Does every set "contain" the empty set?

Question

Define a property "$X$ contains $Y$" iff $Y \in X$ or $Z \in X$ and $Z$ contains $Y$ (meaning in a finite recursive sort of way). Then does every set contain $\emptyset$ (in ZFC maybe $\pm$ some axioms)? For example, every natural number (in any sensible constuction) contain $\emptyset$, and therefore every integer and rational does too. One could show that real numbers contain $\emptyset$, and therefore constructions based on the reals also. I think we probably need an axiom like $V=L$ but is that sufficient?

Answer 1

Your "containment relation" is equivalent to $Y\in\operatorname{tr cl}X$, where $\operatorname{tr cl} X$ is a transitive closure of $X$. We can show that every transitive closure of a non-empty set has the empty set as its element.

Proof. We will use induction on rank of sets. If $x$ is non-empty, then $\operatorname{rank} x\ge 1$.

If $\operatorname{rank} x=1$, so that $x\subseteq V_1=\{\varnothing\}$, then $x=\{\varnothing\}$. We can check that $\operatorname{tr cl}x\supseteq x$ for any $x$, so the transitive closure of $\{\varnothing\}$ contains the empty set.

Now assume that every $x\in V_\alpha$ (i.e. sets $x$ such that $\operatorname{rank} x<\alpha$) satisfies $\varnothing\in\operatorname{tr cl}x$. If $\operatorname{rank}x=\alpha$, then $x\subseteq V_\alpha$ so every elements of $x$ has rank $<\alpha$. Since $\operatorname{tr cl} x\supseteq x\cup \bigcup x$ and $\varnothing\in \bigcup x$, $\varnothing \in \operatorname{tr cl}x$.

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The proof uses the axiom of regularity: the consequence of the axiom of regularity is that the Von Neumann hierarchy covers the class of all sets.

Moreover, the axiom of regularity is necessary for the proof: it is consistent with ZF without regularity that a Quine atom (i.e., a set $x$ satisfying $x=\{x\}$) exists. We can see that the transitive closure of $x$ is just $x$, and $x$ does not contain the empty set.

Answer 2

Well, the empty set doesn't. But every nonempty set does. To be precise, for any nonempty set $X$, there is a finite sequence $X_0,X_1,\dots,X_n$ with $X_0=\emptyset$, $X_n=X$, and $X_i\in X_{i+1}$ for each $i$. This follows from the axiom of regularity (and is essentially a weak variant of it): if there existed a nonempty set $X$ which does not "contain" $\emptyset$ in this sense, then by regularity there would be an $\in$-minimal such set $X$. But then since $X$ is nonempty, it has an element $Y$, and by $\in$-minimality of $X$, $Y$ contains $\emptyset$. But then $X$ contains $\emptyset$ by appending one more term on the end of the sequence for $Y$.

(Note that there is a bit of sleight of hand in how I used regularity here, since regularity says only that a nonempty set has an $\in$-minimal element. See the argument in this answer for how to justify this usage. (Or, as in Hanul Jeon's answer, you can use the cumulative hierarchy, but the same trick is needed to prove every set is in the cumulative hierarchy, so it amounts to the same thing.))