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I could be wrong, but I learned that $∅\in\{∅,\{∅\}\}$ is false but $∅\in\{∅\}$ is true... why is that? I would think that both of these would have to be true. If $∅\in\{∅\}$ is true wouldn't $\{∅\}$ be an element in $\{∅,\{∅\}\}$?

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    $\begingroup$ $a\in\{a,\{a\}\}$ for all $a$, including $a=\emptyset$. $\endgroup$ – Lord Shark the Unknown Sep 1 '18 at 19:10
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    $\begingroup$ Both are true. Maybe you meant $\emptyset \in \{\{ \emptyset \} \}$ is false $\endgroup$ – leibnewtz Sep 1 '18 at 19:11
  • $\begingroup$ In ordinal notation the first statement is $0\in 2$, while the second is $0\in 1$. Both are true, but it's theoretically possible you got the first statement mixed up with one that actually is false, such as $0^+=2$. $\endgroup$ – J.G. Sep 1 '18 at 19:14
  • $\begingroup$ Several similar questions have been posted here; a very detailed answer by Daniel is here. $\endgroup$ – Dietrich Burde Sep 1 '18 at 19:20
  • $\begingroup$ Thank you all for that clarification! I thought I was going mad. $\endgroup$ – maximusg Sep 1 '18 at 19:26
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Both of these statements are correct as you have suggested. In both of your sets the empty set appears as an element so there should not be any confusion.

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since there is one comma in the set it indicates that there are two different objects in the set, hence empty set symbol as one object, and the brace that contains the empty set symbol as another object,therefore, both are elements of the set you constructed and hence true correct, and in similar argument for the second case you said true also I agree true because the set contains the empty set symbol as one object only in it. After you enter inside set only find definite objects separated by comma only to consider which definite objects are the members of the set whatever that definite object is.

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