# Talk:Intersection of Subsets is Subset/Set of Sets

Isn't this exactly the same argument as Intersection is Largest Subset/General Result? The latter starts with $\mathbb S$ being defined as a subset of $\powerset T$, of course, which makes more sense as it establishes the parameters of the result before you start, which makes it easier to follow what it says. --prime mover (talk) 16:58, 18 January 2013 (UTC)

- I have three goals here:
- * When I personally first encountered references to Intersection Largest and Union Smallest, I didn't know what they meant, so I was a bit confused. Since it's easy to understand "intersection of subsets is subset" and "union of subsets is subset", I figured it would be good to factor them out and give them their own names.
- * I personally found the way the "general result" is expressed to be somewhat confusing, but you wanted to preserve it as is so I made a new version.
- * The set of sets version can actually be applied in more situations: you don't have to know of a set including the elements of the family as subsets before you can apply it. Proving that there
*is*such a set actually repeats this proof! --Dfeuer (talk) 17:18, 18 January 2013 (UTC)

- Family schmamily. The only reason that's in there is because someone gave me a hard time about not presenting them in this way.

- That whole indexed family bit is of no interest to me.

- The reason the sets are presented as being a subset of the power set is because that, ultimately, is the most general way of considering a set of subsets of a given set. You're talking about a set of sets, each one of which is a subset of another set. That is exactly what being a subset of a powerset is.

- The point is that in the original, you start with a set, and then you make an assertion about an arbitrary collection of subsets of that set. In this version, you start with a set, and then you take an arbitrary set of sets, and
*then*you make the assertion that oh, yes, each of those sets is a subset of that first set you started with. You have to mentally change gears and back up a step so as to get a handle on what you're talking about.

- The point is that in the original, you start with a set, and then you make an assertion about an arbitrary collection of subsets of that set. In this version, you start with a set, and then you take an arbitrary set of sets, and

- But I'm still intrigued as to what confused you about the idea of a subset of a powerset. I thought it was elementary. Tell me, what is the confusing bit there? --prime mover (talk) 20:17, 18 January 2013 (UTC)

- I think what confused me is that it had nothing to do with my mental image of what this theorem was about. You begin with a set that isn't obviously needed and then eliminate it.

- Proof of your version of Union Smallest using mine: Since $\mathbb S$ is a subset of $\powerset S$, $\mathbb S$ is a set of sets. Thus by my version, your version holds.

- Srsly need me to? Okay, $S \in \mathbb S \implies S \subseteq T \implies S \in \powerset T$ by definition and so $\mathbb S \subseteq \powerset T$ by definition of subset HTR by mine.

$\blacksquare$ --prime mover (talk) 21:05, 18 January 2013 (UTC)

Nope, you only have one implication. Union Smallest calls for two. --Dfeuer (talk) 21:07, 18 January 2013 (UTC)

That's because I was only doing the part that we were discussing. Wake up at the back. --prime mover (talk) 21:16, 18 January 2013 (UTC)

- The advantage of my way of doing it is that it fits into the ZF axioms: given a set it has a power set and by the law of specification we can generate subsets. With your way of doing it you have to first prove that the sets exist before being able to use them in a proof, and so you can't use it at such a fundamental level as mine. Have you never studied axiomatic set theory? --prime mover (talk) 21:16, 18 January 2013 (UTC)

- In the context of Union of Subsets is Subset/Set of Sets, which set(s) do you have to prove the existence of? The existence of $\bigcup \mathbb S$ follows directly from the axiom of unions. --abcxyz (talk) 21:25, 18 January 2013 (UTC)

- $\mathbb S$ itself. Until you have actually defined that each element of $S$ is a subset of $T$, you have to "suppose" the existence of each of them - as a subset of some other ur-set that is the superset of each of those sets. If you start from scratch with the fact that each of these sets is an element of the power set, you lose that extra work. --prime mover (talk) 21:29, 18 January 2013 (UTC)

- Why do you have to prove the existence of $\mathbb S$? All we have to prove is an equivalence; why it is necessary to prove the existence of sets that actually satisfy a side of the equivalence? --abcxyz (talk) 21:43, 18 January 2013 (UTC)

- The whole point of this page is that it is
**not**an equivalence, the original proof was pulled apart into two. --prime mover (talk) 21:48, 18 January 2013 (UTC)

- The whole point of this page is that it is

- Okay fine, but what I said still applies, doesn't it? Why it is necessary to prove the existence of sets that actually satisfy the hypotheses? --abcxyz (talk) 21:53, 18 January 2013 (UTC)

You ALWAYS have to prove the sets exist before using them in practice. One way to do so is to construct them as a power set, but that is not the only way. The axiom of infinity comes to mind rather prominently (especially in combination with the axiom of replacement). Your version takes mine and mashes it together with the power set axiom, making it harder to apply. Your version: given a set, we can make a set of sets whose union has a certain property. My version: given any set of sets, however constructed, its union has a certain property.Yes, that leaves it to the user to produce the set of sets, but I really don't see that as a problem.--Dfeuer (talk) 21:35, 18 January 2013 (UTC)

- Yes but if you've proved the existence of a
**single set**, that is $T$, and that set is uncountable, for example, then you have $\mathbb S$ which is**likewise**uncountable, if you are working in a constructivist context, you then have to posit the construction of an uncountable number of arbitrary sets,**before**you then say, "But it's all right, we didn't need to do that because all the sets I'm going to use are elements of the power set. My way: three constructions: $T$, $\powerset T$, $\mathbb S \subseteq \powerset T$. Your way: $T$, any uncountable number of constructions of the elements of possibly an uncountable $\mathbb S$, a possibly uncountable number of assertions that each of the elements of $\mathbb S$ are subsets of $T$. Come on now, it's a no-brainer. --prime mover (talk) 21:47, 18 January 2013 (UTC)

- All you're saying, in lots of words, is that your way combines the set of sets version with the power set axiom into a package you consider convenient. Fine. A matter of taste, I suppose. That's still not an argument against including my version --Dfeuer (talk) 21:53, 18 January 2013 (UTC)

- No, it's not just a matter of "convenience". See that "uncountable" I posted up there, in conjunction with the word "constructivist" (a.k.a. "intuitionist")? --prime mover (talk) 22:06, 18 January 2013 (UTC)

- It did occur to all of you that this is a result in
*naive*set theory, not*axiomatic*set theory?! Moreover, Z(F)(C) aren't the only axiomatisations possible. Admittedly, the most used, but I'd contend the truth of this theorem does (should?) not depend on what axiomatisation we happen to choose. If it does, I'd be happy to say we're dealing with nonstandard axioms, e.g. constructivist ones. (It be noted that "constructivist" in general may be considered as different from "intuitionist" but I forgot where the fine detail was at.) - On-topic: I recall watching this page and Intersection is Largest Subset, and there is a subtle difference between the two. They are just in a different context, even if the content is almost identical. Both should be retained. --Lord_Farin (talk) 22:18, 18 January 2013 (UTC)

- It did occur to all of you that this is a result in

- PM, I saw the "uncountable". I am well aware that proving the existence of certain sets may be either more difficult or impossible when working in a restricted logic or a restricted axiom system. I think the only context where your approach would work and mine would not is if you accept the power set axiom but not the axiom of union. So I'll admit your version has a role to play, but it's a specialized one. --Dfeuer (talk) 23:00, 18 January 2013 (UTC)

- You can't use the axiom of union
*until you have established that all those sets $S$ are subsets of $T$*.

- You can't use the axiom of union

- The point I'm making is that the version I originally put together is adequate for
**whatever**context you embed the proof in. The original argument, from which this is a spin-off, was that the original version was suboptimal because it was "confusing" or something, which I could not identify with. It's trivial in the grand scheme of things, but I don't want to allow the precedent for changing stuff unless it's actually*wrong*. I'm giving it a rest for the night, and I have commitments tomorrow. --prime mover (talk) 23:26, 18 January 2013 (UTC)

- The point I'm making is that the version I originally put together is adequate for