# How do I formalize the topology generated by a subbasis?

The topology generated by a subbasis $\mathcal{S}$ is defined as the colection $\tau$ of all unions of finite intersections of elements of $\mathcal{S}$. I want to formalize $\tau$ as something similar like

$$\tau = \left\{ \bigcup_{ B \in \mathcal{C}}B \mid \mathcal{C} \subset \mathcal{B} \right\}$$ while in this case, $\tau$ would be the topology generated by the basis $\mathcal{B}$. I tried something like

$$\tau=\left\{\bigcup_{\mathcal{C}\subset \mathcal{S}}\left( \bigcap_{i=1}^mS_i \right)\mid \mathcal{C} \subset \mathcal{S}\right\}$$ but I don't know if it makes much sense. Any help will be higly appreciate.

• As it can be awkward to formalise, it's sometimes handy to use that the topology generated by $S$ as a subbase is the same as $\mathcal{T}_\mathcal{S}=\bigcap\limits_{\mathcal{S}\subset\mathcal{T}} \mathcal{T}$. The more practical meaning is of course what you're after, but this is the easier one to formalize. – snulty Jun 27 '16 at 19:26

Your second expression doesn’t actually work: it doesn’t say that the sets $S_i$ belong to $\mathscr{S}$ or that you’re taking only some of the possible intersections of finite subsets of $\mathscr{C}$. It really is much easier to define $\mathscr{B}$ in terms of $\mathscr{S}$ and then define $\tau$ in terms of $\mathscr{B}$. If you really want to define $\tau$ directly in terms of $\mathscr{S}$, one way is:

$$\tau=\left\{U\subseteq X:\forall x\in U\,\exists\mathscr{F}\subseteq\mathscr{S}\left(\mathscr{F}\text{ is finite and }x\in\bigcap\mathscr{F}\subseteq U\right)\right\}\;.$$

Another is

$$\tau=\left\{\bigcup\mathscr{U}:\forall U\in\mathscr{U}\,\exists\mathscr{F}\subseteq\mathscr{S}\left(\mathscr{F}\text{ is finite and }U=\bigcap\mathscr{F}\right)\right\}\;.$$

Yet another is

$$\tau=\left\{\bigcup\mathscr{U}:\mathscr{U}\subseteq\left\{\bigcap\mathscr{F}:\mathscr{F}\in[\mathscr{S}]^{<\omega}\right\}\right\}\;;$$

here the notation $[A]^{<\omega}$ is one standard notation for the set of finite subsets of $A$.