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We view subsets of the natural numbers as their characteristic functions, which are elements of the Cantor space $2^\mathbb{N}$. We take the uniform probability measure on the Cantor space. Under this view, what is the measure of the family of all productive sets (in the sense of computability theory)? Immune sets? Sets which are neither immune nor productive nor computably enumerable?

edited Apr 14 2011 at 5:01

Andrej Bauer
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asked Apr 14 2011 at 1:51

XL
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thank Mr.Andreas Blass ,Andrej Bauer,Andres Caicedo,Gerry Myerson for their comments and suggests (see discussion of the original question below),and me ,who have never been so patient to rephrase question or articles – XL Apr 14 2011 at 2:05

@Andrej,thank you very much for your editing.:) – XL Apr 14 2011 at 5:07

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You're welcome. I just couldn't stand all those commas and periods with spaces before them and none after ;-) – Andrej Bauer Apr 14 2011 at 6:05

Andreas Blass · Accepted Answer · 2011-04-14 01:59:37Z UTC

Isn't this revised question just the one Andrej Bauer suggested and I answered in the comments on your earlier continued-fraction version? In the uniform measure on subsets of $\mathbb N$ (equivalent, if you replace sets with their characteristic functions, to the product measure on $\{0,1\}^{\mathbb N}$ arising from the uniform measure on $\{0,1\}$ ), the collection of immune sets has measure 1, and therefore the other two collections in your question have measure 0.

What is the measure of productive and immune sets in the Cantor space?

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