8/3/2019 O. Kolman and S. Shelah- A Result Related to the Problem CN of Fremlin

1/5

8/3/2019 O. Kolman and S. Shelah- A Result Related to the Problem CN of Fremlin

2/5

661

r e v

i s

i o n :

1997 -

12 -

23

m o

d i f i

e d

: 1 9 9 7

- 1 2

- 2 3

2 O. KOLMAN AND S. SHELAH

Fact 2. (1) The cardinals and 1 belong to H ( ) for > 1.(2) If > 1 is regular, then H ( ) is a model of all the axioms of ZF C

except possibly the power set axiom.(3) If > 1 is singular, then H ( ) is a model of all the axioms of ZFC

except possibly the power set, union and replacement axioms.

Lemma 3. Suppose that > 1 is regular and N is an elementary submodel of H ( ). Then:

(1) If a H( ) is denable with parameters from N (i.e., there is a formula (x) with one free variable x and possibly parameters in N such that a is the unique element which satises (x) in (H ( ),)),then a N .

(2) The ordinals and 1 belong to N and N .(3) If a,A,B,f N and (in V ) f is a function from A to B then

f (a ) N .(4) For every 1 {1}, N is an ordinal.(5) If N and {A : < } N , then ( N )(A N ).

The proofs of these well-known facts can be found in many places, forexample, in [1] or in the appendix of [3].

Since the proof of Theorem 1 involves some notation, we describe inintuitive terms how it works. First we shall x a family of rational boxeswhich are products of open subintervals of [0 , 1] having rational endpoints.For a given countable family of dense open subsets G n of [0, 1], for eachn , we dene a function f n on rational boxes such that f n (B ) Gn and thediameters of open sets used to dene the box f n (B ) decrease as n increases.For each real , we dene using the f n s an element x

n

8/3/2019 O. Kolman and S. Shelah- A Result Related to the Problem CN of Fremlin

3/5

661

r e v

i s

i o n :

1997 -

12 -

23

m o

d i f i

e d

: 1 9 9 7

- 1 2

- 2 3

A RESULT RELATED TO THE PROBLEM CN OF FREMLIN 3

open subinterval of [0 , 1], and the length g() of the sequence is n . Choose22

ndisjoint open subintervals of I of equal length, say I k , 0 k < 22

n

such that I \k< 22n

I k is nite. This completes the denition of the

family {I : > }.Next for each n , we choose a function f n dened on the family of

non-empty open boxes B =

8/3/2019 O. Kolman and S. Shelah- A Result Related to the Problem CN of Fremlin

4/5

661

r e v

i s

i o n :

1997 -

12 -

23

m o

d i f i

e d

: 1 9 9 7

- 1 2

- 2 3

4 O. KOLMAN AND S. SHELAH

Suppose that x is not injective: so there are i < j < such thatx (i) = x ( j ). We derive a contradiction by showing that is denableusing parameters in N and hence N by Lemma 3.

Let m 0 = min {n : i (n ) = j (n )} (we can calculate m 0 in N since{i : i < } N and i, j N ). It suffices to show that we can dene | nin N for every n > m 0 . We prove this by induction on n . Suppose that wehave dened | n , n > m 0. We show how to calculate (n ) in N , therebydening |( n +1). Note that C i, and C are denable in N (for i < and > 2) and hence f n (C | n ) =

m 0. Since I i I j = ,it follows that g(i ) < g( j ) and j | g(i ) = i , and so there is a uniquenatural number k such that j |( g(i )+ 1) = i k . As we can computekand 2 |{ n 2 : ex i | n }| inside N , it will suffice to show (in H ( ))that:

() (n ) = k

2 |{ n

2 : ex i | n }|.To see this, notice that C i, | ( n +1) = I i k . Why? If not, then C i, | ( n +1) =I i for some = k, and so I i k I i = , and hence

() I j I i = (as I j I i k ). However, x (i) C i, | ( n +1) = I i , and x (i) =x ( j ) I j , so I j I i = , contradicting ( ). By the denition of C i, | ( n +1) = I i k i , it is now immediate that ki = k

. Recalling that

ki = 2 |{ n 2 : ex i | n }| + (n ),

we obtain (), and so | ( n + 1) is denable in N .Thus, =

n

8/3/2019 O. Kolman and S. Shelah- A Result Related to the Problem CN of Fremlin

5/5

661

r e v

i s

i o n :

1997 -

12 -

23

m o

d i f i

e d

: 1 9 9 7

- 1 2

- 2 3

A RESULT RELATED TO THE PROBLEM CN OF FREMLIN 5

open. It is enough to show that every f n