Can someone give an argument, if possible using only the axioms of set theory, because I'm very weak there and have virtually no background, except the usual knowledge of the operation with sets one has to have when doing non-set theoretic non-research mathematics, why $\emptyset \in \emptyset$ or $\emptyset \subseteq \emptyset$ should or should not hold?

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    $\begingroup$ Do you mean $\varnothing \subseteq \varnothing$? $\endgroup$
    – t.b.
    Jul 16, 2011 at 8:35
  • $\begingroup$ No, but this case would also interest me. As I could gather from your question, $\emptyset \in \emptyset$ is false ? (I edited the question accordingly) $\endgroup$
    – temo
    Jul 16, 2011 at 8:37
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    $\begingroup$ By the axiom of the empty set, $\forall x(\neg x\in\emptyset)$. So in particular, it is false that $\emptyset\in\emptyset$. $\endgroup$ Jul 16, 2011 at 8:38
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    $\begingroup$ @Zev: [obligatory nitpicky remark about assuming the consistency of ZF] $\endgroup$
    – user83827
    Jul 16, 2011 at 8:44
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    $\begingroup$ For the answerers: This guy is very weak in set theory. So perhaps answer with $\forall$ symbols and such in them are not helpful? Instead an explanation in words? Maybe? $\endgroup$
    – GEdgar
    Jul 16, 2011 at 13:09

2 Answers 2


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 $y$.

  • $\begingroup$ Nitpick: Regularity gives you an $\in$-minimal element in each non-empty set. $\endgroup$ Jul 16, 2011 at 9:23
  • $\begingroup$ @Brian: True, but we assumed that $\varnothing\in\varnothing$. I will add this clarification. Thanks! $\endgroup$
    – Asaf Karagila
    Jul 16, 2011 at 9:26
  • $\begingroup$ Sorry to be late at the party but I would replace $\forall x\forall y\forall z((z\in x\leftrightarrow z\in y)\leftrightarrow x=y)$ by $\forall x\forall y((\forall z(z\in x\leftrightarrow z\in y))\leftrightarrow x=y)$. And likewise for the definition of $\subseteq$. $\endgroup$
    – Did
    Sep 21, 2011 at 8:57
  • $\begingroup$ @Didier: I tend to agree, I did reduce one parenthesis level from your suggestion though :-) $\endgroup$
    – Asaf Karagila
    Sep 21, 2011 at 9:00

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).

  • $\begingroup$ Why do have to assume, that the ZF axioms are consistent ? $\endgroup$
    – temo
    Jul 16, 2011 at 9:13
  • $\begingroup$ I would say that, if the ZF axioms were inconsistent, answering the question of whether a given statement is true or not using the ZF axioms becomes "pointless" because every statement is both true and false at the same time. But perhaps this explanation is not ideal; I welcome corrections / criticism from people who are more confident in this area than I am. $\endgroup$ Jul 16, 2011 at 9:25
  • $\begingroup$ @temo: I'll answer that since I think I started this mess with my tongue-in-cheek comment. It is always the case that ZF proves the theorem "$\neg(\emptyset \in \emptyset)$" for the reasons discussed. It might also be the case that the theorem "$\emptyset \in \emptyset$" is provable from ZF; this occurs exactly when ZF is inconsistent. But of course this is tangential to the main point and I somewhat regret bringing it up. $\endgroup$
    – user83827
    Jul 16, 2011 at 9:28
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    $\begingroup$ We do not need to assume that ZF is consistent to prove a sentence $\phi$. If ZF happens to be inconsistent, we could also prove $\lnot \phi$, but that is a side remark. $\endgroup$ Jul 16, 2011 at 14:01
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    $\begingroup$ +1 for $\emptyset=\{\}$. Personally, I cannot understand why most people prefer the symbol $\emptyset$ anyway; I always write it as $\{\}$. $\endgroup$ Jul 16, 2011 at 16:02

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