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It may seem that this question is a duplicate of : Axiom of Regularity and Axiom of regularity definition

But, my issue is not with the Axiom itself. I understand its need and use. But the definition has me confused. I've taken the expression from Wikipedia https://en.wikipedia.org/wiki/Axiom_of_regularity

$${\displaystyle \forall x\,(x\neq \varnothing \rightarrow \exists y(y\in x\ \land y\cap x=\varnothing )).}$$

I understand its implication $x \cap \{x\} = \varnothing$.

But, my issue lies with the $(y \in x \land y \cap x)$.

For instance consider a set:

$$R=\big\{\; \{1,2\},\;\{3,4,\{1,2\}\}\;\big\}.$$ Then take the two subsets:

$$r1=\{1,2\} \space \text{ meaning } \space r1\in R$$. $$r2=\{3,4,\{1,2\}\} \space \text{ meaning } \space r2\in R$$.

Would it be wrong to do this?

$$r2 \cap R = \{1,2\} = r1$$.

If not, doesn't this mean that $(r2 \in R \land r2 \cap R) \neq \varnothing$ contradicts with part of the axiom $\exists y(y \in x \land y \cap x) = \varnothing$ ?

Or is the axiom meaning that the instance of $\{1,2\}$ inside $r2$ is not to be considered the same as $r1$ ?

Or am I misunderstanding the semantics of $\exists y(y \in x \land y \cap x) = \varnothing$ ?

Thanks in advance.

EDIT Thanks @RossMillikan/@AsafKaragila I see the error in my understanding $\exists y$. The moment you remove the last disjoint set $r1$ out the $\exists y$ ceases to be true! what was I thinking!?

Brian Pinto
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  • You are correct (under some assumptions about numbers) that $r2 \cap R = {1, 2} = r1$. What is the problem with this, and what does this have to do with regularity? – Mark Saving Jan 25 '22 at 22:55
  • You should read it $\exists y((y \in x )\land (y \cap x= \varnothing))$ – Ross Millikan Jan 25 '22 at 22:59
  • @Mark Saving. How do I justify the part definition ∃y(y∈x∧y∩x)=∅ ? – Brian Pinto Jan 25 '22 at 22:59
  • The $\land$ binds less tightly than $\in, \cap, =$ so the parentheses go where I showed. – Ross Millikan Jan 25 '22 at 23:00
  • @Ross Millikan, Ok I'll have to rethink it from the way you re-arranged the equation. if so Wikipedia's equation must then be wrong - as the null is outside the parenthesis. – Brian Pinto Jan 25 '22 at 23:05
  • @RossMillikan, but My issue of r2∩R={1,2}=r1 still holds. I'm having trouble only with the justification. – Brian Pinto Jan 25 '22 at 23:06
  • But you are just told that some element is disjoint from the set. In fact, $r1 \cap R=\emptyset$ because neither $1$ nor $2$ is an element of $R$. The fact that $r2$ has an intersection with $R$ is not important. – Ross Millikan Jan 25 '22 at 23:09
  • @RossMillikan Yes you are right $\exists y$ means that (atleast one) ... what would happen if we were to remove the subset r1 leaving only r2 in R. Thanks though I can redit and present this case. – Brian Pinto Jan 25 '22 at 23:13
  • @BrianPinto I’m very confused by what you’re saying. We know there is some element $y \in R$ such that $y \cap R = \emptyset$. We also know that $r2 \cap R \neq \emptyset$. There is no contradiction here, since $r1 \cap R = \emptyset$. – Mark Saving Jan 25 '22 at 23:32
  • @MarkSaving. Yes I read it wrong. While learnig ZF I attempted a bunch of practical cases for all axioms and when I created the smallest possible example for regularity the $\exists y$ escaped me I didn't see the introduction/removal of r1 trumping $\exists y$. Thanks – Brian Pinto Jan 26 '22 at 00:06
  • @BrianPinto Actually, $r_2 \cap R = {{1,2}}$. Which is not the point. But still. – Jonathan Schilhan Jan 26 '22 at 00:19
  • @Jonathan, Yes should have worded it out like that - regularity I as I see was built to combat the affect of humans confusing elements and sets which creates the Russels Paradox. I still am new to getting out of naive st methods. I showed the element instead of showing it in a set as you say. – Brian Pinto Jan 26 '22 at 00:44

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Here's a fact: there exists a smallest number in the set $\{3,45\}$.

But wait, $45$ is not the smallest number in that set! Does that mean that there is no smallest number in that set?

The point is that just statements of the form "There exists ..." depend on find a witness, not just finding a witness to the contrary. The problems begin if you cannot find a witness at all.

Asaf Karagila
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