Why is this definition of topological space equivalent to the common one?

Question

While comparing the definition of a topological space among books, I found one book [1] whose definition seems to differ from the others. Here it is:

Definition: Let $X$ be a non-empty set. A class (defined by the book as a "set of sets") $T$ of subsets of $X$ is called a topology on $X$ if it satisfies the following two conditions:

1. The union of every class of sets in $T$ is a set in $T$
2. The intersection of every finite class of sets in $T$ is a set in $T$

This definition omits the usually present condition that the empty set and $X$ itself must be in $T$. It also adds the condition that $X$ is not empty. Shortly after this definition, the author writes:

We observe that the empty set and the full space are always open sets in every topological spaces, since they are the union and intersection of the empty class of sets, which is a subclass of every topology.

Which doesn't make sense to me, given the definition.

1. <deleted>
2. Why is $X$ in $T$?
3. Why does $X$ need to be non-empty?

Update (Much later):

I discovered that the book adopts a convention by which all sets under discussion are restricted to $X$. Not realizing/mentioning that was entirely my fault.

[1] George F. Simmons - Introduction to Topology and Modern Analysis, pp. 92

Answer 1

When dealing with collections of subsets of a common superset $\Omega$, intersections and unions are usually defined as follows:

Let $\mathcal B$ be a collection of subsets of $\Omega$, then \begin{align} \bigcup \mathcal B &:= \bigcup_{A\in\mathcal B} A := \left\{\, x\in\Omega \,\middle|\, \exists A\in\mathcal B\colon x\in A \,\right\} \subseteq \Omega, \\ \bigcap \mathcal B &:= \bigcap_{A\in\mathcal B} A := \left\{\, x\in\Omega \,\middle|\, \forall A\in\mathcal B\colon x\in A \,\right\} \subseteq \Omega. \end{align} For $\mathcal B=\varnothing$ this yields $\bigcup \mathcal B = \varnothing$ and $\bigcap \mathcal B = \Omega$.

For $\mathcal B\neq\varnothing$ the definitions above don't depend on $\Omega$ and agree with the usual intersections and unions of collections of sets. Even the union $\bigcup \varnothing$ doesn't depend on $\Omega$, the only problem is the empty intersection $\bigcap \varnothing=\Omega$, which is a very useful definition in the context of a common superset.

You can think of $\varnothing$ as the identity element with respect to taking unions and of $\Omega$ as the identity element with respect to taking intersections of subsets of $\Omega$. This makes the empty union and empty intersection analogous to the empty sum $\sum_{a\in\varnothing} a = 0$ and the empty product $\prod_{a\in\varnothing} a = 1$, which are the identity elements of addition and multiplication, respectively.

In the case of a topology, $\Omega=X$ and $\tau$ is a collection of subsets of $X$. Now $\varnothing\subseteq \tau$ is vacuously true, since every element of $\varnothing$ is an element of $\tau$: there is no element in $\varnothing$ that could serve as a counter example. Hence, the two axioms imply $\bigcup\varnothing = \varnothing\in\tau$ and $\bigcap \varnothing= X\in\tau$.

Requiring $X$ to be non-empty is indeed unusual, maybe the author was too lazy to exclude the empty space from some of the theorems.

Answer 2

Thanks for all the comments. I've tried (Edit: twice now) to collect them all into a correct, full answer.

Let $(X,T)$ be a topological space. i.e. $T$ is a sets of subsets of $X$. Then the conditions can be written equivalently as:

• (c1-alt): The union of every subclass of T is a set in T.
• (c2-alt): The intersection of every finite subclass of T is a set in T.

Then

• (a1): The empty class is a subclass of T because it is (vacuously) true that every set in it belongs to T. Therefore condition (c1-alt) applies and thus the union of the empty class, which is the empty set, is in T.

• (a2): Updated: (Thanks to helpful comments to the OP).

In axiomatic set theory, intersection of class is defined only for non-empty classes [1] [2]. So when the book suggests that "the intersection of the empty class is the entire class", there appears to be a problem.

However, in the first chapter the author writes (which I overlooked, to my detriment) that

There are certain logical difficulties which arise in the foundations of the theory of sets (i.e. russel's paradox).

We avoid these difficulties by assuming that each discussion in which a number of sets are involved takes place in the context of a single fixed set. This set is called the universal set. it is denoted by U in this section and the next, and every set mentioned is assumed to consist of elements in U.

Having adopted this convention, the definition of class intersection now becomes $$\bigcap S=\{x\in X \ \vert\ \forall A \in S: x \in A\}. $$ That $X \in T$ then follows from this definition since the predicate is vacuously true, and set-theoretic landmines are avoided since we're not in danger of "constructing" the set of all sets.

1. (a3): The author chose to exclude a trivial case from consideration. That's ok for this specific book, but doesn't agree with the "common" definition which includes the case of $X=\emptyset$.

[1] Paul Halmos, Naive set theory, Springer-Verlag 1974, pp15.

[2] Wikipedia:Axiom_of_union#Relation_to_Intersection

Answer 3

1. I think the trick here is how the author talks about classes of sets. Since the union of every class of sets must be in $T$, then the union of an empty class of sets must be in $T$; therefore, $\emptyset \in T$.
2. My initial thought was that the union of all sets in $T$ must be $X$, therefore $X \in T$. But if you consider $T = \{\emptyset, \{1\}, \{1, 2\}\}$, then the union of all open sets that are not $X$ is not $X$.
3. In the standard definition of a topology we make sure that $\emptyset, X \in T$, so we already know it's not empty. Here we need to make it clear.