2

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 $\tau_0 \subseteq \tau_1$?

goblin GONE
  • 67,744

1 Answers1

3

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.

yearning4pi
  • 895
  • I can add details where ever you feel you need them. – yearning4pi Mar 06 '13 at 06:37
  • the intersection of open balls can be described as the union of different open balls – Mr.Guy Mar 06 '13 at 06:51
  • @user18921 Good catch. :) – yearning4pi Mar 06 '13 at 06:54
  • 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 GONE Mar 06 '13 at 06:56
  • @user18921 Refer to my edit. – yearning4pi Mar 06 '13 at 07:16
  • @peoplepower When you say "in the topology generated by the base $\mathcal{O}$," isn't that just $\tau_1$? – goblin GONE Mar 06 '13 at 07:26
  • @user18921 That is correct, I was being clear that we are taking the elements of $\mathcal A$ to be in the topology generated $\mathcal O$ as a base, not the one generated by it as a subbase. – yearning4pi Mar 06 '13 at 07:34
  • Okay I think I kinda get the first paragraph. I still don't really understand how "It follows that $\tau_0 \subseteq \tau_1$" in the second paragraph. In particular, I'm not sure that every element of $\tau_0$ can be written as a union of a finite intersection of elements of $\mathcal{O}$. I'm not saying this is false, just that it isn't obvious to me. – goblin GONE Mar 06 '13 at 07:46
  • @user18921 "I'm not sure that every element of $\tau_0$ can be written as a union of a finite intersection of elements of $\mathcal O$." The phrasing is not correct. It is a union of a family of finite intersections of elements of $\mathcal O$; that is $A\in\tau_0$ implies there exists a set $I$, a collection ${m_i\in\mathbb N:i\in I}$, and a collection ${A_{i,j}\in\mathcal O:i\in I,0\leq j<m_i}$ such that $A=\bigcup_{i\in I}\left(\bigcap_{0\leq j<m_i}A_{i,j}\right)$. – yearning4pi Mar 06 '13 at 08:03
  • I find it hard to fit these explanations into the comments box, do you want to go to chat? – yearning4pi Mar 06 '13 at 08:04
  • Thats an interesting theorem: if you consider the sigma algebra generated by a collection, things are much more complicated. – goblin GONE Mar 06 '13 at 08:07
  • let us continue this discussion in chat – goblin GONE Mar 06 '13 at 08:07