Wikipedia:Reference desk/Archives/Mathematics/2021 February 23

= February 23 =

Ring of sets
In ring of sets it is said that "On the real line $ℝ$, the family of sets consisting of the empty set and all finite unions of half-open intervals of the form ( a, b ], with $a, b ∈ ℝ$ is a ring in the measure-theoretic sense." How does one prove this fact? More precisely, the direct proof of $$\cup(a_i,b_i]\setminus \cup(c_i,d_i]$$ would have lots of cases and portends to be very unwieldy. Is some shorter method possible? Thanks - Abdul Muhsy  talk  01:09, 23 February 2021 (UTC)
 * First, verify that the simple cases $$(a,b\,]\cup(c,d\,]$$ and $$(a,b\,]\setminus(c,d\,]$$ remain within the family, which involves a manageable amount of case distinctions. By combining this with the repeated use of the identities
 * $$(X\cup Y)\setminus Z=(X\setminus Z)\cup(X\setminus Z)$$ and
 * $$X\setminus(Y\cup Z)=(X\setminus Y)\setminus Z,$$
 * we see that computation of a formula in the general form reduces in a finite number of steps to a union of simple cases. --Lambiam 07:46, 23 February 2021 (UTC)


 * Another approach. Consider the extended domain $$\R_{{+}\infty}=\R\cup\{{+}\infty\}$$ and throw the half-open intervals of the forms $$({-}\infty,b\,]$$ and $$(a,{+}\infty]$$ into the mix – including $$\R_{{+}\infty}=({-}\infty,{+}\infty].$$ This gives us again a ring of sets, which is a superring of the construction with only finite endpoints. This time, the ring is also closed under absolute complement, so we can appeal to the identity
 * $$X\setminus Y=\overline{~\overline{X}\cup Y},$$
 * meaning that we can restrict our attention to the easy case of binary unions. The result can be pulled back to the finite-endpoint ring by showing that finite-endpoint-hood is preserved – trivial for unions, since the result endpoints are the extrema of the finite input endpoints and therefore still finite, and trivial for relative complements, since for this operation the segment spanned by the extrema of the left operand will only shrink. --Lambiam 12:50, 23 February 2021 (UTC)
 * Thanks. To generalize this to "semi closed cuboids" in $$\mathbb R^n$$ what do you recommend? To establish $$\Pi(a_i,b_i\,]\setminus\Pi(c_i,d_i\,]$$ does not seem similarly manageable; or is it?  - Abdul Muhsy   talk  15:28, 24 February 2021 (UTC)
 * Not with the second approach, because you would not have closure under complements. Using the first approach, perhaps you can use the $$n$$-dimensional generalization of $$(P{\times}Q)\setminus(R{\times}S)= ((P{\setminus}R)\times Q)\,\cup\,(P\times(Q{\setminus}S))$$ to good effect. I have not checked this, but I expect the identity $$\textstyle{\Pi_i X_i\,\cap~\Pi_j Y_j = \Pi_{i{,}j}~X_i\cap Y_j}$$ to hold, so using the set-theoretic equivalent of De Morgan's laws their complements will be a cubist Emmenthal model. Are you looking for a generalization just for the fun of it, or are you motivated by a potential use? --Lambiam 17:25, 24 February 2021 (UTC)
 * What do you mean by "their complements will be a cubist Emmenthal model"? I am self learning out of personal interest. Both problems are based on similar problems in Halmos' measure theory book. - Abdul Muhsy   talk  17:56, 24 February 2021 (UTC)
 * The set generated by the compliments of $$\Pi(a_i,b_i\,]$$ hypercubes will be a model satisfying the ring axioms, and in the 3D case will be like Emmenthal cheese – as imagined by a cubist sculptor. --Lambiam 19:46, 24 February 2021 (UTC)

Slender groups and Jerzy Łoś
Please see a question by User:Wlod at --CiaPan (talk) 08:50, 23 February 2021 (UTC)
 * Answered there. --Lambiam 12:52, 23 February 2021 (UTC)