# Is the empty set a subset of itself?

Sorry but I don't think I can know, since it's a definition. Please tell me. I don't think that $0=\emptyset\,$ since I distinguish between empty set and the value $0$. Do all sets, even the empty set, have infitinite emptiness e.g. do all sets including the empty set contain infinitely many empty sets?

• Yes, but not a proper subset. – Berci Mar 19 '13 at 11:34
• You refer to the empty set, suggesting that you already know that there is a unique set that is empty, so what do you mean by "contain infinitely many empty sets"? – Trevor Wilson Mar 19 '13 at 18:55
• possible duplicate of Empty set does not belong to empty set – Marc van Leeuwen Mar 20 '13 at 10:26
• Oops, I thought this was a duplicate, but I did not look closely enough. Unfortunately it seems I cannot un-vote to close :-( – Marc van Leeuwen Mar 20 '13 at 10:29
• @NickRosencrantz Regarding your last comment, $\emptyset \notin \emptyset$, because for all $x$ we have $x \notin \emptyset$ by the definition of $\emptyset$. – Trevor Wilson Mar 20 '13 at 19:05

There is only one empty set. It is a subset of every set, including itself. Each set only includes it once as a subset, not an infinite number of times.

• I guess it is a stupid question, but what is a set which includes any set more than 1 time ? – Dominic Michaelis Mar 20 '13 at 10:27
• @DominicMichaelis: A multiset can include multiple copies of elements. Not a stupid question at all. – Ross Millikan Mar 20 '13 at 13:11
• @DominicMichaelis That is a good question. Thank you for asking. Never heard of that before. – BCLC Jul 29 '14 at 23:35

Let $A$ and $B$ be sets. If every element $a\in A$ is also an element of $B$, then $A\subseteq B$.

Flip that around and you get

If $A\not\subseteq B$, then there exists some element $x\in A$ such that $x\notin B$.

If $A$ is the empty set, there are no $x$s in $A$, so in particular there are no $x$s in $A$ that are not in $B$. Thus $A\not\subseteq B$ can't be true. Furthermore, note that we haven't used any property of $B$ in the previous line, so this applies to every set $B$, including $B=\emptyset$.

(From a wider standpoint, you can think of the empty set as the set for which $x\in \emptyset\implies P$ is true for every statement $P$. For example, every $x$ in the empty set is orange; also, every $x$ in the emptyset is not orange. There is no contradiction in either of these statements because there are no $x$'s which could provide counterexamples.)

The empty set is subset of the empty set, as every element of the empty set is an element of the empty set. But $0$ is not in the empty set.

$A \subseteq B$ when $x\in A \implies x\in B$. As $x\in A \iff x\in A$ we see that $A \subseteq A$ is always true, when $A$ is a set.

A value is a value not a set, sometimes $0$ is defined as the empty set but then $0$ is the empty set and not the number.

This happens for example in category theory, as you are only interested in abstract sets, and all sets of the same cardinality are in a sense the same, you just title finite sets by their cardinality.

Any set is a subset of itself. Let be $$A=\{x\,\vert\,\varphi(x)\}.$$ As $$\varphi(x)\implies \varphi(x),$$ we have $$A\subseteq A.$$

Because the reflexivity of $\subseteq$, for all $A$ set $A \subseteq A$ is true. For $A = \emptyset$ we have that $\emptyset \subseteq \emptyset$, so the empty set is a subset of itself.

The intersection of two sets is a subset of each of the original sets. So if $\phi$ is empty set and A is any set then $\phi\cap A$ is $\phi$ which means $\phi$ is subset of A(which is any set) and $\phi$ is a subset of $\phi$ which implies empty set is subset of itself.

By the definition of subset this question means: Is every element of the empty set a member of the empty set? The answer must be yes, since the empty set doesn't contain any elements.

An empty subset can be selected from any set, including the empty set itself, using selection criteria that cannot be satisfied, e.g. selecting those elements $x$ such that $x\neq x$.

See my formal proof (in DC Proof format) at: http://dcproof.com/ExistenseOfNullSet.htm

There is only one empty set. If $\phi_1$ and $\phi_2$ are empty sets, then $\ = \phi_2$.

See my formal proof (in DC Proof format) at: http://dcproof.com/UniquenessOfNullSet.htm

Edit: For any set $X$ we can select an empty subset $S$ such that:

$\forall a:[a\in S\iff a\in X\land a\neq a]$

or

$\forall a:[a\in S\iff a\in X\land a\notin X]$