Talk:Projective hierarchy

message: We seem to have almost the same content on the page for the Borel hierarchy and if one blurs one's eyes, they cannot be told apart. Perhaps some clarifying distinction should be drawn. Well, I mean, the Borel hierarchy starts with \mathbf{\Sigma}^0_1 and the analytic hierarchy does not show up till much later, as \mathbf{\Sigma}^1_1, but it seems that perhaps this should be pointed out in the opening paragraphs. 67.198.37.16 (talk) 20:28, 27 November 2023 (UTC)
thread_title:  Relationship to the [[Borel hierarchy]]? 

message: For example, an important fact from the German version: "Alle Klassen \Sigma^1_n, \Pi^1_n und \Delta^1_n sind abgeschlossen bezüglich abzählbarer Durchschnitte und abzählbarer Vereinigungen, insbesondere ist \Delta^1_n eine σ-Algebra.Y.N. Moschovakis: Descriptive Set Theory, North Holland 1987, ISBN 0-444-70199-0, Corollary 1F.2" (The German version of the pages uses the lightface symbols for both Borel hierarchy and projective hierarchy. ) If I understand it correctly, it says that families \mathbf{\Sigma}^1_n and \mathbf{\Pi}^1_n are all closed under countable union and intersection, so \mathbf{\Delta}^1_n is a σ-algebra.
We know that \mathbf{\Sigma}^1_1 is closed under countable union and intersection from the page analytic set. Since the image of union is the union of images, the implication "\mathbf{\Sigma}^1_n is closed under countable intersection \Rightarrow \mathbf{\Sigma}^1_{n+1} is closed under countable union" would be trivial, but to prove that \mathbf{\Sigma}^1_n is closed under countable intersection seems to be nontrivial for me. A proof would be well appreciated.
Another example: the diagram in the German page says that \mathbf{\Sigma}^1_n, \mathbf{\Pi}^1_n\subset\mathbf{\Delta}^1_{n+1}, which is equivalent to \mathbf{\Sigma}^1_n, \mathbf{\Pi}^1_n\subset\mathbf{\Sigma}^1_{n+1}. This means that every \mathbf{\Sigma}^1_n set and every \mathbf{\Pi}^1_n is the projection of a \mathbf{\Pi}^1_n set; while the latter implication is trivially true, I have no idea about the former even with n = 1. The inclusion gives us the inclusion of σ-algebras \mathbf{\Delta}^1_1\subset\sigma(\mathbf{\Sigma}^1_1)\subset\mathbf{\Delta}^1_2\subset\cdots\subset\mathbf{P}. Perhaps in fact each inclusion is strict in eveyy uncountable Polish space? (We know that this is true for the first inclusion as stated here, of course some choice may be needed; this post may have addressed the second, although I don't know if it would work for eveyy uncountable Polish space.)
What's more, the inclusion \mathbf{\Pi}^1_n\subset\mathbf{\Delta}^1_{n+1} tells us that \mathbf{\Sigma}^1_n sets are precisely the projections of \mathbf{\Delta}^1_n sets, so \mathbf{\Delta}^1_n sets are precisely those sets such that themselves as well as their projections are all \mathbf{\Delta}^1_n sets: by definition \mathbf{\Delta}^1_n\subset\mathbf{\Sigma}^1_n, and the projections of \mathbf{\Sigma}^1_n sets are also in \mathbf{\Sigma}^1_n. Conversely, a \mathbf{\Sigma}^1_n set is projection of \mathbf{\Pi}^1_{n-1} set, and the latter is a \mathbf{\Delta}^1_n set. (This works for n ≥ 2; for n = 1, the proposition "every analytic set is the projection of a \mathbf{\Delta}^1_1 set" is something outside the hierarchy: we have to show that every analytic set is the projection of a Borel set, and \mathbf{\Delta}^1_1 sets are precisely Borel sets (Suslin's theorem).) 129.104.241.214 (talk) 06:07, 12 February 2024 (UTC)
thread_title:  More information need to be mentioned in the article