π-weight

Abstract

If A is a subalgebra of B, then πA can vary either way from πB; for clearly one can have πA < πB, and if we take B = Pω and A the subalgebra of B generated by an independent subset of size 2^ω, then we have πB = ω and πA = 2^ω. Similarly, if A is a homomorphic image of B: it is easy to get such A and B with πA < πB and if we take B = Pω and A = B/Fin, then πB = ω while πA = 2^ω since A has a disjoint subset of size 2^ω. Turning to products, we have π(П _{i ∈ I} A _i ) = max(|I|, supi _{i ∈ I} π A _i ) for any system 〈A _i : i ∈ I〉 of infinite BAs. For, ≥ is clear; now suppose D _i is a dense subset of A _i for each i ∈ I. Let

$$ E = \{ f \in \mathop \Pi \limits_{i \in I} ({D_i} \cup \{ 0\} ):fi \ne 0foronlyfinitelymanyi \in I\} $$

. Clearly E is dense in П _{i ∈ I} A _i , and |E| = max(|I|, sup _{i ∈ I} π A _i ), as desired. The equation π(П ^w _i∈I A _i ) = max(|I|, sup _{i ∈ I} π A _i ) is proved by the same argument.