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This question is related to the previous one I asked.

@Brad proves in that answer that $A \notin A$ by considering the set $A = \{0,1\}$.

What I don't understand in his proof that he takes a single element from Set A and proves that $\{0,1\}\neq 0$

But wasn't the whole proof was about $A \in A$ ?

Replacing A by {0,1} in the above notation, I get $\{0,1\} = \{0,1\}$

So isn't reflexivity proved ? Why does he take a single element when both the sets are same in $A \in A$ ?

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Sibi
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    Generally speaking, you shouldn't accept an answer right away. You should take a bit of time, think about it, ask for clarification in the comments, and when you're done thinking about it, accept it. It will allow more time for receiving more answers too. This second question shows that you were definitely too hasty in accepting the previous question's answer, and I hope you'll consider waiting a bit longer the next time. – Asaf Karagila Apr 22 '14 at 01:09
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    Going further: generally speaking you shouldn't accept an answer until you understand and agree with it. Use comments to ask the person who wrote the answer any questions you need to. – Steve Jessop Apr 22 '14 at 01:52

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The set $A=\{0,1\}$ has two objects in it: $0$ and $1$. Both are numbers, and neither are sets.

The "thing" $A$ is also an object. It is a set. It is an object which is not $0$ or $1$. Since only the objects $0$ and $1$ are in $A$, the object $A$ is not in $A$.

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    Actually, in the standard construction, $0$ and $1$ are sets, namely the empty set, and the set containing (only) the empty set. Indeed in that construction, $A$ is also a number, namely the number $2$. – celtschk Apr 22 '14 at 01:11
  • @NotNotLogical Sorry, but still I'm not able to reason about it. I feel that Object A is in A because A is just the set {0,1} and {0,1} $\in$ {0,1} ? – Sibi Apr 22 '14 at 01:13
  • $A$ is a subset of $A$. But it is not a member of $A$. –  Apr 22 '14 at 01:15
  • @Sibi: $\in$ is "is element of". The set ${0,1}$ has two elements, namely $0$ and $1$. That is, $x\in A$ if and only if either $x=0$ or $x=1$. – celtschk Apr 22 '14 at 01:15
  • @celtschk Then why is the same A used on both sides of A $\in$ A ? – Sibi Apr 22 '14 at 01:16
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    @Sibi: Because one could imagine that there are sets which contain themselves as elements, for example $A={A}$. For that set, if it existed, $A\in A$ would be true. – celtschk Apr 22 '14 at 01:17
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    @Sibi: Maybe you should read this answer of mine. – Asaf Karagila Apr 22 '14 at 01:46