Definition talk:Magma of Sets

Hi LF. I hope you don't mind these questions.

Question 1 (an obvious one)

Is this line:


 * $\phi_i: \mathcal P \left({X}\right)^{J_i} \to P \left({X}\right)$

supposed to be:


 * $\phi_i: \mathcal P \left({X}\right)^{J_i} \to \mathcal P \left({X}\right)$

Question 2

On lines 7 and 8 you present the defining conditions for a magma of sets. On line 9 you give a logically equivalent definition that on the surface seems miles easier to understand. But I've never come across this.

What do you mean by a set being closed under a mapping?

I am trying to understand this concept using finite $\sigma$ algebras. It's very difficult.

Please understand that I'm not making hard work for you. I just seek to see what you see. --Jshflynn (talk) 22:31, 2 October 2012 (UTC)


 * 1. Yes; good catch. 2. I mean with '$S$ is closed under $\phi_i$' that, upon feeding $\phi_i$ with only elements from $S$, it spits out an element of $S$. So if $S$ were to be a $\sigma$-algebra, it would need to be closed under the mappings "intersection", "countable union" and the constant mapping with image $X$.
 * I'd be glad if answering this questions opened at least one person's eyes for this concept (which is like an abstraction of algebraic structure). The condition that the $\phi_i$ are only partial mappings is to ensure that things like $\lim_{n \to \infty}$ could be included as well. --Lord_Farin (talk) 22:16, 3 October 2012 (UTC)


 * Technically, the concept of closure in this specific context has not been created yet. The more specialised ones of closed under an operation and closed for scalar product are just instances of the more general "closed under a mapping", but the latter concept still needs to be specified. Interestingly, although most texts are conscientious about making sure Definition:Closure (Abstract Algebra)/Algebraic Structure is defined as precisely as you like, the associated concept of $\phi \left({S}\right) \subseteq S$ is taken as understood. --prime mover (talk) 05:19, 4 October 2012 (UTC)


 * By looking at the examples provided the concept has just about sunk in. Instances of MoS are everywhere. It describes nearly every space and structure I've seen. Was the partial mapping condition also intended to account for fields and vector spaces? It's very elegant. When you say that an intersection of a magma of sets is a magma of sets what does this say about $\left( X, \circ \right)$ in relation to the induced magma of sets $\left( \mathcal P \left( X \right), \circ_\mathcal{P} \right)$?


 * Also, PM I don't think most people learn the connection between a relation, function and operation. It is curious. Also (just saying) I dislike giving something a name and THEN proving that the name is justified. But that's how it goes... --Jshflynn (talk) 13:51, 4 October 2012 (UTC)


 * I haven't come round to trying to define vector spaces as magmas of sets. The induced operation analogy gives a universal proof that the intersection of substructures is again a substructure. We can define substructures by their power sets being magmas of sets inside the large power set. Conversely, those $A \subseteq \mathcal P \left({X}\right)$ with $\mathcal P \left({\bigcup A}\right) = A$ correspond to power sets of subsets. Coming back to VSp.; I think they can be described as MoS on the disjoint union of space/abelian group and underlying field/ring. I think it also can be employed to encompass the notion of an action of a group/monoid on a set. To me, these constructions appear rather artificial; it may also be that they do not have an if and only if correspondence with the concept they describe. --Lord_Farin (talk) 14:05, 4 October 2012 (UTC)

Topology as Magma of Sets
Couldn't one just take $I = 2$, $J_0 = 1$, and $J_1 = \mathcal P \left({X}\right)$? --abcxyz (talk) 23:25, 4 October 2012 (UTC)


 * No, for it is all too common to construct open covers by taking an open nbhd of each point. But I may have missed your point, since there are three conditions for a topology. Could you please explain it further? --Lord_Farin (talk) 06:04, 5 October 2012 (UTC)


 * Oops, I messed up. How about $I = 3$, $J_0 = 1$, $J_1 = 2$, $J_2 = \mathcal P \left({X}\right)$, $\phi_0 \left({S}\right) = X$, $\phi_1 \left({U, V}\right) = U \cap V$, and $\displaystyle \phi_2 \left({\left\langle{S_j}\right\rangle_{j \in J_2}}\right) = \bigcup_{j \mathop \in J_2} S_j$?


 * By the way, maybe we could write $j$ instead of $j_i$ just to have the notation a bit less cluttered? --abcxyz (talk) 15:24, 5 October 2012 (UTC)


 * Sorry, that's not right either. I guess we additionally need $J_0' = 1$ and $\phi_0' \left({S}\right) = \varnothing$.


 * Alternatively, I think we could have $I = \left\{{0, 1}\right\} \sqcup \mathcal P \left({X}\right)$, $J_0 = 1$, $J_1 = 2$, $J_A = A$ (for $A \subseteq X$), $\phi_0 \left({S}\right) = X$, $\phi_1 \left({U, V}\right) = U \cap V$, and $\displaystyle \phi_A \left({\left\langle{S_j}\right\rangle}_{j \in A}\right) = \bigcup_{j \mathop \in A} S_j$ (for $A \subseteq X$).


 * I hope this is correct. My point is that I'm pretty sure that $I$ can be taken to be a set. --abcxyz (talk) 15:44, 5 October 2012 (UTC)


 * I think you are right. It would be rather cluttered to prove it (passing from $J$ to an equivalence class on $J$), but I think that indeed taking an $\phi$ for each $A \subseteq X$, or even just one for each cardinal less than $\left\vert{X}\right\vert$. I have replaced $j_i$ with $j$ as suggested. --Lord_Farin (talk) 16:50, 5 October 2012 (UTC)
 * Thank you for your constructive input; I will try and amend the notions as appropriate. --Lord_Farin (talk) 16:50, 5 October 2012 (UTC)


 * I don't understand what you mean by "passing from $J$ to an equivalence class on $J$". Could you please clarify? --abcxyz (talk) 17:22, 5 October 2012 (UTC)

Suppose you had a very large index set $J$, and wanted to prove that the union $\bigcup_{j \mathop \in J} U_j$ indexed by $J$ is again in the MoS-topology $\vartheta$ (previously denoted $S$) as you described it. Then it would be natural to pass to $\left\{{U \in \vartheta: \exists j \in J: U = U_j}\right\}$ and take the union of this (and a $\varnothing$ for all $U \in \mathcal P \left({X}\right)$ not in this set). This occurred to me as passing to $J/\sim$ where $j \sim j' \iff U_j = U_{j'}$. The former variant is probably what you had in mind, but the latter is what I had in mind to give a bound on the cardinality; your idea actually does the job. --Lord_Farin (talk) 07:56, 6 October 2012 (UTC)


 * I think I kind of see what you were thinking. As you noted, it is possible to prove it another way (which is indeed what I had in mind). --abcxyz (talk) 19:27, 6 October 2012 (UTC)