Can a collection of subsets be declared a basis for a topology? Something like a "free topology"? I'm an undergraduate reviewing some things about elementary general topology which I don't recall very well because I never really used them and now I do need to get them straight.
Let's recall the definition of a basis: 
Definition. Let $X$ be a topological space with topology $\tau$. We say that a subcollection $B$ of $\tau$ is a basis for $\tau$ if every element in $\tau$ can be written as a union of elements of $B$.
My question is whether given a set $X$ and a collection $C$ of subsets of $X$, can one "declare" that this collection $C$ is a basis and define a topology on $X$ by means of $C$. 
Due to things I've seen that are similar in the past, this is something I would call a "free topology" on a collection $C$ of subsets of $X$.
However, I'm not sure how does this define a topology. 
Given two open sets, their union is certainly an open set, because each one is the union of elements of $C$, and so is their union. How about finite intersections? Can you help me? Thanks
Also, I'd like to know if there is a popular name to this thing I'm refering to.
 A: A basis uniquely defines the topology. In other words, if I tell you what the basis of a topology is, you can construct the whole topology. However, not every collection of subsets is a basis.
In order for a collection $\mathcal B$ of subset of $X$ to form a basis, it needs to satisfy two conditions:

*

*It needs to cover $X$, i.e. $\bigcup_{B\in \mathcal B} B = X$

*For any two base sets $B_1,B_2$ with a non-empty intersection, there must, for every $x\in B_1\cap B_2$, exist some base element $B\in\mathcal B$ such that $x\in B\subset B_1\cap B_2$.

You were right in being worried about intersections. Take a look at this example:
Take $X=\mathbb R$ and $\mathcal B=\{(-\infty, 1), (-1,\infty)\}$. Now, if this is a basis for a topology, then $(-\infty, 1)$ must be an open set, and $(-\infty,1)$ must also be an open set. But that means that $$(-1,\infty)\cap (-\infty, 1) = (-1,1)$$ must also be an open set.
However, all open sets are unions of some basis elements. There (pretty obviously) is no collection of basis elements for which the union is $(-1,1)$. So $\mathcal B$ is not a basis.
A: No, this isn't possible. For example: Consider the space $\{1,2,3\}$ and $C = \{ \{1,2\}, \{2,3\} \}$. If $C$ were a basis for a topology, then $\{1,2\} \cap \{2,3\} = \{2\}$ would have to be a union of elements of $C$ - but that's not the case. 
However, all is not lost. Given a topological space $(X, \tau)$ we say that $\mathcal B$ is a subbase for $\tau$ iff $\mathcal B \subseteq \tau$ and the set $\mathcal C$ of all finite intersections of elements in $\mathcal B$ is a basis for $\tau$.
Fix a nonempty set $X$. Now any $C \subseteq \mathcal P(X)$ such that $\bigcup C = X$ is a subbase for a topology. In particular, for any $D \subseteq \mathcal P(X)$ the set $C = D \cup \{X\}$ is a subbase for a topology on $X$.
A: $\mathcal V\subseteq\wp(X)$ will not in general be a basis of a topology on $X$. 
However, if you define: $$\mathcal B:=\{B\in\wp(X)\mid B \text{ is a finite intersection of elements of }\mathcal V\}$$
then $\mathcal B$ is a basis for the smallest topology on $X$ that contains $\mathcal V$. 
This under the convention that $\cap\varnothing:=X$ so that $X\in\mathcal B$.
In this context $\mathcal V$ is a so-called subbase of this topology.
