Talk:Exists Bijection to a Disjoint Set

Recent edit
I do not have the book, but the recent edit seems incomplete. Where do you take $y$ from? This element has to be an element of some set. Since we are assuming the existence of sets $T$ and $S$, $y$ should belong to one of them, no?--Julius (talk) 21:55, 14 February 2022 (UTC)

For $X$ to be a set we don't have to specify to which set $y$ belongs to, it's defined by a predicate. I don't think it should belong to $S$, that was the problem in the first place! You are right, the proof has to be further edited to show that $X$ is a set. I'm not sure how to do that. Moniker (talk) 22:11, 14 February 2022 (UTC)


 * If you believe a proof to be incorrect, please raise it on the talk page explaining why you think it is wrong.


 * Please note that the proof is not in the book, it is set as an exercise. The proof itself was provided by someone on MathHelpForum quite some time ago. Nobody claimed it was incorrect at the time. --prime mover (talk) 23:47, 14 February 2022 (UTC)

I think is wrong because the step "If $(c, z)\in S$ then $z\in X$" is not justified. Moniker (talk) 23:58, 14 February 2022 (UTC)


 * I agree this proof is currently flawed because the definition of $X$ states that it is a subset of $S$, but the $z$ we found from Exists Element Not in Set may not belong in $S$, so we cannot assert that $z \in X$ (from $\tuple{c,z} \in S$). (In fact, $z$ would belong in $\powerset X$, which is a subset of $\powerset S$.)


 * For example, take $S = \set {\tuple {1, 2}}, T = \set 1$. Then $X = \emptyset$ and we can choose $z = 2$. Then $T' = T \times \set z = S$.


 * If we keep $\in S$ in the definition of $X$, we would need to find $z \in S \setminus X$, but then we need some argument to show $X \ne S$, which seems to be an infinitely deep rabbit hole.


 * Removing $\in S$, or even removing $\in T$ as well (so we would have $X = \set {y: \paren {\exists x: \tuple {x, y} \in S} }$, so $X$ is the set of second coordinates of ordered pairs in $S$) would intuitively work. Is it really necessary to specify their source set(s) (and what would it be?) --RandomUndergrad (talk) 03:20, 15 February 2022 (UTC)


 * I'll leave it to you guys to resolve this, my set theoretical knowledge is self-taught and hence lacking.


 * Incidentally, can everybody agree to use the indentation convention for talk pages? Otherwise it's impossible to tell who wrote what. --prime mover (talk) 06:32, 15 February 2022 (UTC)


 * This guarantees that $X = \set {y: \paren {\exists x \in T: \tuple {x, y} \in S} } \subseteq \text{dom}(S)$ is a set. I think the solution here would be to make a page with the proof that "domain" and "range" of a set is a set, and then reference that proof. And of course, we need to get rid of the condition $y\in S$ in the definition of $X$. Moniker (talk) 16:04, 15 February 2022 (UTC)


 * Or just put $X = \text{dom}(S)$ instead, the $T$ is not needed anywhere. Moniker (talk) 16:25, 15 February 2022 (UTC)


 * The story behind this page is as follows.
 * In Warner there is a remark:
 * [Embedding] may always be done in view of the following fact, which may be proved in any formal development of the theory of sets: If $E$ and $F$ are sets, then there is a bijection from $F$ onto a set $F'$ disjoint from $E$.


 * Then exercise $8.12$ says:
 * Prove that the following statement is equivalent to that given before Theorem $8.1$ [that is, what I put above]: If $G$ is a set, there is a bijection from $G$ onto a set $G'$ disjoint from $G$. (Consider $E \cup F$.)


 * None of the exercises in Warner are proved.
 * I haven't seen that contentious statement made in any of the works of set theory that I've worked through, but then I haven't got very far with most of them. I get bogged down in the details and move on to something else. Hence I have been reliant upon others who perhaps have seen this theorem proven in some set theory text, or had it demonstrated, or whatever. I asked the question in MathHelpForum which I don't even know whether it's active any more, most of the contributors moved on to other places (e.g. me, I moved here). Hence I can't vouch even whether I presented it correctly.
 * Have fun. --prime mover (talk) 18:22, 15 February 2022 (UTC)


 * I've made a new page Domain_and_Range_of_a_Set_is_a_Set to fix the proof of this theorem. Cheers. Moniker (talk) 18:31, 16 February 2022 (UTC)


 * Thank you. I'll go in and tidy it up and put it into house style in due course. Incidentally, please exclusively use links to internal pages without the underscores in them, for obvious reasons. I also remind you of the indentation convention for talk pages. --prime mover (talk) 19:11, 16 February 2022 (UTC)