# Is a weak basis of a topological space a basis and vice versa? (Kechris' book)

I'm reading Kechris' book "Classical Descriptive Set Theory" and the author gives the following definition (pp. $$49$$, row $$3$$):

A weak basis of a topological space $$X$$ is a collection of nonempty open sets s.t. every nonempty open set contains one of them.

My question is: is this definition equivalent to that of a basis for a topology?

The fact that the author gives a specific name to such a family suggests that it is not, but for every $$x\in X$$ and for every open nhbd $$U(x)$$ there exists $$V(x)$$ in the weak basis contained in $$U$$. This means that a weak basis is also a covering and hence satisfies the conditions for being a basis.

• Even a weak basis that is a cover of $X$ need not be a base for $X$. And e.g. $\beta \omega$ has a countable pseudobase (the singletons of $\omega$) but its smallest base has size $2^\mathfrak{c}$. – Henno Brandsma Mar 29 at 22:37

While it is true that for every $$x,$$ any neighborhood $$U(x)$$ contains an element of the weak basis, say $$V(x),$$ we don't know that $$V(x)$$ is a neighborhood of $$x$$! All we know is that it is a subset of $$U(x)$$ and that it is open and nonempty. Thus, a weak basis need not cover the space, so need not be a basis.
For example, consider the topology of the empty set together with the cofinite sets (sets whose complement is finite) on the set of non-negative integers. A weak basis would be the set of cofinite sets of positive integers, but this cannot be a basis, having no neighborhood of $$0.$$
In general--among $$T_1$$ spaces, anyway--I suspect that if a space has the property that every weak basis is a basis, then the space is discrete. (The converse trivially holds.)