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Hello,

We know that A is Wadge reducible to B if there is a continuous map $f$ such that A is the preimage of B via $f$, and the Wadge order is defined by $A\leq_{w}B$ if $A$ is Wagde reducible to $B$ or $A$ is Wadge reducible to $B^c$. Martin proved this is a prewellordering on sets of reals. The Wadge rank of $A$ is the order-type of $A$ in this prewellordering.

My question is if $A\equiv_{w}B$, what can we know about the Wadge rank of $A\cup B$? The union can be weaker than $A$ and can be stronger than $A$. But is there some bound? For example, is it true that the Wadge rank of $A\cup B$ is smaller than (the Wadge rank of $A)+\omega_1$? And how about a countable union of sets with the same Wadge rank?

Thank you!

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Martin's proof that the Wadge hierarchy is pre-well-ordered assumes the Axiom of Determinacy. en.wikipedia.org/wiki/Wadge_hierarchy –  Joel David Hamkins Jun 25 '12 at 19:26
    
Thanks Joel. I forgot to mention that the whole argument is under AD. –  Grace Jun 26 '12 at 1:47
    
I don't know the answer to your question, but important examples of pointclasses that are not closed under positive Boolean operations (which you may already be aware of) are given in Steel's paper "Closure properties of pointclasses." These are the Steel pointclasses that are discussed in Jackson's handbook chapter "Structural consequences of AD." –  Trevor Wilson Jul 17 '12 at 18:57
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