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Mathematical Logic Quarterly, 24 August 2009

On long increasing chains modulo flat ideals

Saharon Shelah 1,2

1 The Hebrew University of Jerusalem, Einstein Institute of Mathematics, Edmond J. Safra Campus, Givat

Ram, Jerusalem 91904, Israel2 Department of Mathematics, Hill Center-Busch Campus, Rutgers, The State University of New Jersey, 110

Frelinghuysen Road, Piscataway, NJ 08854-8019 USA

Received xxx, revised xxx, accepted xxx

Published online xxx

Key words increasing chain, order modulo an ideal

MSC (2000) 03E05 03E10

We prove that, e.g., in (3)(3) there is no sequence of length 4 increasing modulo the ideal of countable sets.

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This note is concerned with the depth of the partial order of the functions in modulo the ideal of the formI = []

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4 Saharon Shelah: Increasing Chains

Theorem 4 Assume + < andJ = []

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() fs(i) j < fs+1

(i)

and therefore

() f(i) j < f+1(i).

But |I()| + |s| = < +, so for some pair (j, i) s I() we may choose 1 < 2 < + such that

()11 j1 = j2 = j and i1 = i2 = i.

But the sequence f(i) : < + is increasing by ()6(b) (see the choice of), so

f1 (i) < f1+1(i) f2 (i) < f2+1(i).

It follows from ()10()+()11 that the ordinalj belongs to [f1 (i), f1+1(i)) and to [f2 (i), f2+1(i)),which are disjoint intervals, a contradiction.

Similarly,

Theorem 5 Assume that

(a) J is an ideal on ,

(b) I [], I / J for < ,

(c) = || + andcf([], ) < ,

(d) ifu J for < +, then for some < the setI is disjoint from