# Is the word “empty” in set theory different from the word “empty” in ordinary language? [duplicate]

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I skimmed over this question Is the empty set a subset of itself?, and I'm currently under the impression that it's a widespread belief that the empty set contains itself. But a contradiction seems to arise: if the empty set contains itself, then it's NOT empty. After all, if we call a set non-empty just because it contains the empty set, then why should we treat the empty set itself differently?

Then it occurred to me that maybe mathematicians define "empty" differently in set theory. Maybe by "empty set" they mean a "set that contains only itself", instead of a "set that contains absolutely nothing".

Is my surmise correct?

## marked as duplicate by Christoph, Jendrik Stelzner, José Carlos Santos, Strants, Lord Shark the UnknownSep 10 '18 at 16:09

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• "Contains" is an ambiguous word that people should stop using with regard to sets. – Malice Vidrine Sep 10 '18 at 10:48
• "contained" $\in$ is different (in set theory) from "contained" $\subseteq$. – Mauro ALLEGRANZA Sep 10 '18 at 11:17
• @MauroALLEGRANZA If the only type of objects that we can have is sets, then what is the difference? Every element would be a subset. – user161005 Sep 10 '18 at 11:21
• NO. The basic relation between sets is : $A \in B$. Empty set is defined as the set satisfying the formula : $\exists E \ \forall x \ \lnot (x \in E)$. – Mauro ALLEGRANZA Sep 10 '18 at 11:23
• Subset is defined as : $A \subseteq B \leftrightarrow \forall x \ (x \in A \to x \in B)$. – Mauro ALLEGRANZA Sep 10 '18 at 11:24

## 3 Answers

I think you are confusing two possible meanings of "contains". Often we say $A$ contains $B$ to mean $B$ is an element of $A$ (but see below). In this sense, $\varnothing$ doesn't contain itself, because it has no elements. (Also, no set is an element of itself.)

However, $\varnothing$ is a subset of itself, because it doesn't contain any elements outside itself. By the same logic, every set is a subset of itself.

To see the distinction in non-empty sets, $\{1\}$ is a subset of $\{1,2\}$ but $\{1,2\}$ does not contain $\{1\}$. It does contain $1$, but that is not the same thing.

On the usage of "contains", wikipedia says:

The expressions "A includes x" and "A contains x" are also used to mean set membership, however some authors use them to mean instead "x is a subset of A". Logician George Boolos strongly urged that "contains" be used for membership only and "includes" for the subset relation only.

• "It does contain 1, but that is not the same thing." I wonder, how it could look like to a set contain set {1} as element? And would be it possible for {1} to be an element without being a subset? – user161005 Sep 10 '18 at 11:36
• Sure, the power set (set of all subsets) of $\{1,2\}$ is $P=\{\{1,2\},\{1\},\{2\},\varnothing\}$. So $\{1\}\in P$, but $1\not\in P$, meaning that $\{1\}\not\subset P$. – Especially Lime Sep 10 '18 at 11:47
• The opening paragraph has the contains relationship backwards. – jaxad0127 Sep 10 '18 at 12:02
• @user161005: By definition something is an element of $\{1,2\}$ if and only if it equals $1$ or it equals $2$. We cannot have $\{1\}=1$ due to the Axiom of Regularity. Whether $\{1\}=2$ depends on your definition of "$2$". One convention that has been proposed in the past is that $2$ does indeed mean $\{1\}$, but for the last many decades it has been nearly universal to define $2$ to mean $\{0,1\}$. In that case $\{1\}=2$ is false because $0\in 2$ but $0\notin\{1\}$. Therefore, $\{1\}\notin\{1,2\}$. – Henning Makholm Sep 10 '18 at 13:42
• @HenningMakholm I feel like the whole "numbers are really sets" thing makes everything even more confusing for beginners asking questions like "why is $\emptyset \subseteq \emptyset$?" – Vincent Sep 10 '18 at 15:34

The problem here is you are using the word "contains" for two different things:

• When $x$ is an element of a set $A$, we write $x\in A$ and sometimes say "$A$ contains $x$".
• When $B$ is a subset of $A$, we write $B\subseteq A$ and also sometimes say "$A$ contains $B$".

This is a common problem and usually we can determine from the context whether which of the two types of containment is being referred to.

With the empty set, it is always false that $x\in\varnothing$, i.e., the empty set does not have elements, it is empty.

However, it has a subset: $\varnothing\subseteq\varnothing$ is true. Why is that? Well, $B\subseteq A$ means that whenever $x$ is an element of $B$ it also is an element of $A$, or in other words, $A$ contains (as elements) at least all elements of $B$. In particular $\varnothing\subseteq X$ is true for any set $X$, since there are no elements in $\varnothing$ that have to be contained in $X$ at all.

$A\subseteq B$ is not the same as $A\in B$. The former one says that all elements of $A$ are also in $B$. Whereas the latter one means the set $A$ is an element of $B$. So $B=\{\,A,\dots\,\}$.

Therefore it is correct to say $\emptyset\subseteq\emptyset$ but not $\emptyset\in\emptyset$ since the empty set has no elements.

• Along with the ambiguity of "contains", I really wish people would stop using $\subset$ to mean $\subseteq$. Nobody would ever write $<$ to mean $\leq$ so why do we do it with sets? – David Richerby Sep 10 '18 at 12:09
• @DavidRicherby Thank you. I have changed it. In university we have used $\subset$ and $\subsetneq$ so far. – EuklidAlexandria Sep 10 '18 at 12:30