# If a collection of sets is a subbase for a topology $\tau_0$ and a base for a topology $\tau_1$, can we conclude $\tau_0 = \tau_1$?

Let $X$ denote a set, and let $\mathcal{O}$ denote a collection of subsets of $X$.

Then $\mathcal{O}$ is necessarily a subbase of a unique topology, call it $\tau_0$.

And it may or may not be the case that $\mathcal{O}$ is the base of a topology, in which case this topology is unique. Suppose $\mathcal{O}$ indeed the base of a topology, call it $\tau_1$.

Does $\tau_0$ necessarily equal $\tau_1$?

Edit: Since $\tau_1$ is built using arbitrary unions only, while $\tau_0$ is built using not only arbitrary unions but also finite intersections, it follows that $\tau_1 \subseteq \tau_0$. The question is, is $\tau_0 \subseteq \tau_1$?

The assumption that $\mathcal O$ is a base means that any finite intersection of elements of $\mathcal O$ is a union of elements of $\mathcal O$. For, let $\mathcal A$ be a finite family of elements of $\mathcal O$. Regard the elements of $\mathcal A$ as open sets in the topology generated by the base $\mathcal O$ to see that their intersection is also open.
Thus, an arbitrary union of finite intersections of elements of $\mathcal O$ can be rewritten, by reindexing, as an arbitrary union of elements of $\mathcal O$. It follows that $\tau_0\subseteq \tau_1$.
As you aptly noted $\tau_1\subseteq \tau_0$, so we are done.
• Thanks; it seems more reasonable now. How does one show that if $\mathcal{O}$ is a base and $A_0,...,A_n \in \mathcal{O}$, then there exists $\mathcal{A} \subseteq \mathcal{O}$ such that $A_0 \cap \cdots \cap A_n = \bigcup \mathcal{A}$? – goblin Mar 6 '13 at 6:56