# Why must a topology on a set contain the empty set?

I have just had my first week of topology, and I have a question that is rather basic.

Why must the empty set be an element of any given topology?

(For reference, the definition of a topology T I am working with, for a set X:

1. X and the empty set must be in T.
2. the union of elements of any subset in T, is also in T.
3. the intersection of elements of any finite subcollection in T, is also in T.)

• If a finite intersection of open sets is open, what if the open sets in question are disjoint? Dec 12, 2015 at 21:46
• It would be rather awkward if we had to say "the intersection of finitely many elements of $T$ is in $T$ provided that the intersection is not empty", etc. Also, if open sets are the complement of closed sets, then we would not be able to declare $X$ closed if $\emptyset$ was not open.
– user169852
Dec 12, 2015 at 21:46
• @user4894 what if there are no disjoint open sets, e.g. if $X$ is the only open set. I think that it is the union (rather than intersection) that makes the empty set necessary. Every member of the empty family of sets is open, hence its union (the empty set) must be open too. Dec 12, 2015 at 22:13
• @Mirko There is a great counterexample to both our ideas called the particular point topology. en.wikipedia.org/wiki/Particular_point_topology Consider $\mathbb{R}$. Let $T = \{A \subset \mathbb{R} : 0 \in A\}$. $T$ is closed under intersections (arbitrary in fact) and unions, and $\mathbb{R} \in T$. To make it into a topology we are forced to arbitrarily add $\emptyset$ to $T$. So the empty set axiom is necessary. I don't see how unions would help. Dec 12, 2015 at 22:29
• @user4894 $T$ is not closed under unions, as the empty set which is the union of the empty family of open sets does not belong to $T$. Dec 12, 2015 at 22:32

It's more convenient this way. $\def\less{\smallsetminus}$

Usually in math when you have definitions that involve the empty set, the reason they do is that it's more elegant, convenient, economical to phrase things this way.

Suppose you didn't want to deal with the empty set. You could define topology to be a collection of nonempty subsets of X that satisfy a few axioms. But then your axioms would have to be more complicated and include special cases.

You'd have to say that only a nonempty finite intersection of open sets is open. You could modify the 3rd axiom to say that. But the reason you need that axiom in the first place is that very frequently in proofs you take a few open sets, and rely on the fact that their intersection is still open (and therefore has some qualities useful to you). In the new version, every time you need to intersect two open sets in a proof, you'd have to list two cases: a) if they're disjoint, then their intersection is empty and therefore... b) if they're not disjoint, then their intersection is open and therefore... Your proofs would be longer and more tedious.

Similarly, normally you define a subset $Y$ of $X$ to be closed if $X \less Y$ is open. Then under the usual definition the whole space $X$ is a closed set. There're many reasons why you want that to be the case and why that corresponds to the intuition of "closed set". But if you don't consider the empty set to be open by definition, you don't get $X$ to be closed. Again, you can patch it up by modifying the definition of closed set: "A subset $Y$ of $X$ is closed if $Y=X$ or $Y$ is the complement of an open set". This will work. But, again, your proofs will become more complicated: when you have a closed set $Y$ about which you know little, you can't just assume anymore that $X \less Y$ is open and work with that, now you have to consider two cases.

These are just two examples, but there're more. Basically, it turns out that while making the empty set open by definition may not look a priori very natural to you, it makes all kinds of objects and properties "click together" more naturally and economically than otherwise.

• I'm reminded of the Q of why 1 is not a prime :Convenience.We need a short name for positive indecomposable non-invertible integer, and prime (Latin primus : first or foremost) is the chosen word. Dec 12, 2015 at 22:31
• The question was “why do we require the empty set”, not “why do we allow the empty set”. Not requiring the empty set needn't imply a change in the third axiom, it's just that usually the empty set will be required because of the third axiom and a pair of disjoint open sets. Dec 13, 2015 at 17:51

Note that the fact that $X$ and $\emptyset$ are open can be seen as a consequence of the two other axioms if one allows the empty intersection and the empty union, where $X$ is the empty intersection (that is, the intersection over the empty family) and $\emptyset$ is the empty union (that is, the union over the empty family) .

Seeing things this way is useful not only in this aspect. For instance, every collection of subsets of a given set is a subbasis for some topology (in the sense that the unions of finite intersections is a topology) if we take this approach, instead of having to require that the collection covers the space.

One reason the empty set is open is due to vacuous truth, which is a reason that has nothing to do with topology at all, but rather logic.

The implication $p \implies q$ is always true if the statement $p$ is false (regardless of statement $q$ -- statement $q$ can be any random statement). This is called vacuous truth.

Now, one definition or characterization of openness is that $A \subseteq X$ is open if for each $x \in A$, we can find an open neighborhood $U$ with $x \in U$ and $U \subseteq A$.

In other words, $A \subseteq X$ is open if for every $x$, the statement $$x \in A \implies \exists U \text{ open with } x \in U \subseteq A$$ is true.

Let's check the truth value of this statement if $A = \emptyset$. For $\emptyset$ to be open, we need for every $x$ that the following statement is true: $$x \in \emptyset \implies \exists U \text{ open with } x\in U \subseteq \emptyset$$

Now, for every $x$, the "if" statement in the implication above, which is the statement $x \in \emptyset$, is always false. There is absolutely no $x$ satisfying $x \in \emptyset$, so $x \in \emptyset$ is false. But if $p$ is false, then for any statement $q$, $p \implies q$ is true (this is vacuous truth). Thus for every $x$, the above implication holds, so by definition of open, $\emptyset$ is open.

Thus, to recap, the empty set $\emptyset$ is open because it satisfies the definition of being open. But it satisfies this definition vacuously.

So, in short, one reason $\emptyset$ is open has nothing to do with topology at all, but with logic and why we take the statement $p \implies q$ to be true no matter what statement $q$ is as long as statement $p$ is false.

(Extra non-topological example of vacuous truth: "If $1 + 2 = 20$, then $5 = 16$" is a true statement (vacuously) since the "if" part is false.)

• Good explanation. Also, the empty set is open because it contains none of its boundary points.
– bof
Dec 13, 2015 at 10:51

Two important concepts in topology are interior and closure of a set $A\subseteq X$. (Interior is the largest open subset, closure is the smallest closed super-set.) If $\emptyset$ would not be open, then interiors and closures would be undefined in general. Indeed, if the full space $X$ is not closed (in itself), what would be the closure $\overline{X}$?

Topological open sets are abstractions of open sets in metric spaces. Recall, an open set in a metric space $(X,d)$ is a set $U$ such that for all $u\in U$ there exists an open neighborhood of $u$ contained in $U$.

It is the "for all" in the definition that compels us to include the empty set. This is because of the excluded middle in mathematical logic, i.e. a statement is true or false. If the empty set is not an open set it means there exists an element in the empty set that serves as a counter-example to the definition. Since no elements exist in the empty set, all elements have the property of an open set.

Actually, the following definition also "works", perhaps justifying your doubts.

Let $X$ be a set and $T\subseteq\mathcal P(X)$. Then $T$ is called a topology on $X$ if

1. For any $U,V\in T$ we have $U\cap V\in T$
2. If $S\subseteq T$ then $\bigcup S\in T$
3. $X\in T$

Note that I used only intersection of two open sets, not finitely many. Of course the claim for the intersection of $n>2$ open sets follows from the above by induction. The fact that $\emptyset\in T$ is a conseqeunce of 2 with $S=\emptyset$. So we necessariy have $\emptyset \in T$, wbut we do not need to require it explicitly (given suitable axioms). So "my" axioms are equivalnt to "yours", but allow simplified checking if a certain $T$ is a topology, while "yours" allow simplified application, one might say.

Contrary to some other comments, I believe that it is union (rather than intersection) that makes the empty set necessary. Indeed we could conceivably define a topology on a (non-empty) $X$ such that $X$ is the only open set. Then there are no disjoint open sets and we could not get the empty set as the intersection of open sets, so no contradiction or any problem with intersections.

On the other hand the requirement that the union of every family of open sets is open, is essential in topology. Now, every member of the empty family of sets is open, hence its union (the empty set) must be open too.

It is also often a convention that the intersection of the empty family is $X$ (the whole space), but again, considering the empty set, I think we get that as the union of the empty family.

One must admit though, that for all practical purposes, the empty set is often the result of the intersection of two disjoint non-empty open sets in some interesting and natural topology, so this is a reason on its own why the empty set should be considered open.

• It's not necessary, just more convenient. Just as for intersections, you could just rewrite the requirement, in this case to "every nonempty family of open sets has an open union". That's not inconsistent, just more complicated. Dec 13, 2015 at 9:03

A reason why I think we should consider $\emptyset$ open is so that the whole space $X$ is closed.

Note that in most topological spaces, every element of the space $X$ is the limit point of set $X$. Hence we should consider whole $X$ to be closed.