This is just a sharing of an interesting paradox. I am definitely not an expert in Set Theory. If you want to learn more about the paradox, you may refer to the formal text on Set Theory.

Consider the barber paradox in the picture above. The question asks “who shaves the barber?”. There are two possibilities, the barber shaves himself or he does not shave. If he shaves, he should not shave since he only shaves those who do not shave. If he does not shave, he should shave himself since he should shave everyone who does not shave himself. So, in either case, there is a contradiction. The barber cannot shave or not shave himself.

The Russell Paradox is similar. Consider a set A. Whatever set A may be, if B = \{x \in A: x \notin x\}, then for all y, y \in B \iff (y \in A \land y \notin y).

The question is can B belong to A? So, the proof goes like this (it requires some knowledge in proof writing). Suppose B \in A. Let’s just assume the statement is true. Then, we proceed to prove two cases. Case 1: B \in B, and Case 2: B \notin B. Why do we have these two cases? If you look at the definition of B, and take into account the assumption that B \in A (replace y with B), we can find that there are only two possibilities. If B \in B, then B \notin B. If B \notin B, then B \in B. In both cases, we have contradictions. That means B \notin A.

What is the implication? If A is everything, a set of everything, then we have proved that we can always find a set B that is not in A. Then is A the set of everything?

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