- #1

Seydlitz

- 263

- 4

Afterwards comes this excerpt,

It follows that, whatever the set A may be, if ##B = \{x \in A \mid x \not \in x \}## then,

for all ##y##,

##(*)## ##y \in B## if and only if ##(y \in A \text{ and } y \not \in y)##

Can it be that ##B \in A##? We proceed to prove that the answer is no. Indeed, if ##B \in A## then either ##B \in B## also (unlikely, but not obviously impossible), or else ##B \not \in B##.

How does one infer that if ##B \in A## then either ##B \in B## or ##B \not \in B##. Is it like when I am shopping to the market ##A## and there's a basket ##B## there. If a product is in the market it is possible to have a product either in the basket or not in the basket i.e in the shelves.

Or is it by inferring from the definition like this:

##(*)## ##B \in B## if and only if ##(B \in A \text{ and } B \not \in B)##

But then clearly there's something wrong since it cannot be both true that ##(B \in B## and ##B \not \in B)##? I mean yes if ##B \in A## then according to the definition ##B \not \in B## but then if we accept ##B \in A## such that ##B \not \in B##, clearly according to the definition ##B \in B##.

Could you guys explain this part further? I also think that I haven't fully appreciated this example. What does the writer want to show to us? That a set cannot have everything and there must be at least something that is not in the set, hence the Axiom of Specification?

Thank You