Base for a topology? I am having some trouble understanding the definition of a base, according to wikipedia.
According to wikipedia, a base is defined as follows:
A base is a collection B of subsets of X satisfying these two properties:
1) The base elements cover X.
2) Let B1, B2 be base elements and let I be their intersection. Then for each x in I, there is a base element B3 containing x and contained in I.
I thought I understood this but then wikipedia states the following sentence:
If a collection B of subsets of X fails to satisfy either of these, then it is not a base for any topology on X.
Wikipedia's definition defines a base for a set X. So what do they mean by 'a base for a topology on X'? Say we have a topological space (X,T), what does a base B have to do with T? My understanding of the definition of base leads me to believe that B is fully defined with respects to X. 
 A: The main point you may be missing is that there can be more than one topology for $X$. For example, if $X=\{1,2\}$, then one possible topology on $X$ is $\tau_1 = \{\emptyset, \{1,2\}$, while another topology is $\tau_2=\{\emptyset, \{1\},\{2\}, \{1,2\}\}$.
Note that the two topologies are not the same, and, for example, while $B=\{\{1\}, \{2\}\}$ is a basis for $\tau_2$, it is not the basis for $\tau_1$. 

So, what the sentence 

If a collection B of subsets of X fails to satisfy either of these, then it is not a base for any topology on X.

is trying to (rather awkwardly) say is that there exists no topology on $X$ for which $B$ is a basis.
A: There is an important distinction to be made, so please read my answer carefully.
Let $(X,T)$ be a topological space.  Now, if we have a collection $\mathcal{B}$ of subsets of $X$ which satisfy properties $1$ and $2$ in your question, then these are called a "base to the set $X$", and they form their own topology, call it $T_{\mathcal{B}}$.  This topology is the set of all possible unions of elements of $\mathcal{B}$.
Now, a natural question (which is one that you have) is: how is this new topology, $T_{\mathcal{B}}$, related to the original topology $T$?
Well, notice that when we defined $\mathcal{B}$ above, we didn't say all of the subsets in $\mathcal{B}$ have to be from $T$.  If they are all from $T$, then since $T$ is closed under unions, it follows that $T_{\mathcal{B}} \subseteq T$.  If in addition you want $T \subseteq T_{\mathcal{B}}$, you will need to show every element $A \in T$ can be written as a union of elements from $\mathcal{B}$, since this will imply $A \in T_{\mathcal{B}}$.
In general, though, you can have $T_{\mathcal{B}}$ be a different topology than $T$.  You only get $T_{\mathcal{B}} \subseteq T$ if, as we said, the base $\mathcal{B}$ comes from sets in $T$.  And you only get $T_{\mathcal{B}} = T$ if in addition to $\mathcal{B}$ coming from $T$, we also can write each element of $T$ as a union of elements of $\mathcal{B}$.  
In general, $\mathcal{B}$ is always a basis for the topology $T_{\mathcal{B}}$ (the set of all possible unions of elements of $\mathcal{B}$) -- you should check that all possible unions of elements of $\mathcal{B}$ forms a topology.  In the event that $T_{\mathcal{B}} = T$, we would then say $\mathcal{B}$ is a basis for the topology $T$.
A: The base $\mathcal{B}$ corresponds to the topology $\tau$ in the following way, for any $U\in \tau$, and for any $x\in U$, $\exists B\in \mathcal{B}$ such that $x\in B\subseteq U$. I hope this should clear the confusion. 
