Timeline for A recursive description of the smallest divisor-closed subsemigroup containing a set
Current License: CC BY-SA 4.0
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| Feb 21 at 11:15 | comment | added | Salvo Tringali | @PaceNielsen I agree with your last comment. But overall, this is not really the kind of ref I was looking for. I'm still hoping for something closer to the spirit of what I have in mind, that is, a ref where the "iterative description" in the OP is proved, possibly in a generalized form, without any need for the reader to work out too many additional details (before getting to see what's going on in the case of interest). Many thanks anyway, the exchange was very fruitful for me. | |
| Feb 20 at 17:14 | comment | added | Pace Nielsen | I believe "Exercise 6.3:4(ii)" would be an appropriate reference for the claim about the joint closure stabilizing at the $\omega$th stage. [Writing the semigroup operation multiplicatively, take the set $G$ in that exercise to consist of all pairs $(\{x,y\},xy)$ (forcing closure under being a semigroup) together with all pairs $(\{x\},d)$ whenever $d$ is a divisor of $x$ (forcing closure under divisors).] | |
| Feb 20 at 16:08 | comment | added | Salvo Tringali | @PaceNielsen Right, $\Gamma$ need not be a closure op (otherwise, we would have had $\Gamma^{\circ n} = \Gamma$ for all $n = 1, 2, \ldots,$ and life would have been easier). I'll have a look at the PDF linked in your last comment. Thanks! Maybe I'm wrong again, but in hindsight all this looks like an elegant rewording of the same proof that Pedro Garcia-Sanchez and I have in our work (which motivated the question in the OP). The main difference is that we never mention closures. | |
| Feb 20 at 15:00 | comment | added | Pace Nielsen | For a reference, you might try Bergman's introduction to general algebra. There is a lot on closure operators...don't have time right now to look through it all...math.berkeley.edu/~gbergman/245/3.2.pdf | |
| Feb 20 at 14:58 | comment | added | Pace Nielsen | It is not true that $\bigcap_{Y\in C(X)}Y = \bigcap_{X\subseteq Y\subseteq S}\Gamma(Y)$ because $\Gamma$ itself might not be a closure operator. (The composition of closure operators is not necessarily a closure operator.) The set on the left is $\Gamma$-closed (and hence both $\Pi$- and $\Delta$-closed), and hence equals $\bigcup_{n}\Gamma^{\circ n}(X)$ [since this is $\Gamma$-closed by the argument I gave, and clearly minimal]. | |
| Feb 20 at 6:00 | comment | added | Salvo Tringali | @PaceNielsen Let $C(X)$ be the set of all $\Gamma$-closed subsets of $S$ that contain $X$ (I keep the notation from my previous comment). It seems clear to me that $Y\in C(X)$ iff $Y$ is a divisor-closed subsgrp of $S$ containing $X$. So, $$[\![X]\!]_S=\bigcap_{Y\in C(X)} Y = \bigcap_{X \subseteq Y\subseteq S} \Gamma(Y).$$ To my understanding, your 1st comment is suggesting that, by some general properties of closure operators, the intersection $\bigcap_{X \subseteq Y\subseteq S} \Gamma(Y)$ and the union $\bigcup_n \Gamma^{\circ n}(X)$ are equal. What's a ref? That's my question. | |
| Feb 20 at 5:44 | comment | added | Salvo Tringali | @PaceNielsen I continue to miss something. AFAICS, what you show in the last two comments is that, for each subset $X$ of the sgrp $S$, the set $T:=\bigcup_n\Gamma^{\circ n}(X)$ is "$\Gamma$-closed" (i.e., $\Gamma(T) = T$). But this is not my problem with your "high-tech argument". My problem is that I don't see why $T$ equals $[\![X]\!]_S$ (the intersection of all divisor-closed subsgrps of $S$ containing $X$). Actually, I know how to prove it. The point is rather that I don't understand how this should follow from the kind of "general principles" mentioned in your 1st comment [tbc]. | |
| Feb 19 at 18:28 | comment | added | Pace Nielsen | I don't know a reference off the top of my head for that last fact, The proof is super easy. Let $T=\bigcup_n \Gamma^{\circ n}(S)$. It suffices to show closure under $\Pi$ (as the argument is similar for $\Delta$). Let $x\in \Pi(T)$. Then $x\in \Pi(T_0)$ for some finite subset $T_0\subseteq T$. Thus, $T_0\subseteq \Gamma^{\circ m}(S)$ for a sufficiently large $m$. Then $x\in \Pi(T_0)\subseteq \Gamma^{\circ m+1}(S)\subseteq T$. | |
| Feb 19 at 18:24 | comment | added | Pace Nielsen | In general, it might take longer than an $\omega$-length union to achieve stability. But your operators are special. A closure operator $C$ is called finitary when $x\in C(S)$ exactly when $x\in C(S_0)$ for some finite subset $S_0\subseteq S$. So, when talking about algebras with operations of finite arity, the "subalgebra generated by" operator is a finitary closure operator. Clearly, the "set of divisors" operator is also finitary. Now, a general fact is that if $\Pi$ and $\Delta$ are finitary, and $\Gamma=\Pi\circ \Delta$, then joint closure happens at the union you wrote. | |
| Feb 19 at 5:40 | comment | added | Salvo Tringali | @PaceNielsen I had never realized that "the subgroup generated by" and "the set of divisors" can be viewed as finitary closure operators. That's cool! Let me denote the 1st operator by $\Pi$, the 2nd by $\Delta$, and the composition "$\Pi$ after $\Delta$" by $\Gamma$. I agree that, in the notation of the OP, $D_n(X)=\Gamma^{\circ n}(X)$ for all $n\in\mathbb N$, where $\Gamma^{\circ n}$ is the $n$th iterate of $\Gamma$. How does this show that $[\![X]\!]_S=\bigcup_n\Gamma^{\circ n}(X)$ from general principles? Not very familiar with closure operators and clearly missing something. Any refs? | |
| Feb 18 at 21:33 | comment | added | Pace Nielsen | There are actually two closure operators in play. The "subgroup generated by" operator, and the "set of divisors" operator. I'll leave it to you to check that both of these operators satisfy the usual three axioms of a closure operator en.wikipedia.org/wiki/Closure_operator They are also both finitary. Now, the joint closure under these two operators is just the alternating application of one operator after the other (your iteration) [or the intersection you described]. It only takes an $\omega$-chain of alternations because of the finitariness. | |
| Feb 18 at 17:21 | comment | added | Salvo Tringali | @PaceNielsen Do you have in mind any candidate for the closure operator alluded to in your comment? I'm not sure whether I would be satisfied with this approach, but I'm a big fan of high tech. I guess that much will ultimately depend on the details. | |
| Feb 18 at 16:08 | comment | added | Pace Nielsen | Note an answer, just a comment: Whenever there is a closure operator in play, you have a top-down construction (the intersection you mentioned) and a bottom-up construction (closing, then closing again, etc...). So, if you are satisfied with showing that there is a natural closure operator, you could just refer to the literature on closure operators for the "iterative description" you want. (In your case it stabilizes in countable many steps because the operator is finitary.) | |
| Feb 18 at 12:20 | history | edited | Salvo Tringali | CC BY-SA 4.0 |
fixed further details
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| Feb 18 at 12:12 | history | edited | Salvo Tringali | CC BY-SA 4.0 |
fixed a few mistakes
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| Feb 18 at 9:02 | history | asked | Salvo Tringali | CC BY-SA 4.0 |