Consider the sets A and B , where

$A=\{3,|B|\}$

and $B=\{1,|A|,|B|\}$

What are the sets?

Answer: We need to be a little careful here. If B contains 3 elements, then A contains just the number 3 (listed twice). So that would make $|A|=1$, which would make $B=\{1,3\}$, which only has 2 elements. Thus $|B|\ne 3$. This means that $|A|=2$ so B contains at least the elements 1 and 2. Since $|B|\ne 3$, we must have $|B|=2$ which agrees with the definition of B . Therefore it must be that $A=\{2,3\}$ and $B=\{1,2\}$.

I do not understand how the solution ended. If at the end A now has 2 elements, B should be updated as well. Previously B knows that A is just {3} with $|A|=1$, but since A now has two elements 2 and 3, then $|A|=2$ and B must be $\{1,2,3\}$. The last 3 is the new cardinality of B .

Of course, if I go this route, then A must be updated as well because now $|B|=3$, making A go back to $\{3\}$. Thus we go back to the original forms of the sets, namely $A=\{3\}$ and $B=\{1,2\}$ and so on.

It seems to me, finding both cardinalities leads to never ending transformations. There must be something that I am missing. Why is it correct to end the sets such that $A=\{2,3\}$ and $B=\{1,2\}$.

It seems to only respect the cardinality of B but did not continue to respect the cardinality of A. But continuing on this reasoning will lead to an infinite loop. Can someone explain what is wrong in my thinking?

$A=\{3,|B|\}$

and $B=\{1,|A|,|B|\}$

What are the sets?

Answer: We need to be a little careful here. If B contains 3 elements, then A contains just the number 3 (listed twice). So that would make $|A|=1$, which would make $B=\{1,3\}$, which only has 2 elements. Thus $|B|\ne 3$. This means that $|A|=2$ so B contains at least the elements 1 and 2. Since $|B|\ne 3$, we must have $|B|=2$ which agrees with the definition of B . Therefore it must be that $A=\{2,3\}$ and $B=\{1,2\}$.

I do not understand how the solution ended. If at the end A now has 2 elements, B should be updated as well. Previously B knows that A is just {3} with $|A|=1$, but since A now has two elements 2 and 3, then $|A|=2$ and B must be $\{1,2,3\}$. The last 3 is the new cardinality of B .

Of course, if I go this route, then A must be updated as well because now $|B|=3$, making A go back to $\{3\}$. Thus we go back to the original forms of the sets, namely $A=\{3\}$ and $B=\{1,2\}$ and so on.

It seems to me, finding both cardinalities leads to never ending transformations. There must be something that I am missing. Why is it correct to end the sets such that $A=\{2,3\}$ and $B=\{1,2\}$.

It seems to only respect the cardinality of B but did not continue to respect the cardinality of A. But continuing on this reasoning will lead to an infinite loop. Can someone explain what is wrong in my thinking?