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I am very confused about whether $\{∅\} ∈ \{∅\}$ or not. I thought it's right because in curly brackets the phy is also a member.

Can anyone help me understand this?

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No, this isn't the case. The set $\lbrace \emptyset \rbrace$ contains one and only one element: the empty set $\emptyset$. It does not contain $\lbrace \emptyset \rbrace$.

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    $\begingroup$ You need to argue that $\varnothing\neq\{\varnothing\}$ as well. That's not very hard, though. $\endgroup$ – Asaf Karagila Jul 9 '15 at 21:06
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In general, no set is an element of itself. It is true that:

$\emptyset \in \{\emptyset \}$

$\emptyset \subseteq \{\emptyset \}$

$\{\emptyset \} \subseteq \{\emptyset \}$

but

$\{ \emptyset \} \notin \{ \emptyset \}$, and more generally,

$A \notin A$ where $A$ is any set whatsoever.

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The set $\{x\}$ contains just one element, that is, $x$.

This is the same as saying that the statement $y\in\{x\}$ is true if and only if $y=x$.

Now let's try with $x=\emptyset$ and $y=\{\emptyset\}$. Can it be $\{\emptyset\}=\emptyset$? No, because the set in the left-hand side has an element, whereas the set in the right-hand side has none: two sets are equal if and only if they have the same elements.

Therefore, $\{\emptyset\}\ne\emptyset$ and so $\{\emptyset\}\notin\{\emptyset\}$.

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You may find this question (which distinguishes $\in$ from $\subseteq$) to be useful and this answer to a separate question which details several claims concerning the empty set and gives the veracity of these clams as well.

In the linked to answer, you will see the following:

(j) $\{\varnothing\}\in\{\varnothing\}\qquad\qquad$ [False]

(k) $\{\varnothing\}\in\{\{\varnothing\}\}\qquad\qquad$ [True]

That is, you really need to understand the difference between $\in$ and $\subseteq$ to understand why Theo Bendit's answer makes sense.

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