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.

  • $\begingroup$ I can add details where ever you feel you need them. $\endgroup$ – peoplepower Mar 6 '13 at 6:37
  • $\begingroup$ the intersection of open balls can be described as the union of different open balls $\endgroup$ – Mr.Guy Mar 6 '13 at 6:51
  • $\begingroup$ @user18921 Good catch. :) $\endgroup$ – peoplepower Mar 6 '13 at 6:54
  • $\begingroup$ 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 $A_0 \cap \cdots \cap A_n = \bigcup \mathcal{A}$? $\endgroup$ – goblin Mar 6 '13 at 6:56
  • $\begingroup$ @user18921 Refer to my edit. $\endgroup$ – peoplepower Mar 6 '13 at 7:16

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