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I'm also interested in the converse of this: does $A \in B \implies B \not\subset A $?

I can give examples where the first one holds: (e.g. $ A = \{1,2,3\}, B= \{1,2,3,4\}$), or the second one holds ($A=\{1\}, B=\{\{1\}\} $), but I can't come up with counterexamples for either of them.

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  • $\begingroup$ Can you resolve whether it's possible that $B\in B$? (i.e. what sort of axioms for set theory are you familiar with? This is fairly easy in ZFC, but maybe less clear if you're used to working informally) $\endgroup$ – Milo Brandt Feb 4 at 5:57
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You can't have a set contain itself (see this post). Assume $B \in A$. Then $A = \{B,...\} \subset B$ is a contradiction. To answer your other question, $A \in B \implies B = \{A,...\} \not\subset A$.

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  • Suppose A is included in B .

  • It means that : all the elements of A are also elements of B.

  • Now, in case B is a member of A, then, by the rule stated above, B is a member of B ( for the rule holds for all the elements of A, and B is supposed to be one of these).

  • So, you question amounts to : does the relation " being a member of" possibly hold between an object and itself, or , is rather this relation absolutely irreflexive?

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You should understand the difference between the two symbols and .
[1] A ⊂ B⟹B ∉ A

  1. When you say A ⊂ B, this implies that set A is a subset of set B; which implies that all the elements of set A, are also elements of set B.
    For example, let set A = {1, 2} and set B = {1, 2, 3}. Here all the elements of set A are also in set B, so A⊂B.
  2. When you say B ∈ A, this implies that set B is an element of set A.
    For example, let set B = {1, 2, 3} then set A would contain set B, along with the possibility of other elements, then A = {{1, 2, 3}, …}.
    So, if A⊂B, it cannot contain set B as a member, hence in this case B∈A is not possible.

[2] A ∈ B ⟹ B ⊄ A
When A ∈ B this implies B = {A, …}. In this case, it holds good that B⊄A.

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