Existence of a Topology with a Base

Theorem: Existence of a Topology with a Base

Let $X$ be a non-empty set and let $\mathcal{B}$ be a collection of subsets of $X$.

There exists a topology ${\tau }_{\mathcal{B}}$ on $X$ such that $\mathcal{B}$ is a base for the topological space $(X,{\tau }_{\mathcal{B}})$ if and only if

• $X$ is the union of $\mathcal{B}$, i.e. $X=\bigcup \mathcal{B}$, and

• for each $B_1,B_2\in \mathcal{B}$, the intersection $B_1\cap B_2$ is a union of a subset of $\mathcal{B}$.

PROOF

We need to prove the following things separately:

• If there exists a topology ${\tau }_{\mathcal{B}}$ on $X$ such that $\mathcal{B}$ is a base for the topological space $(X,{\tau }_{\mathcal{B}})$, then $X=\bigcup \mathcal{B}$ and for each $B_1,B_2\in \mathcal{B}$, the intersection $B_1\cap B_2$ is a union of a subset of $\mathcal{B}$.
• If $\mathcal{B}$ is such that $X=\bigcup \mathcal{B}$ and for each $B_1,B_2\in \mathcal{B}$, the intersection $B_1\cap B_2$ is a union of a subset of $\mathcal{B}$, then there exists a topology ${\tau }_{\mathcal{B}}$ on $X$ such that $\mathcal{B}$ is a base for the topological space $(X,{\tau }_{\mathcal{B}})$.

Part 1:

Suppose that there exists a topology ${\tau }_{\mathcal{B}}$ on $X$ such that $\mathcal{B}$ is a base for the topological space $(X,{\tau }_{\mathcal{B}})$.

By the definition of a topology, we have that $X$ is an open set. Since $\mathcal{B}$ is a base for ${\tau }_{\mathcal{B}}$, the set $X$ can be expressed as a union of a subset of $\mathcal{B}$. In particular, since $\mathcal{B}$ is comprised only of subsets of $X$, then $X=\bigcup \mathcal{B}$. Similarly, since $\mathcal{B}$ contains only open sets by definition, the intersection $B_1\cap B_2$ is in ${\tau }_{\mathcal{B}}$, which completes the proof, since every open set can be expressed as a union of elements of $\mathcal{B}$ by definition.

Part 2:

Suppose $\mathcal{B}$ is such that $X=\bigcup \mathcal{B}$ and for each $B_1,B_2\in \mathcal{B}$, the intersection $B_1\cap B_2$ is a union of a subset of $\mathcal{B}$. Define ${\tau }_{\mathcal{B}}$ as the collection of all subsets of $X$ which are unions of subsets of $\mathcal{B}$.

We need to show that ${\tau }_{\mathcal{B}}$ is a topology on $X$. To do this we need to show the following three things:

• $X\in {\tau }_{\mathcal{B}}$ and $\varnothing \in {\tau }_{\mathcal{B}}$;
• ${\tau }_{\mathcal{B}}$ is closed under arbitrary unions;
• The intersection $U_1\cap U_2$ of each $U_1,U_2\in {\tau }_{\mathcal{B}}$ is in ${\tau }_{\mathcal{B}}$.

The assumption that $X=\bigcup \mathcal{B}$ entails that $X\in {\tau }_{\mathcal{B}}$. The empty set is a subset of $\mathcal{B}$ and since $\varnothing =\bigcup \varnothing$, we have that $\varnothing \in {\tau }_{\mathcal{B}}$.

Consider now an arbitrary subset $T$ of ${\tau }_{\mathcal{B}}$. Since each $U\in T$ is a union of a subset of $\mathcal{B}$, then $\bigcup T$ is also a union of a subset $\mathcal{B}$, meaning that $\bigcup T\in {\tau }_{\mathcal{B}}$.

Finally, we consider two sets $U_1,U_2\in {\tau }_{\mathcal{B}}$. By definition of ${\tau }_{\mathcal{B}}$, there exists some subset $\{B_i{\}}_{i\in I}\subseteq \mathcal{B}$ such that $U_1={\bigcup }_{i\in I}{B}_i$. Similarly, $U_2={\bigcup }_{j\in J}{B}_j$. We thus have

$U_1\cap U_2=\left({\bigcup }_{i\in I}{B}_i\right)\cap \left({\bigcup }_{j\in J}{B}_j\right)={\bigcup }_{i\in i,j\in J}(B_i\cap B_j)$

By assumption, for all $i\in I$ and $j\in J$, the intersection $(B_i\cap B_j)$ is a union of a subset of $\mathcal{B}$. Therefore, the union ${\bigcup }_{i\in i,j\in J}(B_i\cap B_j)$ is also a union of a subset of $\mathcal{B}$ and is thus in ${\tau }_{\mathcal{B}}$, Q.E.D.