# Subbasis elements are open?

So the basis elements of a topology are open, and the topology generated by a subbasis is the collection of union of finite intersections of subbasis elements. My question is: Can we claim that subbasis elements are open? Because if we let $U \in \mathcal{T_x}$, then $U = \bigcup({S_{i_1}\cap\ S_{i_2}\cap...\cap \ S_{i_n})}$, then $S_{i_j}\in \mathcal{T_x}$ because the union of finite intersection of $S_{i_j}$ is $U\in \mathcal{T_x}$?

• Well, $S_i$ is a union (of one set) of an intersection (of one set) of elements of the subbasis. – Daniel Fischer Feb 14 '16 at 22:02

## 1 Answer

In your definition of a subbase (or subbasis, but I prefer base and subbase), the set of finite intersections of subbase elements is a base for the topology. And base elements are by definition open (otherwise we'd call it a "network", which is a collection $\mathcal{N}$ of (arbitrary) subsets of $X$ such that we can write any open subset $O$ of $X$ as a union of members of $\mathcal{N}$). Or we could say that base elements are open, because unions of them form all open sets, including the empty union ($\emptyset$) and unions of subfamilies of size $1$ (which are exactly the elements of the base).

A finite intersection of elements of $\mathcal{S}$ is the intersection of a finite subfamily $\mathcal{S}'$ of $\mathcal{S}$. This $\mathcal{S}'$ can consist of one element, so be of the form $\{S\}$ for $S \in \mathcal{S}$. The intersection is then just $S$, so every element of $\mathcal{S}$ is itself in the set of finite intersections from $\mathcal{S}$. In particular, all $S \in \mathcal{S}$ are in the base so are open.

My definition of a subbase is just "generating set", so $\mathcal{S} \subseteq \mathcal{T}$ is a subbase for the topology $\mathcal{T}$ if the smallest topology that contains $\mathcal{S}$ as a subset equals $\mathcal{T}$. This comes down to the same description as you have (finite intersections form a base) if we adopt the convention that the intersection of the empty subfamily of $\mathcal{S}$ equals $X$ (for which there are good logical arguments, based on the Morgan's laws etc.), so that the finite intersections always form a base for a topology. In this definition $\mathcal{S}$ elements are also open (but then basically (pun intended) by definition).