Topological space with countable basis, confusion in the definition. I didn't really understand something : Let $(X,\mathcal T)$ a topological space with countable basis. Let $\mathcal B$ a basis of $\mathcal T$. Does it mean that $\mathcal B$ is countable or that every element of $\mathcal T$ can be written as a countable union of element of $\mathcal B$ ? If $\mathcal B$ is countable, then of course every element of $\mathcal T$ can be written as a countable union of element of $\mathcal B$, but I really don't think that the converse is true, that's why I ask.
 A: OK so if $(X, \tau)$ has a countable basis then there may still be some bases that are not countable. If we let $X$ be the real line and $\tau$ the usual set of open sets then $\tau$ is a basis for the reals, as is the set of all open intervals, but the set of all open intervals with rational end points is a countable basis. 
In a general topological space, every set can be written as a union of members of the basis. This need not be a countable union. In the case of the basis given above for the real line (the set of all open intervals with rational end points) then is so happens that every set can be written as a countable union but that is a happy accident.
And answering your last point, you are right, the converse is not true. If we take the set of all open sets as the basis then every set can be written as the union of a single set (itself) but the basis is not countable.
A: If a space $X$ is hereditarily Lindelöf and $\mathcal{B}$ is any base for the space $X$, then any open set $O$ can be written as a countable union of elements from $\mathcal{B}$. 
A space can be hereditarily Lindelöf without there being a countable base for $X$, an example of this is the Sorgenfrey line (or the lower limit topology on $\mathbb{R}$, as Munkres calls it).
A: It doesn't mean that a basis $\cal B$ of $\cal T$ is countable. It means that there exists a basis $\cal B$ of $\cal T$ that is countable.
