# 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

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No typo here.${}$ –  Michael Greinecker Feb 13 '13 at 11:08
Herbert? Herbert B. Enderton? –  Martin Feb 13 '13 at 11:11
@Martin, yes, Herbert Enderton –  Vikram Feb 13 '13 at 11:12

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.

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(It is important to recognize that "nothing" plays an unusual role in English! And should not be confused with the idea of an "empty set") –  Hurkyl Feb 13 '13 at 11:13
@Hurkyl, but that is how I remember/understand an "empty set" –  Vikram Feb 13 '13 at 11:27
@Vikram: Then you are utterly confused. The empty set is not some essence of nothing. It is a set. Specifically, it is a set that has no elements. –  Chris Eagle Feb 13 '13 at 11:29
Oh, I get it, thanx, –  Vikram Feb 13 '13 at 11:36
• 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.

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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.

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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 $\varnothing$.

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