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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.

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    $\begingroup$ 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 $2^\mathfrak{c}$. $\endgroup$ – Henno Brandsma Mar 29 at 22:37
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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.)

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A base covers the space.
A weak base may not cover the space.
The set of all not empty, open subsets of R that exclude 0 is a weak base.

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