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

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.

• This is not an answer but a counter-question. – rexkogitans Jan 21 at 9:36
• @rexkogitans Also known as a "hint" – 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).