Question about cardinality of a base of a topology I am having trouble expressing in mathematical notation for the hint given in the following question:
Prove that if a space $(X,\mathcal{T})$ has a base $\mathcal{B}$ of cardinality $\alpha$ then the cardinality of $\mathcal{T}$ can not exceed$\ 2^{\alpha}.$  (Hint: Define a surjective function from the set of all sub-families of $\mathcal{B}$ onto $\mathcal{T}.$ Note that the cardinality of the set $X$ itself is of little importance here.)
I am not sure if the surjective function written in the following manner would suffice
Let $\mathcal{T}$ be a topology on X, $\mathcal{B}$ be a base for topology $\mathcal{T}$, and let $\ C=\{\mathcal{B}:\mathcal{B}\subset\mathcal{T}\}$ be defined as the sub-families of $\mathcal{B},$ and let $\ D=\{\mathcal{V}:\mathcal{V}\in\mathcal{T}\}.$ 
The surjective function from $f:C\rightarrow\ D$ be defined as $f(\mathcal{B})=\mathcal{V}$
Thank you in advance.
 A: Your map $f$ should be defined on $\mathcal{B},$ not on the power set of $\mathcal{T}$. As is written, you've just proven that the cardinality of $\{\mathcal{B}\}$ is the same as the cardinality $\{\mathcal{V}\},$ whatever the latter symbol refers to, but both of these sets have one element, so that isn't even an interesting statement.
So let's take $U\in \mathcal{T}$ and let's figure out what to do with it. By definition of a base, for every $x\in U$ there exists some $B_x\in \mathcal{B}$ such that $x\in B_x\subseteq U$ (similar to how a ball works in a metric space). Then, $U=\cup_{x\in U} B_x$. This gives us a clue as to what to do, since $\{B_x|x\in U\}$ is clearly an element of the power set $2^{\mathcal{B}}$.
So let $f:2^{\mathcal{B}}\to \mathcal{T}$ be defined by $f(A)=\cup_{B\in A} B$. By what we've just proven, then $U=f(A),$ where $A=\{B_x|x\in U\}$. Hence, $f$ is surjective and the cardinality of $\mathcal{T}$ cannot be greater than that of $2^{\mathcal{B}}$.
I don't know if you can prove this without using the axiom of choice (which featured quite heavily in our construction of $f$), but my gut feeling says probably not.
A: If $\mathcal{B}$ is a base for $(X,\mathcal{T})$, of size $\kappa$, then there is a standard injection $F:\mathcal{T} \to \mathscr{P}(\mathcal{B})$ defined by
$$F(O) = \{B \in \mathcal{B}: B \subseteq O\} $$
where the right hand side is a well-defined subset of $\mathcal{B}$, so an element of $\mathscr{P}(\mathcal{B})$ which has size $2^\kappa$. 
$F$ is injective:  
By the definition of a base, $\bigcup F(O)=O$ (all members of $F(O)$ are subsets of $O$ so the union is too, and if $x \in O$ there is at least some $B \in \mathcal{B}$ such that $x \in B \subseteq O$ so that $B \in F(O)$ and $x \in B \subseteq \bigcup F(O)$ which shows the reverse inclusion)
and so $F(O)=F(O')$ implies $O=O'$ showing injectivity of $F$.
This shows that $|\mathcal{T}| \le 2^\kappa = 2^{|\mathcal{B}|}$ as required.
Alternatively, the argument above also shows that the map $\mathcal{B'} \to \bigcup \mathcal{B}'$ from $\mathscr{P}(\mathcal{B})$ to $\mathcal{T}$ is surjective ($F(O)$ is a pre-image for $O$), which is another way to show the same inequality of cardinals.
