Is the empty set a subset of itself?

Question

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 infinite emptiness e.g. do all sets, including the empty set, contain infinitely many empty sets?

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
@TrevorWilson Since set members should be unique there can be only one empty subset for every set. I was thinking something like {}∈{}∈{}∈{}∈{}∈{}∈... that you can always get a new empty set. — Niklas Rosencrantz, Mar 20 '13 at 10:24
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
@MarcvanLeeuwen The link helped, thank you. ∅∉∅ and {}∉{} I'm now not absolutely sure which is which from the 2: ∅ and {} but I think that ∅∉∅ and {}∉{} while {}⊂{}⊂{}⊂{}...(?) — Niklas Rosencrantz, Mar 20 '13 at 19:46
One has $\emptyset={}$, it is just a shorthand. And indeed $\emptyset\notin\emptyset$ but $\emptyset\subseteq\emptyset$. And there is no point in making long chains, saying $\emptyset\subseteq\emptyset\subseteq\emptyset\subseteq\emptyset$ is as pointless as saying $2\leq2\leq2\leq2$. — Marc van Leeuwen, Mar 20 '13 at 20:00
Related: Is {∅} a subset of {{∅}}? and Is ∅∈∅ or ∅⊆∅? — Scott - Слава Україні, Mar 24 '16 at 02:55
It seems to be a difference between the set {} and the subset. If the empty set has a subset then it is not "empty" because the cardinality is 1 isn't it? — Niklas Rosencrantz, Mar 24 '16 at 16:49

score 55 · Accepted Answer · answered Mar 19 '13 at 16:13

55

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.

answered Mar 19 '13 at 16:13

Ross Millikan

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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
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@DominicMichaelis: A multiset can include multiple copies of elements. Not a stupid question at all. – Ross Millikan Mar 20 '13 at 13:11
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@DominicMichaelis That is a good question. Thank you for asking. Never heard of that before. – BCLC Jul 29 '14 at 23:35
Then could not one make a multiset containing multiple copies of the empty set? M = {Ø, Ø} ? – Dion Silverman May 22 '21 at 22:14
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@DionSilverman: yes, you can. A multiset can contain multiple copies of any element. The empty set is not at all special here. – Ross Millikan May 22 '21 at 22:42
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The empty set, $\emptyset = {}$, has $\emptyset$ as a subset of itself once. What about the set $S = {\emptyset} = {{}}$? How many times does $S$ have $\emptyset$ as a subset of itself? Once? Twice? – joseville Nov 18 '21 at 21:35
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@joseville: sets can only contain things once, so it doesn't make sense to ask how many times $S$ contains the empty set. $S$ has only one subset that is the empty set. It also has one subset that is $S$ itself. That one is not the empty set. – Ross Millikan Nov 18 '21 at 21:38
@RossMillikan, thanks. Just want to point out that I'm strictly differentiating between contains $x$ and has $x$ as a subset. As some other comments point out, it's possible for $x \notin S$ whilst still having $x \subset S$ (example, $\emptyset \notin \emptyset$, whilst $\emptyset \subset \emptyset$). So asking "how many times $S$ contains $\emptyset$?" is different than asking "how many times $S$ has $\emptyset$ as a subset?". – joseville Nov 18 '21 at 21:55
Now that I think about it, $S = {\emptyset}$ means $\emptyset \in S$, $\emptyset \subset S$, and ${\emptyset} \subset S$. – joseville Nov 18 '21 at 21:55
@joseville: yes to the last. That is spot on. – Ross Millikan Nov 19 '21 at 04:33

Alexander Gruber · Answer 2 · 2017-10-07T20:48:29.437

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.)

score 6 · Answer 3 · answered Dec 04 '14 at 08:19

6

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

answered Dec 04 '14 at 08:19

Martín-Blas Pérez Pinilla

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Dominic Michaelis · Answer 4 · 2013-03-19T11:32:58.447

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.

score 3 · Answer 5 · answered Dec 07 '14 at 17:09

3

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.

answered Dec 07 '14 at 17:09

user153012

12,240

score 2 · Answer 6 · answered Mar 08 '16 at 05:53

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.

score 1 · Answer 7 · answered Feb 11 '18 at 05:46

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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.

answered Feb 11 '18 at 05:46

ANorell

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Dan Christensen · Answer 8 · 2021-01-22T04:03:52.037

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 $\ \phi_1 = \phi_2$.

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

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

That is correct. Thanks. – Dan Christensen Jan 22 '21 at 04:04 — Dan Christensen, Jan 22 '21 at 04:04

Is the empty set a subset of itself?

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