# How is an empty set truly "empty"?

In a related question, an answerer says:

an empty bag is a bag with nothing inside it.

Makes sense, but I'm reading a textbook right now that says:

The empty set has only one subset (namely, itself).

Which seems odd to me. If a set is empty, why would it contain a subset? Regardless of whether or not the subset is also an empty set, the "empty set" still contains a subset, so how is it really "empty"?

To go back to the empty bag analogy, it just sounds like my text book is saying:

an empty bag is a bag with nothing but an empty bag inside it.

Which doesn't make sense because then the bag isn't empty. I can't wrap my head around this concept.

• While it's true that $\emptyset \subset \emptyset$, this does not imply that $\emptyset \in \emptyset$. Sep 1, 2015 at 0:50
• @littleO You mean $\emptyset \subseteq \emptyset$? Sep 1, 2015 at 13:02
• @immibis: There are different conventions on whether $\subset$ denotes a strict subset or not. Obviously, littleO used the convention where it is a non-strict subset. Sep 1, 2015 at 13:37
• If you alter the first quote, perhaps the problem goes away: the empty set is the contents of an empty bag. Sep 1, 2015 at 16:05
• Possible duplicate of What is an Empty set? Feb 5, 2016 at 14:01

The use of the word "contains" is a bit misleading, there. When we talk about a set $A$ "containing" a subset $B$, what we really mean is that $A$ contains all elements of the set $B$ (that is, for every $x\in B,$ we have $x\in A$). In that sense, we're saying that an empty bag is a bag that contains exactly what an empty bag contains: nothing at all.

It is possible (though not in the case of the empty set) that a set $A$ contains a set $B$ both as an element and as a subset. For example, the set $$A=\bigl\{1,\{1\}\bigl\}$$ has $$B=\{1\}$$ as an element (obvious, hopefully) and as a subset (because $A$ contains every element of $B$). Here's (one place) where the grocery bag analogy breaks down, however, since the grocery bag analogy would suggest that the above $B$ is a subset of the above $A$ by virtue of being contained in $A$ as an element. This is not so. Indeed, if we consider $$C=\bigl\{\{1\}\bigl\},$$ then we find that $B$ is an element of $C,$ but not a subset of $C,$ since $1$ is not an element of $C$!

So, given two arbitrary sets $A$ and $B$, $B$ may be:

• a subset of $A$ but not an element (e.g.: for any set $A$ such that $\emptyset\notin A$, let $B=\emptyset$),
• an element of $A$ but not a subset,
• an element and a subset of $A$,
• neither an element nor a subset of $A$ (consider $A=\{1\},$ $B=\{3\}$).
• Yes! I for one think that especially in pedagogical contexts, far too many metaphors are presented (such as sets as bags, or fractions as slices of cake that one is somehow supposed to multiply together) without the senses in which they are inapplicable and those in which they are appropriate. Given only the analogy you can't really be expected to know when it's safe by osmosis. Sep 1, 2015 at 6:26
• @Vandermonde The "bag" analogy perfectly works to explain why $\varnothing \subset \varnothing$: every item in the "first" empty bag is also in the "second" empty bag. Sep 1, 2015 at 9:41
• @Najib: That's a good point. On the other hand, when one tries to think about the bag that contains only my drivers license and keys, and the bag that contains only my drivers license and passport, it becomes difficult to think about these two bags existing simultaneously. The analogy certainly has its pedogogical applicability, but it also has stark limits, which aren't often explored. Sep 1, 2015 at 11:44
• In regard to Najib's comment, what I meant by the bag metaphor not (generally) working was not that it never worked.... Anyway, the model can be improved if one just thinks instead about potential bags or dotted lines surrounding things (rather than actual ones, which due to their walls can't nontrivially intersect on/at a driver's license but can be related only by (physical, not necessarily in the sense of set containment or elementhood) containment or disjointness), and now that I think about it, it actually works rather well after such an improvement with the understanding that Nov 28, 2015 at 5:35
• elementhood is not transitive (since in passing from a set to one of its elements, one goes down one 'level'/depth, so to speak, as in a file tree; I think the axiom of regularity or similar formalises this?), although containment of sets is, and so long as one really understands what $x \in y$ is supposed to mean; e.g., $1$ (let's say it's an ur-element) is not in your set $C$ since you can't get $1$ by emptying out (i.e., removing one layer of curly braces from) the bag $C$ and choosing one item so obtained, but would rather have to open another bag, which isn't allowed for this purpose. Nov 28, 2015 at 5:36

Do you know what it really means for one set to be a subset of another set? If the set $A$ is a subset of the set $B$, then we write $A\subseteq B$; now, to show that $A\subseteq B$, we must show that $x\in A\to x\in B$.

The fact that $\varnothing\subseteq\varnothing$ is really not that surprising if you are aware of what a so-called "vacuous truth" is. That is, $x\in\varnothing$ is clearly not true; thus, we can freely conclude (i.e., "vacuously") whatever we want. Hence, $x\in\varnothing\to x\in A$, where $A$ is any set whatever, including the empty set.

Does that make things any clearer? It seems like you want a materialistic or intuitive way of seeing that $\varnothing\subseteq\varnothing$ when really the key lies in understanding what it means for one set to be the subset of another set and also what a vacuous truth is and how you can use that.

You are confusing "subset of" ( $\subseteq$ ) with "element in" ( $\in$ ).

This is possibly because you are confusing the set of elements (marbles) with the container (the bag) which holds the set.   The bag is not the set, the set is the collection held by the bag.

The set is empty, as there are no marbles held by the bag.   This empty set cannot be subdivided into smaller sets; as none is as few marbles as you can get.

Thus the empty set is the only subset of the empty set, $\varnothing \subseteq \varnothing$, but the empty set is not an element of itself. $\varnothing \notin\varnothing$.

Conversely, the set of an empty set is not an empty set. $\{\{\}\}\neq \{\}$

Your last statement "an empty bag is a bag with nothing but an empty bag inside it." is not correct.

Because, A is a subset of B does not mean that B contains the set A. It simply means that A is another bag(container) like B(which is also a container) which contains the items taken from inside the bag B.

As other posters have noted you are confusing "element of" with "subset of".

To apply your bag/marble analogy to this situation, if we say that A is a subset of a set B then we are saying the following things:

1. A and B are the same kind of thing. That is, both A and B are bags and neither of them are marbles.
2. x ∈ A implies x ∈ B. That is, all of the marbles in bag A are also to be found in bag B. For example, perhaps A contains a red marble and a blue marble, while B contains a red marble, a blue marble, and a green marble.
• In standard set theory, there is no distinction between "bags" and "marbles". Such a distinction would only make sense with ur-elements. Sep 1, 2015 at 16:08
• @ZevChonoles Okay, but in the bag analogy, this is actually a pretty reasonable distinction to make. Sep 1, 2015 at 17:00

In my opinion, OP's confusion arises from the fact that the bag analogy suggests that if bag $B$ is in bag $A$, then bag $B$ is a subset of $A$ (because all things in bag $B$ are, physically, in bag $A$ as well). This is where the bag analogy breaks down. The fact that a set $B$ is in a set $A$ does NOT imply that $B$ is a subset of $A$. For example, let $B=\{2\}$ and $A=\{B, 1\}$. Then $B\in A$ but $B\nsubseteq A$. The reason why $B\nsubseteq A$ is because there's at least one thing that is in B but is NOT in A. One such thing is $2$ (in this case, it's the only one). $A$ contains only $B$ and $1$. The fact that $B$ itself contains $2$ doesn't, in Mathematics at least, imply that $2$ is in $A$. $A$, again, contains only $B$ and $1$, not $2$.