The Union and Intersection of Two Topologies

Recall that if $X$ is a set then a topology $\tau$ is a collection of subsets of $X$ that satisfy three conditions - $\emptyset, X \in \tau$, given any arbitrary collection of subsets from $\tau$ we have that the union of this arbitrary collection is in $\tau$, and given any finite collection of subsets from $\tau$ we have that the intersection of this finite collection is in $\tau$.

Now suppose that we have two distinct topologies on $X$, say $\tau_1$ and $\tau_2$. What can we necessarily say about $\tau_1 \cup \tau_2$ and $\tau_1 \cap \tau_2$? Are these collections of subsets of $X$ necessarily topologies?

For example, consider the finite set $X = \{ a, b, c \}$ and the topologies $\tau_1 = \{ \emptyset, \{ a \}, \{ a, b \}, X \}$ and $\tau_2 = \{ \emptyset, \{ b, c \}, X \}$. The union of $\tau_1$ and $\tau_2$ is given as:

(1)

\begin{align} \quad \tau_1 \cup \tau_2 = \{ \emptyset, \{ a \}, \{ a, b \}, \{ b, c \}, X \} \end{align}

However, $\tau_1 \cup \tau_2$ is NOT a topology on $X$ since $\{a, b \} \cap \{ b, c \} = \{ b \} \not \in \tau_1 \cup \tau_2$. With this counterexample, we see that $\tau_1 \cup \tau_2$ need not be a topology.

Now, the intersection of $\tau_1$ and $\tau_2$ is given as:

(2)

\begin{align} \quad \tau_1 \cap \tau_2 = \{ \emptyset, X \} \end{align}

In this case, we see that $\tau_1 \cap \tau_2$ is merely the indiscrete topology on $X$, i.e., $\tau_1 \cap \tau_2$ is a topology. In fact, as we will see in the following theorem, if $\tau_1$ and $\tau_2$ are both topologies on $X$ then $\tau_1 \cap \tau_2$ is also a topology on $X$.

Theorem 1: Let $X$ be a set and let $\tau_1$ and $\tau_2$ both be topologies on $X$. Then $\tau_1 \cap \tau_2$ is a topology on $X$.

• Proof: To show that $\tau_1 \cap \tau_2$ is a topology on $X$ we must verify that all three conditions hold for $\tau_1 \cap \tau_2$ to be a topology.

• Since $\emptyset, X \in \tau_1$ and $\emptyset, X \in \tau_2$ by the definition of $\tau_1$ and $\tau_2$ being topologies, we see that $\emptyset, X \in \tau_1 \cap \tau_2$, so the first condition is satisfied.

• Now let $\{ U_i \}_{i \in I}$ be a collection of sets such that $U_i \in \tau_1 \cap \tau_2$ for each $i \in I$ for some index set $I$. Since each $U_i \in \tau_1 \cap \tau_2$ for each $i \in I$ we see that $U_i \in \tau_1$ and $U_i \in \tau_2$ for each $i \in I$. So $\bigcup_{i \in I} U_i \in \tau_1$ and $\bigcup_{i \in I} U_i \in \tau_2$. Hence $\bigcup_{i \in I} U_i \in \tau_1 \cap \tau_2$. Therefore, the second condition is satisfied.

• Now let $U_1, U_2, ..., U_n \in \tau_1 \cap \tau_2$. Then $U_i \in \tau_1$ and $U_i \in \tau_2$ for each $i \in \{1, 2, ..., n \}$. Therefore $\bigcup_{i=1}^{n} U_i \in \tau_1$ and $\bigcup_{i=1}^{n} U_i \in \tau_2$. So $\bigcup_{i=1}^{n} U_i \in \tau_1 \cap \tau_2$. Therefore the third condition is satisfied.

• Hence $\tau_1 \cap \tau_2$ is a topology on $X$. $\blacksquare$

Corollary 1: Let $X$ be a set and let $\tau_1, \tau_2, ..., \tau_m$ be topologies on $X$. Then $\bigcap_{i=1}^{m} \tau_i$ is a topology on $X$.

• Proof: Let $\tau_1, \tau_2, ..., \tau_m$ be topologies on $X$. By Theorem 1, we have that $\tau_1 \cap \tau_2$ is a topology on $X$. Inductively we have that the following set is a topology on $X$:

(3)

\begin{align} \quad (\tau_1 \cap \tau_2) \cap \tau_3 \end{align}

• And similarly, we have that the following set is a topology on $X$:

(4)

\begin{align} \quad (\tau_1 \cap \tau_2 \cap ... \cap \tau_{m-1}) \cap \tau_m = \bigcap_{i=1}^{m} \tau_i \quad \blacksquare \end{align}