Is $\emptyset \in \emptyset$ or $\emptyset \subseteq \emptyset$?

Solution 1:

The empty set $\varnothing$ is the only set which satisfies $\forall x(x\notin y)$ (that means the formula is true if and only if $y=\varnothing$)

There are many ways to define the empty set (the set of all $x$ such that $x\neq x$ - we will use this formula later on) but by the axiom of extensionality it is unique.

The axiom of extensionality is, in simple words, two sets are equal if and only if they have the same elements. Or formally: $$\forall x\forall y\bigg(\forall z(z\in x\leftrightarrow z\in y)\leftrightarrow x=y\bigg)$$ So since $\varnothing=\{x\mid x\neq x\}$ we can deduce that $\varnothing\notin\varnothing$, otherwise $\varnothing$ would be such $x$ for which $x\neq x$.

Even worse, if $\varnothing\in\varnothing$ then we contradict another axiom of ZFC - the axiom of foundation (or regularity) which asserts that there is a $\in$-minimal element in every non-empty set. Since $\varnothing\in\varnothing$, we will have that $\varnothing$ is not empty.

What does that mean? It means that if I consider a set, there is someone in that set that no one else in that set is its element. A quick corollary is that $\forall x (x\notin x)$, in particular for $\varnothing$.

Lastly, $\varnothing\subseteq\varnothing$. Let us consider the meaning of $\subseteq$: $$\forall x\forall y\bigg(\forall z(z\in x\rightarrow z\in y)\leftrightarrow x\subseteq y\bigg)$$ Informally we have that $x$ is a subset of $y$ if and only if all the elements of $x$ are elements of $y$, with this we can reformulate the axiom of extensionality as $\forall x\forall y(x=y\leftrightarrow(x\subseteq y\land y\subseteq x))$.

In particular, $\varnothing=\varnothing$ so $\varnothing\subseteq\varnothing$.

A stronger conclusion, however, can be drawn from the definition of $\subseteq$ by the idea of vacuous truth, since for all $z$ we have $z\notin\varnothing$ the clause $z\in\varnothing\rightarrow z\in y$ is always true, therefore $\varnothing\subseteq y$, for every $y$.

Solution 2:

An axiomatic argument (as ccc points out, we must assume that the ZF axioms are in fact consistent) would proceed as follows: By the axiom of the empty set, $\forall x(\neg x\in\emptyset)$. So in particular, it is false that $\emptyset\in\emptyset$.

Here is a more intuitive explanation. $\emptyset\in\emptyset$ is false because $\emptyset$ has no elements (by definition). In other words, $$\emptyset=\{\}.$$ A set containing the empty set is a perfectly fine set, for example, $$\{\emptyset\};$$ but it is evident that $\emptyset$ itself is not a set containing $\emptyset$ as an element.

However, it is true that $\emptyset\subseteq\emptyset$. For any sets $A$ and $B$, we say that $A\subseteq B$ precisely when $\forall x(x\in A\Rightarrow x\in B)$. There are two ways this is clear for the case $A=B=\emptyset$; firstly, it is true that $P\Rightarrow P$ for any statement $P$ (here the statement $P$ is "$x\in \emptyset$"), so $A\subseteq A$ for any set $A$, and in particular $\emptyset\subseteq\emptyset$. Secondly, for all $x$ it is false that $x\in\emptyset$, so the implication $x\in\emptyset\Rightarrow P$ is true for any statement $P$ (see here).