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June 6

Banach-Tarski paradox quantifier depth

I'm trying to figure out the formal statement of the Banach-Tarski (BT) paradox. Something like: there exist sets A,B,C,D,E in R³ whose union A+B+C+D+E is the unit ball B1, and there are rigid motions S,T,U,V so that A+S(B)+T(C)=B1 and U(D)+V(E)=another copy of B1. A rigid motion is just a rotation and translation so it can be written as a 5-tuple of reals. Meanwhile, to say two sets of real triples are equal (like G=H) uses a universal quantifier over real triples, $\forall x.(x\in G\land x\in H)\lor (x\notin G\land x\notin H).$

Since the outer (existential) quantifier is over sets of reals, would we say it is a third-order arithmetic quantifier? The inner one is (nested) universal quantifiers over reals so that would be second-order arithmetic. Does the whole thing become $\Sigma _{2}^{2}$ ?

At first I didn't notice that the outer quantifiers were over sets of reals, so the BT paradox seemed to conflict with Schoenfeld's absoluteness theorem. But I think the set quantifiers mean BT is not analytic, so no issue there. Thanks.

173.228.123.207 (talk) 20:48, 6 June 2019 (UTC)[reply]

June 8

3x3 arrangements of D&D alignments...

Sparked by a puzzle in a D&D game. There are 9! ways that the numbers 1-9 can be placed into a 3x3 grid. In how many of them are *none* of the opposites in the "normal" placements placed side by side (or just above below). Or to put it another way, the number where 1 is not adjacent (H or V) to 9 and similarly 2 to 8, 3 to 7 and 4 to 6? (my guess is that the easiest way to start is to split it into 3 cases based on where 5 is (center, side, corner), but I'm not sure where to go from there.Naraht (talk) 13:19, 8 June 2019 (UTC)[reply]

If you have one arrangement then you can get others by swapping opposites, for 2⁴=16, and by permuting the sets {1, 9}, {2, 8}, {3, 7}, {4, 6}, for another factor of 24. In other words it's 16⋅24 times the number of near-perfect matchings of the graph of allowable placements. The graph of adjacent placements is P₃×P₃, so the problem reduces to finding the near-perfect matchings of the complement of P₃×P₃ where P₃ denotes the Path graph with three vertices. For the case where the unmatched vertex (the 5) is in the center then you have to find the the number of perfect matchings of C₈ where is the Cycle graph with 8 vertices. Finding the number of perfect matchings on C_2n in general is not that hard; you can combine the inclusion–exclusion principle with the fact that the number of k-edge matchings of C₈ is given by the coefficients of the Lucas polynomials. For C₈ the result is 7⋅5⋅3⋅1 - 8⋅5⋅3⋅1 + 20⋅3⋅1 - 16⋅1 + 2 = 31, so the number of arrangements with 5 in the center is 16⋅24⋅31 = 11904. There are more for the 5 on a side or in a corner but there is less symmetry with them so their enumeration will probably be more tedious, but I assume the easiest way is to find the matching polynomial for the corresponding graphs. As a guestimate I'd say 100,000 for the total. Another approach might be to start by computing the matching polynomial for the complement of P₃×P₃. There is a well-known formula for the number of perfect matchings of P_m×P_n where mn is even, but I don't know about imperfect matchings. (Note, this is all assuming I've understood the question correctly.) --RDBury (talk) 17:09, 8 June 2019 (UTC)[reply]

Actually the problem is much easier if you compute the matching polynomial for the complement of P₃×P₃; I get 1 + 12x + 44x² + 56x³ + 18x⁴, from which the number of near-perfect matchings is 9⋅7⋅5⋅3⋅1-12⋅7⋅5⋅3⋅1+44⋅5⋅3⋅1-56⋅3⋅1+18=195 and the total number of arrangements is 16⋅24⋅195 = 74880. --RDBury (talk) 17:54, 8 June 2019 (UTC)[reply]

(Note: I made a correction to the previous computation.) --RDBury (talk) 18:30, 8 June 2019 (UTC)[reply]

The 3x3 grid is small enough for easy hand enumeration (via backtracking) of all solutions modulo opposite swapping, set permutation, and (for other than the five-in-the-center case) rotation. In addition to the 31 solutions with five in the center (mentioned by RDB above), there are 22 solutions with five in the middle of a given side and 19 solutions with five in a given corner. This yields the 16⋅24(31+4⋅22+4⋅19) = 16⋅24(195) = 74880 from RDB's elegant solution. -- ToE 03:09, 10 June 2019 (UTC)[reply]

June 9

Categorical Peano axioms

I'm interested in Peano axioms#Interpretation in category theory but there are no sources listed. Does anyone know where I can find a more complete exposition? --RDBury (talk) 12:17, 9 June 2019 (UTC)[reply]

Maybe here. See also the links/xrefs from that article. 173.228.123.207 (talk) 19:19, 9 June 2019 (UTC)[reply]

I've since found Natural number object as well. The link is interesting and the nLab site in general is worth bookmarking, but all the talk about topoi and sheafs (sheaves?) is a bit out of my comfort zone. I guess what I'm specifically looking for is a proof, not necessarily referencing category theory, that axioms 7-9 in the Formulation section are equivalent to something like:

10: Given a set X, p∈X and a map f:X→X, here is a unique map r:N→X such that r(0)=p and r(S(n)) = f(r(n)) for all n.

In category jargon 10 says (N, 0, S) is an initial object in the category of Pointed unary systems (if I'm using the jargon correctly). My understanding is you need set theory to prove the 7-9⇒10 direction even if you assume X=N; nothing too fancy but it's not trivial. Not sure about the 7-9⇐10 direction. --RDBury (talk) 02:20, 10 June 2019 (UTC)[reply]

June 13

Wikipedia standard notation for set of complex numbers: ℂ or $\mathbb {C}$ or C?

Will changing C to ℂ or $\mathbb {C}$ uniformly throughout an article raise objections?

@@ Line 44: / Line 44: @@
 = June 13 =
+== Wikipedia standard notation for set of complex numbers: ℂ or <math>\mathbb{C}</math> or '''C'''? ==
+Will changing '''C''' to ℂ or <math>\mathbb{C}</math> uniformly throughout an article raise objections?