Topology, First and Second Countable

First Countable

You need to be familiar with the term countable, before we proceed.

If every point p in a topological space has a countable base at p, the base is first countable.

this does not mean there are countably many base sets containing p. Rather, we mean this. Each open set in the base is "assigned" to a point p in the space. The most common example is a metric space, where base sets are open balls, and each base set is assigned to its center. Assign base sets to multiple points if you like, but in some fashion, base sets are assigned to points in the space, and for any point p, countably many base sets are assigned. Furthermore, any open set containing p contains one of the base sets assigned to p. This is an indispensable property of first countable, and it is used in various proofs. Something like this: p is part of an open set V - let W be a base set in V containing p - there are countably many ways to select W - etc. Once again the metric space leads by example. If p is part of an open set, there is an open ball centered at p, i.e. assigned to p, inside our open set.

A topology is first countable if it has a first countable base.

As you might expect, a metric space is always first countable. Base open sets are open balls, which can be assigned to their centers. Restrict radii to rational values, and the balls centered at p are countable. We already showed such a ball is part of any open set containing p, so the base, and the topology, is indeed first countable.

Second Countable

A base is second countable if it contains a countable number of open sets.

A topology is second countable if it has a second countable base.

Some Examples

The reals with half intervals [a,b) acting as a base is first countable, but not second countable. Given p, choose [p,p+s) for s rational. Every half interval containing p contains one of these base sets, which contains p. The rationals are countable, so this is a countable base at p. However, the entire base is not countable. If it were, we could always find an irrational p that does not start some half interval in the base, whence [p,p+1) cannot be the union of base sets.

Perhaps we just selected a bad base. Perhaps some other base would be countable, and would make the space second countable. Let B be a collection of open sets from this topology. Suppose B forms a base. Each set X in B is the union of disjoint half intervals. In fact X includes a countable number of disjoint half intervals. Let s(X) be the start set of X, the left end points of its half intervals. Consider all the start sets over every X in B. The result is a countable set of points along the real line. Let p be a point that is not in this set, and [p,p+1) is not covered by the base B. Thus the topology itself is first countable, but not second countable.