Mathematics · Topology

The Intersection of Topologies result in a Topology

Topology is the study of invariant properties of mathematical objects. Through the study of topology, mathematicians to translate properties between mathematical objects. There are many different subfields of topology, e.g. general topology (a.k.a. point-set topology), algebraic topology, etc. Unfortunately (or fortunately?), I have only encountered general topology in my undergraduate degree.

Perhaps one interesting (basic) result in topology is as follows

Given a space X and a collection of topologies \tau_i, where i \in I and I is an index set, the intersection \bigcap_{i \in I} \tau_i is also a topology on X.

Given a collection of objects that share property a property, let’s say property a, we do not expect the union and intersection of this collection to share the same properties. For instance, the set \{n\}, where n is an integer, is finite. But when you take the union of all singleton set (namely \{n\}), we have an infinite set.

So why is the above statement true? To prove this, we need need at least the definition of a topology, so here it goes:

Let X be a set, and \mathcal{P}(X) be its power set. A subset \tau \subseteq \mathcal{P}(X) is called a topology if the following conditions hold:

  1. \emptyset, X \in \tau
  2. Arbitrary union of elements of \tau is also an element of \tau, i.e. \bigcup_{\lambda \in \Lambda} U_{\lambda} \in \tau where U_{\lambda} \in \tau is arbitrarily chosen.
  3. Finite intersection of elements of \tau is also an element of \tau, i.e. if n \in \mathbb{Z}, then \bigcap_{i=1}^n U_{i} \in \tau where U_{i} \in \tau is arbitrarily chosen

So far so good. Now that we know the definition of a topology, we can easily see that the union of topologies is NOT a topology; consider X = \{a,b,c\}, and we have two topologies \tau_1 = \{\emptyset, \{a\}, X\} and \tau_2 = \{\emptyset, \{b\}, X\}, then the union is

\tau_1 \cup \tau_2 = \{\emptyset, \{a\},\{b\},X\}

This violates the arbitrary union property; \{a\} and \{b\} are elements of the union, but \{a\} \cup \{b\} = \{a,b\} is not. Note that this is not to say that all unions of topologies does not result in a topology; it is only the counterexample to the claim that all unions of topologies gives a topology.

We have shown above that the union of topologies itself need not be a topology. So why is the intersection of of topologies is itself a topology? Clearly, by definition of topology, each topology includes the sets \emptyset and X, thus the empty set \emptyset and X is in the intersection. Now let I be an index set, and for i \in I, U_{i} \in \bigcap_{\lambda \in \Lambda} \tau_{\lambda} be an arbitrary family of elements in the intersection of the topologies. Then by definition of topology, arbitrary unions of elements in \{U_i\}_{i \in I} is also in \tau_{\lambda} for all \lambda \in \Lambda. Thus the arbitrary union is also an element of \bigcap_{\lambda \in \Lambda} \tau_{\lambda}. By similar reasoning, we can also conclude that finite intersection of elements in \bigcap_{\lambda \in \Lambda} \tau_{\lambda} is also an element of \bigcap_{\lambda \in \Lambda} \tau_{\lambda} itself. Thus \bigcap_{\lambda \in \Lambda} \tau_{\lambda} is a topology.

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