Is $∅ ∈ \{ \{∅\} \}$ true? If $ \{\emptyset\} ∈ \{\emptyset,\{\emptyset\}\} $ is true, does it mean this $ \emptyset \in \{\{\emptyset\}\} $ true ? If it is not, why it is false?
Also, does $ \{\{\emptyset\}\}$ mean $\{\emptyset,\{\emptyset,\{\emptyset\}\}\}$ ?
 A: The notation $a \in A$ says that among the elements of $A$ there is one element that is exactly equal to $a.$
The notation $\{\emptyset, \{\emptyset\}\}$ describes a set with exactly two elements.
The first element is $\emptyset.$ The second element is $\{\emptyset\}.$
Is one of those two elements exactly equal to  $\{\emptyset\}$?
The notation $\{ \{\emptyset\}\}$ describes a set with one element.
That element is $\{\emptyset\}.$
Which element of $\{ \{\emptyset\}\}$ do you think is exactly equal to $\emptyset$?
Hint: there's only one element you have to check.
The notation $\{\emptyset, \{\emptyset, \{\emptyset\}\}\}$ again describes a set with two elements.
One element is $\emptyset$ and the other is
$\{\emptyset, \{\emptyset\}\}.$
So this is definitely not the same thing as any set that has only one element.
A: The set $\{\{\emptyset\}\}$ is a set with one element: $\{\emptyset\}$ (which in turn is a set with one element, the empty set, which in turn is a set with no elements).
Therefore the empty set is not an element of the set you described. The way of thinking about it is: the more external (outer) brackets determine which elements compose the set.
This also explains why $\{\{\emptyset\}\} \not\equiv \{\emptyset,\{\emptyset,\{\emptyset\}\}\}$.
A: Let us begin with naming the sets in the question: The set $B=\{a\}$ has a single element, which is $a$. Now, let $a=\emptyset$, then we have $B=\{\emptyset\}$, which is of course also a set with a single elementy, namely $\emptyset$.
If we put $A$ into a set $C=\{B\}$, then we have $C=\{\{\emptyset\}\}$. It can be easily seen that $a \not\in C$. Hence, $\emptyset \not\in \{\{\emptyset\}\}$.
The set $A=\{\emptyset, b\}$ has two elements. Now, let $b=\{\emptyset\}$. Then we have the set $A=\{\emptyset, \{\emptyset\}\}$. If we now take a look at $b$, then, since $b\in A$, also $\{\emptyset\} \in A$, which is $\{\emptyset\} \in \{\emptyset, \{\emptyset\}\}$.
To answer if $\{\{\emptyset\}\}$ means $\{\emptyset,\{\emptyset,\{\emptyset\}\}\}$:
Look at $C$, it has only one element, which is $B$. It follows that $\emptyset \not\in C$, so these sets are different (the second set has two elements, which are the empty set and set set containing the empty set).
