The last lesson started with a topology and found a family inside it. This one runs the other way: start with a family of subsets and ask whether unions of them form a topology. Two conditions are enough, and meeting them is how most topologies are actually defined.

A builder is handed a pile of bricks and asked whether a wall can be made of them alone. What matters is that the bricks cover the ground and that where two overlap a brick fits in the overlap.

The bricks of the lower-limit topology: half-open intervals $[a,b)$, drawn solid at the left end and hollow at the right. They cover the line, and where two of them overlap the overlap is another one — so their unions form a topology, finer than the standard one because $[0,1)$ is now open.

Let $\mathcal{B}$ be a collection of subsets of a set $X$ — not open sets yet, since there is no topology. Declare open exactly the unions of members of $\mathcal{B}$, together with $\varnothing$. This collection is a topology precisely when the two conditions below hold, and $\mathcal{B}$ is then a basis for it.

• The first condition makes the whole space a union of members, so the space itself comes out open. The empty union gives the empty set.
• The second is what makes finite intersections open. Without it two basic sets could overlap in a region that no union of basic sets can reach.
• The half-open intervals $[a,b)$ satisfy both, and the topology they generate on $\mathbb{R}$ is the lower-limit topology, written $\mathbb{R}_\ell$ and often called the Sorgenfrey line. It is the standard counterexample of chapter 8.
• In $\mathbb{R}_\ell$ the set $[0,1)$ is open, so the topology is strictly finer than the standard one; and every standard open set is still open, since $(a,b)=\bigcup_{a<c<b}[c,b)$.
• A family can fail. On $\mathbb{R}$ the two sets $(0,2)$ and $(1,3)$ alone do not cover the line, and even restricted to their union they give no basic set inside $(1,2)$ around the point $$ unless one is added.

The empty union is the empty set and the first condition makes the whole space a union, so the forced sets are there. A union of unions of basic sets is again a union of basic sets, so the union axiom holds with nothing to prove. Only the intersection needs the second condition: given a point in the intersection of two generated open sets, it lies in a basic set inside each, and the condition supplies a third basic set inside the overlap. Collecting one such set for each point writes the intersection as a union of basic sets.

The empty union gives the empty set, and the covering condition gives the whole space.

A union of sets each of which is a union of basic sets is again one, so the union axiom is free.

For an intersection, take a point and use that each side is a union of basic sets.

The second condition hands over a basic set inside the overlap.

Collecting one for each point writes the intersection as a union of basic sets, so it is open.

A family of subsets generates a topology exactly when it covers the set and has a member inside every overlap around each point of it.

• A family generates a topology when it covers the set and fits inside every overlap around each point.
• Its unions are then the open sets, and the family is a basis for the result.
• The half-open intervals generate the lower-limit topology $\mathbb{R}_\ell$, strictly finer than the standard one.
• Most topologies are defined this way rather than by listing their open sets.

Great work! You now know how to:

• test a family of subsets against the two conditions;
• build the topology it generates;
• describe the lower-limit topology and compare it with the standard one;
• recognise the two conditions behind the metric and product topologies.