A basis has to fit inside every overlap. Drop that demand and ask only that the family cover the space: take all the finite intersections first, and a basis appears. Such a family is called a subbasis, and it is how a topology is described when only a few conditions matter.

Two rules, "after nine" and "before five", cut the day at one end each. Combining them gives the working day, and the pair of half-days is a cheaper description than the interval it produces.

Two rays and what they have in common. $(-\infty,b)$ and $(a,\infty)$ are the members of the subbasis; their intersection is the interval $(a,b)$, and intervals are a basis. So the rays alone generate the standard topology of the line.

A subbasis for a topology on $X$ is any collection $\mathcal{S}$ of subsets whose union is $X$. The finite intersections of members of $\mathcal{S}$ form a basis, and the topology that basis generates is called the topology generated by $\mathcal{S}$ — the coarsest one in which every member of $\mathcal{S}$ is open.

• No overlap condition is needed: taking finite intersections first makes the second condition hold automatically, since an intersection of two finite intersections is a finite intersection.
• The rays $(-\infty,b)$ and $(a,\infty)$ form a subbasis for the standard topology on $\mathbb{R}$, because $(a,b)$ is the intersection of two of them.
• A subbasis is the natural way to define a topology by demanding that certain sets be open and nothing more. Chapter 5 defines the product topology exactly so, from the preimages of open sets under the projections.
• Every basis is a subbasis; the converse fails, since a subbasis need not fit inside its own overlaps.
• A map is continuous as soon as the preimage of every subbasic set is open, because preimages respect both intersections and unions. That is the cheapest continuity test the course has.

The covering condition is inherited: every point lies in some member of the subbasis, which is itself a one-fold intersection and so a member of the collection. For the overlap condition there is nothing to construct: the intersection of two finite intersections of subbasic sets is again a finite intersection of subbasic sets, so it is itself a member of the collection and can serve as the set sitting inside the overlap.

The subbasis covers the space, which gives the first condition.

A single subbasic set is a finite intersection of one term, so it already belongs to the collection.

For the second condition take two members and intersect them.

The result is again a finite intersection of subbasic sets, so it lies in the collection.

The overlap itself can be taken as the set sitting inside it, so both conditions hold.

A subbasis is any family covering the space. Its finite intersections are a basis, and their unions are the open sets.

• A subbasis is any covering family; its finite intersections form a basis.
• It generates the coarsest topology in which its members are open.
• The rays generate the standard topology on $\mathbb{R}$.
• A map is continuous as soon as the preimages of the subbasic sets are open.

Great work! You now know how to:

• say what a subbasis is and how it generates a topology;
• build the basis of finite intersections;
• use the rays and the open half-planes as subbases;
• state the continuity test a subbasis provides.