Empty set does not belong to empty set

Herbert in his book "Elements of set theory" on page no 3 says that

we can form the set $ \{ \emptyset \} $ whose only member is $\emptyset $. Note that $ \{ \emptyset \} \neq \emptyset $, because $ \emptyset \in \{ \emptyset \} $ but $\underline{ \emptyset \notin \emptyset} $·

By the last argument $\emptyset \notin \emptyset$, is he saying that empty set is not a member of or does not belong to empty set

OR

it is a typo and he wanted it to be $ \{\emptyset \} \notin \emptyset $, that set containing empty set is not a member of empty set

Solution 1:

Of course the empty set is not an element of the empty set. Nothing is an element of the empty set. That's what "empty" means.

Solution 2:

• No set is member of the empty set
• In the usual axiomization of set theory (ZF or ZFC) no set is member of itself.

By either argument the empty set is not member of the empty set.

Maybe you were confused by the fact that the empty set trivially contains (as a subset, not as a member) the empty set; every set contains as a subset the empty set, and it also equally trivially contains itself as a subset. For the empty set these two cases are in fact the same, and indeed the empty set is the unique subset of the empty set.

Solution 3:

The empty set has no elements. So he is really saying that, as a particular case, the empty set is not a member of the empty set.

Solution 4:

The general form of the argument he makes is

$A\ne B$ because $C\in A$ but $C\notin B$.

That, by definition, is how you prove that two sets are different from each other -- you prove that there is something that is member of one but not of the other.

What you're reading is simply this argument where $A$ is $\{\varnothing\}$, and $B$ and $C$ both happen to be $\varnothing$.