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.
Any comment is appreciated. Thank you in advance for your help.
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.)