# Generating Topologies from a Collection of Subsets of a Set

So far we have described all of the topologies we have looked at somewhat explicitly in that we describe what exactly the open sets for the topology are. Sometimes this is not that easy or convenient. Instead, sometimes it is easier to describe a topology in terms of a basis.

On the A Sufficient Condition for a Collection of Sets to be a Basis of a Topology page we saw that if $\tau$ is a topology on $X$ then we can verify whether or not $\mathcal B$ is a basis of $\tau$ if for every $U \in \tau$ and for every $x \in U$ there exists a $B \in \mathcal B$ such that $x \in B \subseteq U$.

What if we don't know what $\tau$ is though? Then $\mathcal B$ is just a collection of subsets of $X$ and the collection may form a basis for SOME topology on $X$ or may form a basis for no topology on $X$.

In the following theorem, we will see that if the collection of sets $\mathcal B$ satisfies certain conditions then we can guarantee that $\mathcal B$ is a basis for *SOME* topology on $X$!

Theorem 1: Let $X$ be a set. A collection of sets $\mathcal B$ is a basis for SOME topology $\tau$ if and only if the following two conditions hold:1) $X = \bigcup_{B \in \mathcal B} B$.2) For every pair $B_1, B_2 \in \mathcal B$ and for every $x \in B_1 \cap B_2$ there exists a $B \in \mathcal B$ such that $x \in B \subseteq B_1 \cap B_2$. |

- $\Rightarrow$ Suppose that $\mathcal B$ is a basis of $\tau$. Then for all $U \in \tau$ there exists a $\mathcal B^* \subseteq \mathcal B$ such that $U = \bigcup_{B \in \mathcal B^*} B$.

- Since $X \in \tau$ (by the definition of $\tau$ being a topology) we therefore have that:

- So the first condition is satisfied. Now, let $B_1, B_2 \in \mathcal B$. Then $B_1, B_2 \in \tau$. So $B_1 \cap B_2 \in \tau$ (since the intersection of a finite collection of open sets is open by the definition of a topology). Let $U = B_1 \cap B_2 \in \tau$. By the theorem on the A Sufficient Condition for a Collection of Sets to be a Basis of a Topology page, then for all $x \in U = B_1 \cap B_2$ there exists a $\mathcal B \in U = B_1 \cap B_2$ such that:

- Therefore the second condition is satisfied. $\blacksquare$

- $\Leftarrow$ Let $\mathcal B$ be a collection of sets from $X$ such that conditions (1) and (2) hold. We need to show that there exists a topology $\tau$ such that $\mathcal B$ is a basis of $\tau$. Define $\tau$ to be the collection of all unions of the sets $B \in \mathcal B$:

- We will prove that $\tau$ is indeed a topology.

- Clearly $X \in \tau$ by (1), and also, $\emptyset \in \tau$ if we take an empty union of elements from $\mathcal B$.

- Let $\{ U_i \}_{i \in I}$ be an arbitrary collection of sets $U_i \in \tau$ for each $i \in I$. But $\tau$ is defined to be the collection of all unions of the sets $\mathcal B$, i.e., for each $U_i$ there exists a $\mathcal B_i \subseteq \mathcal B$ such that $U_i = \bigcup_{B \in \mathcal B_i} B$. So:

- But then $\bigcup_{i \in I} U_i$ is simply a union of sets in $\mathcal B$. So $\bigcup_{i \in I} U_i \in \tau$.

- Lastly, let $U_1, U_2, ..., U_n \in \tau$ be a finite collection of sets in $\tau$. We will inductively show that $\bigcap_{i=1}^{n} U_i \in \tau$.

- First consider the union $U_1 \cap U_2$. Since $U_1, U_2 \in \tau$ there exists a $\mathcal B_1, \mathcal B_2 \subseteq \mathcal B$ such that $U_1 = \bigcup_{B \in \mathcal B_1} B$ and $U_2 = \bigcup_{B \in \mathcal B_2} B$. So:

- Let $B_1 = \bigcup_{B \in \mathcal B_1} B$ and $B_2 = \bigcup_{B \in \mathcal B_2} B$. Then by (2), for every $x \in B_1 \cap B_2$ there exists a $B_x \in \mathcal B$ such that $x \in B_x \subseteq B_1 \cap B_2$. So for every $x \in U_1 \cap U_2$ there exists a $B_x \in \mathcal B$ such that $x \in B_x \subseteq B_1 \cap B_2$. So:

- So $U_1 \cap U_2$ is equal to the union of elements from $\mathcal B$, so $U_1 \cap U_2 \in \tau$. By the same argument, $(U_1 \cap U_2) \cap U_3 \in \tau$, …, $(U_1 \cap U_2 \cap ... \cap U_{n-1}) \cap U_n \in \tau$ so $\bigcap_{i=1}^{n} U_i \in \tau$.

- Therefore $\tau$ is a topology on $X$ whose basis is $\mathcal B$. $\blacksquare$