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How can it be that the empty set is a subset of every set but not an element of every set?

I've understood that the empty set must be a subset of every set because if it were not a subset of every set then the statement $\exists x : x \in \emptyset \land x \notin M$, would need to be ture, but since the empty set has no elements this statement is a contradiction.

But how can it be that the empty set is not element of every set, if it is a subset of every set?

I also understood that it cannot be an element of itself, otherwise it wouldn't be the empty set anymore, but why is it not an element of every non-empty set? If the empty set is subset of every set, than it also would need to be a subset of itself ($\emptyset \subset \emptyset$), how can that be?

Eric Wofsey
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NilsK
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    It seems like there is a fundamental misunderstanding you have between the definition of an element of a set and a subset of a set. For example, do you recognise that the sets ${1}, {{1}}, {{{1}}}$ are three different and distinct sets, each being an element of but not a subset of the next? – Dan Rust Dec 04 '20 at 15:20
  • $\varnothing$ is a subset of e.g. set $A$ because there are no elements in $\varnothing$ that are not also elements of $A$. Even stronger: there are no elements in $\varnothing$ at all. That is purely an observation. – drhab Dec 04 '20 at 15:21
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    A "bag analogy" is useful here. The empty set is like an empty bag. It is vacuously true that the things in an empty bag are in all bags whatsoever (this is the "bag version" of $\emptyset$ being a subset of every set); however, of course it is not the case that every bag has an empty bag inside it (this is the "bag version" of $\emptyset$ not being an element of every set). The "bag analogy" is of limited use, but this is a situation where it works well. – Noah Schweber Dec 04 '20 at 15:22
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    Think of sets as boxes. The empty set is an emtpy box. Box a is a subset of box b if after you dump the contents of box a into box b, but not putting in things that are already there, box b 's contents do not change. c is an element of a box if the box has c inside it. So, the empty set is a subset of any set (any box contains all elements of an empty box); but not all boxes contain an empty box (not all sets contain the empty set). There is a difference between an empty box and a box containing an empty box. Perhaps this is a silly analogy... – David Mitra Dec 04 '20 at 15:24
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    Because it satisfy the definition of subset of. – Mauro ALLEGRANZA Dec 04 '20 at 15:28
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    The set of real numbers is "made of" real numbers. The set of rationals is a subset of the set of real but it is not itself a number. Thus, element and subset are two concepts that do not coincide. – Mauro ALLEGRANZA Dec 04 '20 at 15:38
  • Related: https://math.stackexchange.com/questions/2620616/what-is-the-difference-between-x-and-x-when-x-itself-is-a-set/2620621#2620621 – Ethan Bolker Dec 04 '20 at 15:40
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    Note that while you can't have a set containing each subset as element (because that set would contain itself), you can take any $X$ and construct $X':={Z:\exists Y\in X:Z\subseteq Y}$ (which is the closure of $X$ under $\subseteq$) and this $X'$ is either empty (when $X$ is the empty set) or contains the empty set. – fweth Jul 24 '22 at 06:10
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    A little bit more subtle, but you can also take the transitive closure $\text{TC}(X)$ of any set $X$ and if $X$ is non-empty then $\text{TC}(X)$ contains the empty set, otherwise it would contradict the axiom of regularity. – fweth Jul 24 '22 at 06:22

3 Answers3

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  • There might be versions of set theory where the requirement " the empty set is an element of every set" is satisfied. What I mean is that it does not seem absurd prima facie. For example, in the set theoretic consruction of natural numbers, number zero ( that is : the empty set) is an element of every ( natural) number greater than 0 , and these numbers are sets. ( for example , $1=\{\emptyset\}= \{0\}, 2= \{\emptyset, \{\emptyset\}\}=\{0,1\} , 3= \{0,1,2\} $.

  • However, the question " is every set a member of every set ( different from itself)?" can be settled as a pure matter of fact. Any counter-example would do; Consider, for example, the set : $\{1, 2,3\}$.

  • I think the question is : why does it seem plausible that, if a set is a subset of every set, then it should also be an element of every set? Maybe one could try to reconstruct the reasoning that produces this false appearence :

(1) The empty set s a subset of every set, say, of set S

(2) Therefore, all the elements of $\emptyset$ are also elements of S.

(3) Therefore the totality of the elements of $\emptyset$ is an element of S.

(4) But this totality is $\emptyset$ itself .

(5) Therefore $\emptyset\in S$.

  • The mistake is hidden in steps (3) and (4).

a) as to (3) : though it is true distributively that every element $\emptyset$ is an element of S, it is not true collectively

b) as to (4) the totality of the elements of $\emptyset$ is nothingness ( nil, nothing) , and therefore is not the same thing as the empty set which is something ( namely, a set).

Floridus Floridi
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This is a crude analogy but I think it gives a concrete explanation of the idea.

Think about a set as a bowl full of things.

A subset is what you get when you take some things out of the bowl.

Take everything out of the bowl, and the bowl is empty. This corresponds to the empty set.

Every bowl can be emptied. So the empty set is a subset of every set.

A = {} Set A is the empty set. Since it's empty, it can't contain the empty set.

B = { 1, {} } Set B actually contains the empty set.

C = { {1}, {1, {}} } Set C can also be written as C = {{1}, B}. Set B contains the empty set but set C does not.

Steven Alexis Gregory
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  • Element:

An element x is said to be an element of a set A if x is contained within the set A. This is denoted as: x ∈ A.

  • Subset:

A set A is considered a subset of another set B if every element of A is also an element of B.

Mathematically, this is denoted as: A ⊆ B.

The empty set (∅) is a subset of every set because it contains no elements that are not in the other set, but it is not an element of every set because it is not an object that can be contained within a set.

Sanaullah Khan
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