 = U Z2 = Z

1

(g) if lg() = lg() = n + 1 and n = n then Z2 = Z2 but

(h) if lg() = lg() = n + 1, n = n but = then Z1

= Z1

.

Why this is sufficient? Let Zdf=

{Z2 : n

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n n + 1. Then choose any V T, so we know U V (byclause (f)) and V T(n+1) (by clause (b)), and V Z

2(n+1) = Z

1(n+1) (by

clause (f)), together finishing.

Case 3: n. By clause (d), Z2(n+1) is disjoint from U.

So we have finished to prove sufficiency, but we still have to carry theinduction. For n = 0 try to apply (c), the main point being |T| > which

holds by the choice of T

(which was possible by the assumption that theclaim fails). Suppose we have defined for n and let


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(a) |Y| and(b) if V, W are elements of T with V X = W X then there is some

element y Y which is in exactly one of V, W (possible as X I).

Now if V, W are elements of T which differ on X =


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Observation 11 Suppose is strong limit of cofinality 0, I is a linearorder of cardinality , < 0, and I has > Dedekind cuts, then ithas 0(= 0) Dedekind cuts.

Remark: This observation does not relay on the assumptions of Theorem 2.

PROOF: We define by induction on when does rkI(x, y) = for x < yin I.

for = 0 rkI(x, y) = iff (x, y)I = {z I : x < z < y} has cardinality<

for > 0 rkI(x, y) = if: for < , [rkI(x, y) = ] but for any

(xi, yi) (i < ), pairwise disjoint subintervals of (x, y), there is i such that 0 and x < y from I, rkI(x, y) = iff for < , [rkI(x, y) =] and for some < for any (xi, yi) (i <

), pairwise disjointsubintervals of (x, y) there are i < and < such that rkI(xi, yi) =

[Why? the demand in ()2 certainly implies the demand in the definition, forthe other direction assume that the definition holds but the demand in ()2fails, and we shall derive a contradiction. So for each n < there are pairwisedisjoint subintervals (xni , y

ni ) of (x, y), for i < n such that [rkI(x

ni , y

ni ) = ]

(when < and i < n). As we can successively replace {(xni , yni ) : i < n}

by any subfamily of the same cardinality (when the ns are finite - by asubfamily of cardinality n1) wlog: for each n, all members of{x

ni : i < n}

realize the same Dedekind cut of {xm

j , ym

j : m < n, j < m} and similarlyfor all members of {yni : i < n}. So for m < n, i < n, the interval (xni , y

ni )

cannot contain a point from {xmj , ymj : j < m} (as then the same occurs

for all such is, for the same point contradicting the pairwise disjoint) soeither our interval (xni , y

ni ) is disjoint to all the intervals (x

mj , y

mj ) for j < m

or it is contained in one of the intervals (xmj , ymj ); as j does not depend on i

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we denote it by j(m, n); if = 0, by the Ramsey theorem wlog for m < nj(m, n) does not depend on n; now the family {(xmi , ymi ) : m < , i < m

and for every n < which is > m we have i = j(m, n)} contradicts thedefinition]

IfrkI(x, y) is not equal to any ordinal let it be . Let = sup{rkI(x, y)+1 :x < y in I and rkI(x, y) < }. Clearly rkI(x, y)

{} for every x < yin I (and in fact < + ). As we can add to I the first and the last elementsit suffices to prove:

(A) if rkI(x, y) = < then (x, y)I has Dedekind cuts and(B) ifrkI(x, y) = then it has

0 Dedekind cuts

(B) is straightforward.Proof of (A): We prove this by induction on . If is zero this is trivial.So assume that > 0, hence by ()2 for some < there are no pairwisedisjoint subintervals (xi, yi) for i <

such that < implies [rkI(xi, yi) =]. Let J be the completion of I, so each member of J \ I realizes on Ia Dedekind cut with no last element in the lower half and no first elementin the upper half, and |J| > |I|. Let J+

def= {z J : z I and if

x I, y I and x


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(2) wlog for all x we get the same case above, in fact it is case (b).(3) wlog for any two distinct members x, y of for some U B we have

x U iff y / U.

PROOF: (0) By claim 10 and the proof of claim 5 (if E has equiv-alence classes we can repeat the proof of claim 5 and get contradiciton toclaim 10).

(1), (2), (3) Let X : < list the E-equivalence classes, so < n().

As I, and I is +-complete (claim 8) for some , X I. Let = X,T = {U : U T}, B = {U : U B}; so , B, T has all the

properties we attribute to , B , T and in addition now E has one equivalenceclass. So we assume this.

Fix any x0 , let B0 = {U B : x0 U}, T

0 = {U T : x0 U} {}, B1 = {U B : x0 U}, T1 = {U T : x0 U} { }. For some {0, 1}, |T| > , and then , B, T satisfies the earlier requirements andthe demands in (1) and (2). For (3) define an equivalence relation E on :xEy iff (U B)[x U y U], let be a set of representatives,B = {U : U B} and finish as before. The only thing that is left isthe second phrase in (2). But if it fails then for every U T \ {} choose anonempty subset V[U] from B. As the number of possible V[U] is |B| ,

for some V B \{}, for > members U ofT, V = V[U] and hence V U.Choose x V; so for x clause (a) of (2) fails and hence for all y clause(b) of (2) holds, as required. 213

PROOF 14 (of Theorem 2 (MAIN)):

Consider for n = n() (from claim 13(0) and as in claim 10) the following:() there are an open set V and a subset Z of V and for each < n

Z Z and open subsets V, U of V such that:(a) for < < n the sets V Z, V Z are distinct(b) U Z = Z

(c) the number of sets U T satisfying U Z = V Z and U U is>

So by claim 10 we know that this fails for n.Let be large enough and let N = Ni : i < + be an elementary

chain of submodels of (H(), ) of cardinality (and B, , T belong to N0of course) increasing fast enough hence e.g.: if X Ni is a small set, U T

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then there is U

Ni T with U X = U

X (you can avoid the nameelementary submodel if you agree to list the closure properties actuallyused; as done in [Sh 454]). For x let i(x) be the unique i such that xbelongs to Ni+1\Ni or i = 1 if x N0 (remember || =

+).

Definition 15 We define : x is N-pertinent if it belongs to some smallsubset of which belongs to Ni(x) (and i(x) 0) and N-impertinent other-wise.

Observation 16 ip = {x : x is N-impertinent } is not small (seeDefinition 7).

PROOF: As N0 is small by claim 8, for some U, T

def= {U T :

U N0 = U N0 } has cardinality > . So it suffices to prove:

() U1 = U2 T U1 ip = U2 ip.Choose x (U1 \ U2) (U2 \ U1) with i(x) minimal. As U1, U2 T

,i(x) = 1 ( i.e. x N0) is impossible, so x (Ni+1 \ Ni) for i = i(x).If x ip we succeed so assume not i.e. x is N-pertinent, so for some smallX Ni x X. Hence by the choice of N: for some U

1, U

2 Ni T we have:

U1 X = U1 X, U2 X = U2 X so U

1 X, U

2 X Ni are distinct

(as x witness) so there is x Ni X, x U1 x

U2; but this implies

x

U1 x

U2, contradicting i(x)s minimality. 216

We define a binary relation on ip by:

x y for all U B, if y U then x U.

Claim 17 The relation is clearly reflexive and transitive. It is antisymet-ric [why antisymetric? by claim 13(3)].

Observation 18 If J ip is linearly ordered by then J is small.

PROOF: For each U1, U2 B such that U1 J U2 J choose yU1,U2 J(U1\U2). Let I = {yU1,U2 : U1, U2 B & U1J U2J}. Clearly |I| .We claim that I is dense in J (with respect to , i.e. I has a member in everynon empty interval of J). Suppose that x,y,z J, x y z. By 13(3)we find U1, U2 B such that x U1, y / U1, and y U2, z / U2. ConsideryU2,U1 I. Easily x yU2,U1 z. Thus if (x, z) = then (x, z) I = .

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Now note that each Dedekind cut of I is an restriction of at most 3Dedekind cuts of J (and the restriction of a Dedekind cut of J to I is aDedekind cut of I). For this suppose that Y1, Y2, Y3, Y4 are lower parts ofdistinct Dedekind cuts of J with the same restriction to I, wlog Y1 Y2 Y3 Y4. For i = 2, 3, 4 choose yi Yi such that Y1 y2, Y2 y3 and Y3 y4.As (y2, y4) = we find x (y2, y4) I. Since y2 x we get x / Y1 and sincex y4 we obtain x Y4. Consequently x distinguishes the restrictions ofcuts determined by Y1 and Y4 to I.

To finish the proof of the observation apply observation 11 to I (whichhas essentially the same number of Dedekind cuts as J). 218

Continuation 19 (of the proof of theorem 2)

Now it suffices to prove that for each x ip, i = i(x) > 0 there is no membery of ip Ni such that x, y are -incomparable.[Why? then we can divide ip to sets such that any two in the same partare -comparable contradicting 16+18 and 8; How? By defining a functionh : ip such that h(x) = h(y) x y y x. We define h(ip Ni)by induction on i, in the induction step let Ni+1\Ni = {xi, : < }. Chooseh(xi,) by induction on : for each there are || < forbidden values sowe can carry the definition.]

So assume this fails, so we have: for some x ip, i = i(x) > 0 there isy0 Ni ip which is -incomparable with x; so there are U0, V0 B suchthat x V0, x / U0, y0 U0, y0 / V0. Now U

=

{U T : y0 / U} isin T Ni and x U (as V0 witnesses it) but by 13(2) we know that U issmall, so it contradicts x ip. This finishes the proof of theorem 2. 22

Concluding Remarks 20 Condition (b) of Theorem 2 holds easily for =. Still it may look restrictive, and the author was tempted to try to eliminateit (on such set theoretic conditions see [Sh 420,6]). But instead of workinghonestly on this the author for this purpose proved (see [Sh 460]) that itfollows from ZFC, and therefore can be omitted, hence

Conclusion 21 (Main) If is strong limit, cf = 0, and T a topologywith base B, |T| > |B| then |T| 2 and thew conclusion of 2(2) holds.

Theorem 22 1. Under the assumptions of Theorem 2, if the topology Tis of the size 2 then there are distinct x for

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such that letting Z = {x : n


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PROOF Let U (for l l) be given by one of the clauses. For

l


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Let U

=

{U T I : (n < m)(xn / U)}. As || < = cf we get()3 |{V T : (n < m)(xn / V) & U

\ (V n


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(e) like (d) but

() U {y : < n} = V {y : < n} = {y : < n}

(f) like (d) but

() U {y : < n} = V {y : < n} = {y}

(g) like (d) but

() U {y : < n} = V {y : < n} = {y : < n, = }

(h) there are V, y for < n such that V U are open, y Y arepairwise distinct, (U \ V)

=V I

+ and

() U {y : < n} = V {y : < n} = {y : < }

(i) like (h) but

() U {y : < n} = V {y : < n} = {y : < n}

(j) like (h) but

() U {y : < n} = V {y : < n} = {y}

(k) like (h) but

() U {y : < n} = V {y : < n} = {y : = , < n}

PROOF: First note that if n < m < , Y1 Y0, Y1, Y0 I+ and one ofthe cases (a)(k) of 1 occurs for Y1, m then the same case holds for Y0, n.

Consequently, 1 implies that for each Y I+

one of (a)(k) occurs for Y, nfor every n . Moreover, if1 then for some x {a,b,c,d,e,f,g,h,i,j,k}and Y0 I+ we have() for every Y1 Y0 from I

+ and n case (x) holds.If x = a, clause (a) of 22(1) holds. For this we inductively define open

sets V, V for

n

l


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1. V

V, (V \ V

) Y0 I+

, (V+1 \ V) Y0 I+

2. if < n+1 then V V V, V+1.

Let U : < 0 be the increasing sequence of open sets given by (a) forY0, n = 0. Put V = U2+1, V

= U2 for < 0. Suppose we have

defined V, V for lg() m. Given

l


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Now for n


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E if and only if( < )( U U) and( < )( Z Z).

The relation E has 2||+ < equivalence classes. Consequently for each

< |{V T : V is a union of E-equivalence classes}| < .

As |T| > we find a nonempty open set U which for no < is a union ofE-equivalence classes. This U is as needed.

Now let (n, n) be a pair (, ) satisfying (a)(c) for = n and let

{n, n} Zn

. Then w

= {n : n < }, U contradict the choice of P andTw : w P. 227

PROOF 28 of 1:

For the notational simplicity we assume that Y = . Let B =n .

Let y for < n+6 be pairwise distinct. For each < < n+6 thereis = (, ) {, } such that T0,

def= {U T : {y, y} U = {y}} has

cardinality > . For each U T0, there is V[U] B, y V[U] U. As|B| for some V, B we have that the set

T1, = {U T : {y, y} U = {y} and y V, U}

has cardinality > . For U T let fU, gU be functions such that:

1. fU : n+6 , gU : n+6 B,

2. gU() = 0 iffy / U,3. ify U then y gU() U,

4. fU() = min{n : gU() Bn}.

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For each < < n+6 we find f, : n+6 such that the setT2, = {U T

1, : fU = f,}

has the cardinality > . By Erdos-Rado theorem we may assume that foreach < < n+5, < n+5 the value of f,() depends on relationsbetween , and only. Consequently for some n < , if < n+5, U T2, , < < n+5 then gU() Bn. As |Bn| < we find (for each < < n+5)a function g, : n+6 Bn such that the set

T3, = {U T2, : gU = g,}

is of the size > . Let

U, =

T3, , V, =


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Now, A,,n is either included in U1,2 or is disjoint from it (uniformly forn); similarly for V1,2.Case A: A3,4,n U1,2 =

Let U =U2,2+1. Then U

: < n is an increasing sequence of open

sets andn A2+2,2+3,n U

+1 \ U

, which witnesses that the last set is in

I+. Thus we get clause (a).Case B: A1,2,n U3,4 =

Let U =


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[J2] I. Juhasz. Cardinal functions in topology ten years later. Math.Center. Tracts. Amsterdam, 1980

[JuSh 231] I. Juhasz and S. Shelah, How large can a hereditary separable orhereditary Lindelof space be. Israel J. of Math., 53 (1986) 355364.

[KR] K. Kunen and J. Roitman. Attaining the spread of cardinals ofcofinality , Pacific J. Math. 70 (1977). 199205

[Ku] Casimir Kuratowski, Topologie I, Panstwowe WydawnictwoNaukowe, Warszawa 1958.

[R] J. Roitman. Attaining the spread at cardinals which are not stronglimit. Pacific J. Math. 57 (1975). 545551.

[RuSh 117] M. Rubin and S. Shelah, Combinatorial problems on Trees: Par-titions, -systems and large free subsets, Annals of Pure and Ap-plied Logic, 33 (1987) 43-82.

[Sh 36] S. Shelah. On cardinal invariants in topology, General topology andits applications, 7 (1977) 251 259.

[Sh 92] S. Shelah. Remarks on Boolean algebras, Algebra Universalis, 11

(1980) 7789.

[Sh 95] S. Shelah. Canonization theorems and applications, J. of Symb.Logic, 46 (1981) 345353.

[Sh 233] S. Shelah. Remarks on the number of ideals of Boolean algebrasand open sets of a topology, Springer-Verlag Lecture Notes Vol-ume, , vol 1182 (1982) 151187.

[Sh 262] S.Shelah. Number of pairwise non-elementarily embeddable mod-els, Journal of Symbolic Logic, vol. 54 (1989) 1431-1455.

[Sh 355] S. Shelah. +1 Has a Jonsson Algebra, Cardinal Arithmetic,OUP.

[Sh 420] S. Shelah. Advances in Cardinal Arithmetic, Proceedings of theConference in Banff, Alberta, April 1991.

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[Sh 430] S.Shelah. Further cardinal Arithmetic, Israel Journal of Mathe-matics, accepted.

[Sh 454] S. Shelah. Cardinalities of countably based topologies, Israel Jour-nal of Mathematics, in press.

[Sh 460] S. Shelah. The generalized continuum hypothesis revisited,preprint.

[Sh C1] S.Shelah Remarks on General topology (1/78), preprint.

[Sh E6] S.Shelah. If 1+ there is an 1-Kurepa tree with branches

then some B.A. of power 1 has filters and 0-ultrafilters.Mimeographed notes from Madison, Fall 77.

[Sh G.1] S.Shelah. P-points, () and other results in general topology,Notices of the AMS 25 (1978) A-365.

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Saharon Shelah- Cardinalities of topologies with small base