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 $\{\emptyset,\{\emptyset,\{\emptyset\}\}\}$ ?


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.

  • $\begingroup$ This is not an answer but a counter-question. $\endgroup$ – rexkogitans Jan 21 at 9:36
  • 2
    $\begingroup$ @rexkogitans Also known as a "hint" $\endgroup$ – TreFox Jan 21 at 19:36

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 $\{\{\emptyset\}\} \not\equiv \{\emptyset,\{\emptyset,\{\emptyset\}\}\}$.


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).


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