# Consider the sets A and B , where A = <mo fence="false" stretchy="false">{ 3 , <mrow c

Consider the sets A and B , where
$A=\left\{3,|B|\right\}$
and $B=\left\{1,|A|,|B|\right\}$
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 $B=\left\{1,3\right\}$, 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 $A=\left\{2,3\right\}$ and $B=\left\{1,2\right\}$.
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 $\left\{1,2,3\right\}$. 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 $\left\{3\right\}$. Thus we go back to the original forms of the sets, namely $A=\left\{3\right\}$ and $B=\left\{1,2\right\}$ 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 $A=\left\{2,3\right\}$ and $B=\left\{1,2\right\}$.
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?
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Dustin Durham
Step 1
Previously B knows that A is just 3 with $|A|=1$.
No, previously an assumption was made from which it followed that $A=\left\{3\right\}$. It was shown that that led to a contraction. It was then concluded that the assumption that led to A being {3} was false. So we're now no longer making that assumption. The author is engaging in proof by contradiction: they are provisionally making an assumption, showing that assumption leads to a contradiction, and then concluding that the assumption is false. Everything after that is proceeding from that conclusion that the assumption is false.
It might help (or maybe make things more confusing, but it's worth a try) to consider the riddle "Alice and Bob are both from an island where everyone either always tells the truth or always lies. Alice says 'Bob said he's a liar'. What you can conclude?" If Alice is telling the truth, then Bob apparently is a liar. But if he is a liar, then he was telling the truth when he said he was a liar, so he's not a liar. So he's a truth-teller. Which means he lied about being a liar. Which means he's a liar. But he can't be. The assumption "Alice told the truth" leads to a paradoxical loop, so our assumption was wrong. We then conclude that Alice was lying, and the paradox disappears.
Step 2
It seems to only respect the cardinality of B but did not continue to respect the cardinality of A.
It's completely consistent with all that was stated about A and B.

tr2os8x
Step 1
You don’t go in circles: a set has fixed elements and a fixed cardinality, so you don’t go back and forth on what its members are or what its cardinality is. The solution shows that if B has 3 elements, then B has 2 elements. From this, we can conclude that B cannot have 3 members, because it having 3 members would lead to a contradiction (and not to some kind of infinite loop). And, once you have established that, the rest follows.
Step 2
And no, with A being $\left\{2,3\right\}$, B does not become $\left\{1,2,3\right\}$: We know that B has to contain 1 and 2 (the 2 being yhe cardinality of A), so if the cardinality of B is either 1 or 2, then that is not a new element. And indeed, with the cardinality of B being 2, that works out exactly right.