Empty set

The empty set is the set hav­ing no mem­bers. It is usu­ally de­noted as \(\emptyset\). What­ever ob­ject is con­sid­ered, it can’t be a mem­ber of \(\emptyset\). It might be use­ful in the be­gin­ning to think about the empty set as an empty box. It has noth­ing in­side it, but it still does ex­ist.

For­mally, the ex­is­tence of the empty set is as­serted by the Empty Set Ax­iom:

$$\exists B \forall x : x∉B$$

The empty set ax­iom it­self does not pos­tu­late the unique­ness of \(\emptyset\). How­ever, this fact is easy to prove us­ing the ax­iom of ex­ten­sion­al­ity. Con­sider sets \(A\) and \(B\) such that both \(\forall x : x∉A\) and \(\forall x: x∉B\). noteThat is, sup­pose we had two empty sets. Re­mem­ber that the ex­ten­sion­al­ity ax­iom tells us that if we can show \(\forall x : (x ∈ A \Leftrightarrow x ∈ B)\), then we may de­duce that \(A=B\). In this case, for ev­ery \(x\), both parts of the state­ment \((x ∈ A \Leftrightarrow x ∈ B)\) are false: we have \(x \not \in A\) and \(x \not \in B\). There­fore the iff re­la­tion is true.

The ex­is­tence of the empty set can be de­rived from the ex­is­tence of any other set us­ing the ax­iom schema of bounded com­pre­hen­sion, which states that for any for­mula \(\phi\) in the lan­guage of set the­ory, \(\forall a \exists b \forall x : x \in b \Leftrightarrow (x \in a \wedge \phi(x))\). In par­tic­u­lar, tak­ing \(\phi\) to be \(\bot\), the always-false for­mula, we have that \(\forall a \exists b \forall x : x \in b \Leftrightarrow (x \in a \wedge \bot)\). Since \(x \in b \Leftrightarrow (x \in a \wedge \bot)\) is log­i­cally equiv­a­lent to \(x \in b \Leftrightarrow \bot\) and hence to \(x \notin b\), the quan­tified state­ment is log­i­cally equiv­a­lent to \(\forall a \exists b \forall x : x \notin b\), and as soon as we have the ex­is­tence of at least one set to use as \(a\), we ob­tain the Empty Set Ax­iom above.

It is worth not­ing that the empty set is it­self a sin­gle ob­ject. One can con­struct a set con­tain­ing the empty set: \(\{\emptyset\}\). \(\{\emptyset\} \not= \emptyset\), be­cause \(\emptyset ∈ \{\emptyset\}\) but \(\emptyset ∉ \emptyset\); so the two sets have differ­ent el­e­ments and there­fore can­not be equal by ex­ten­sion­al­ity. noteIn terms of the box metaphor above, \(\{\emptyset\}\) is a box, con­tain­ing an empty box, whilst \(\emptyset\) is just an empty box

Another way to think about this is us­ing car­di­nal­ity. In­deed, \(|\{\emptyset\}| = 1\) (as this set con­tains a sin­gle el­e­ment - \(\emptyset\)) and \(|\emptyset| = 0\) (as it con­tains no el­e­ments at all). Con­se­quently, the two sets have differ­ent amounts of mem­bers and can not be equal.

the empty set is of­ten used to rep­re­sent the or­di­nal 0

Punc­tu­a­tion can be weird in this edit, as the au­thor is not a na­tive English speaker. Might need to be improved

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  • Set

    An un­ordered col­lec­tion of dis­tinct ob­jects.