confusion over definition of weight of a topological space I would like some clarifications over the definition for the weight of a topological space.  According to Engelking's General Topology text, 2nd edition, page 12, it states:
"Every set of cardinal numbers being well-ordered by $<,$ the set of all cardinal numbers of the form $|\mathcal{B}|,$ where $\mathcal{B}$ is a base for a topological space $(X,\mathcal{O}),$ has a smallest element; this cardinal number is called the weight of the topological space $(X,\mathcal{O})$ and is denoted by $\mathcal{w}((X,\mathcal{O})).$"
I am a bit confused over this definition.  The definition is defined in terms of a base of a topological space.  So when this cardinal number stating that a base $\mathcal{B}$ has a smallest element, is it referring to the smallest open set within a particular base for a topology, or is it referring to all possible bases which has the smallest number of open sets for a topology. Since for any given topology defined over an underlying set say $X$, there could be many different bases that could be defined for that particular topology.
Thank you in advance.
 A: The definition you cite says the weight is the smallest cardinal from all bases, not of an open set for a given base. This removes any ambiguity. Concretely, given a space $(X,\tau)$ we could define
$$
\mathfrak{B} = \{B : \text{ $B$ is a basis for $X$}\}
$$
and then
$$
\omega(X,\tau) := \min_{B \in \mathfrak{B}} |B|.
$$
The justification for 'well definedness', as you cite, relies on the fact that these cardinals are well ordered.
Edit: as per the comments, here's a small example. If $X$ has a countable basis (for example, if $X$ is metrizable) then its weight is either finite or $\aleph_0$ (i.e. countable). 
If additionally we ask for $X$ to be Hausdorff, then for each $x,y \in X$ we have $U_{x,y} \ni x$ and $V_{x,y} \ni y$ open and disjoint. Inductively given $x_1, \dots, x_n$ we can find a pariwise dijsoint collection $U_i \ni x_i$ of open sets. So if $X$ is infinite, then there are infinite basic open sets. Reciprocally, if $X$ is finite, then the amount of open sets (in particular basic ones) is bounded by $\mathcal{P}(X)$ which is finite. 
This shows that

Proposition. Let $(X,\tau)$ be a second countable Hausdorff space. Then$$\omega(X,\tau) = \min\{|X|,\aleph_0\}.$$

