Show two sets $A\notin B$ or $B \notin A $

Question

See this question Prove that for any two sets $A$ and $B$, $A\notin B$ or $B\notin A$

I am having some doubt in it

Let two sets A and B be such that

A ={ B } and B = { A } . Then both contain each other . What is wrong here ?

Principal of regularity says atleast one element is disjoint from A and A contains a set which contains A . So different objects and I guess disjoint .

Also if we keep using principal of substituting here then sets

A = { B }

A = { { A } }

A = { { { B } } }

.

I think these all sets are different because they contains different objects but then they all are equal to A .

I think I found it because$$A\displaystyle\cap B = A\displaystyle\cap {A}\neq \phi $$ — RKK, Nov 15 '21 at 07:47
But what if B = { { A } } and A = { { B } } now we won't have the problem . I still can't get that infinite substitution sequence out of my head — RKK, Nov 15 '21 at 07:53
Consider $A= { a, b , c }$; this means that the only elements of $A$ are $a, b$ and $c$. But ${ A }$ has only one element: $A$, and thus there are no common elements between the two sets. — Mauro ALLEGRANZA, Nov 15 '21 at 07:54
"I think these all sets are different because they contains different objects..." CORRECT, "... but then they all are equal to A." WRONG: they are different, and thus they are not all equal to A. — Mauro ALLEGRANZA, Nov 15 '21 at 08:13
Then doesn't it defies law of substitution when two sets A = B then we can substitute B at place of A . — RKK, Nov 15 '21 at 13:50
Yes we can... If $A = { B }$, then we can use this "identity" in substitution; thus e.g. ${ A } = { { B } }$. — Mauro ALLEGRANZA, Nov 15 '21 at 14:48
But consider again an example: $B$ is the empty set. Thus, $B$ has no members, while $A= { B } = { \emptyset }$ has one member: the empty set. Thus, $A \ne B$. — Mauro ALLEGRANZA, Nov 15 '21 at 14:50

score 2 · Answer 1 · answered Nov 15 '21 at 08:24

2

Here is a claim: there are real numbers $x, y$ such that $x=y+1$ and $y=x+1$.

There, I wrote it down, so there are two such real numbers! Now we can simplify and see $x=y+1=x+1$ and so $0=1$. Mathematics is inconsistent, we are all working too hard, let's go home and take a nap.

Or, more likely, writing down equations does not mean that there is a solution to these equations.

In the context of set theory $A=\{B\}$ is an equation, which we would normally posit has a solution. But this requires proof, after all, $R=\{x\mid x\notin x\}$ is an equation, and Russell's paradox is exactly the statement that it has no solution.

So the short answer here is that $A=\{B\}, B=\{A\}$ simply doesn't admit a solution, assuming $\sf ZFC$.

answered Nov 15 '21 at 08:24

Asaf Karagila

393,674

So sir I think adding principal of regularity should get rid off all of such problem so how this defies this rule ? – RKK Nov 15 '21 at 14:02
Also why we can't use law of substitution here? – RKK Nov 15 '21 at 15:48
I'm not sure that I follow these questions. – Asaf Karagila Nov 17 '21 at 10:45

Show two sets $A\notin B$ or $B \notin A $

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