Question about a basis for a topology vs the topology generated by a basis? This is a really basic (no pun intended......no?  Ok...) question about what it means to be a basis for a topology.
Here is what I know:  If $(X, \mathcal{T})$ is a topological space, and $\mathcal{B} \subseteq \mathcal{T}$ is a basis for $\mathcal{T}$, then we know by definition of basis that the following are true:

  
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*For every $x \in X$, $x \in B$ for some $B \in \mathcal{B}$.
  
*If $x \in B_{1} \cap B_{2}$ for some $B_{1}, B_{2} \in \mathcal{B}$, then $\exists B_{3} \in \mathcal{B}$ such that $x \in B_{3} \subseteq B_{1} \cap B_{2}$.
  

Also, I know that if we have a basis $\mathcal{B}$ for a topology $\mathcal{T}$, then the topology generated by the basis, $\mathcal{T}_{\mathcal{B}}$, consists of all possible unions of elements of $\mathcal{B}$.
Here is my question:  I want to prove that if $\mathcal{B}$ is a basis for $\mathcal{T}$, then $\mathcal{T} = \mathcal{T}_{\mathcal{B}}$.  Showing $\mathcal{T}_{\mathcal{B}} \subseteq \mathcal{T}$ is really easy.  To show $\mathcal{T} \subseteq \mathcal{T_{\mathcal{B}}}$ relies on some "fact" that I don't know to prove: that if $U \in \mathcal{T}$, and $x \in U$, $\exists B \in \mathcal{B}$ such that $x \in B \subseteq U$.
How do I prove that for any $U \in \mathcal{T}$, $\exists B \in \mathcal{B}$ such that $x \in B \subseteq U$?  I don't see how this fact follows from the definition of a basis.  And without this fact, I can't prove that $\mathcal{T} = \mathcal{T_{\mathcal{B}}}$.
 A: I just checked Munkres - he made everything fine - but just to clarify:

In principle there are two problems:
a. Given a collection $\mathcal{B}$. Then $\langle\mathcal{B}\rangle$ is a topology iff it satisfies the characteritation:
$$\forall x\in X\exists B_x\mathcal{B}:\quad x\in B_x$$
$$\forall B,B'\in\mathcal{B}\forall x\in B\cap B'\exists B_x\in\mathcal{B}:\quad x\in B_x\subseteq B\cap B'$$
b. Given a collection $\mathcal{B}$ and a topology $\mathcal{T}$. Then $\langle\mathcal{B}\rangle=\mathcal{T}$ iff it fulfills the criterion:
$$\forall U\in\mathcal{T}\forall u\in U\exists B_u\in\mathcal{B}\subseteq\mathcal{T}: u\in B_u\subseteq U$$
Note that though both problems are conceptually similar they are solved quite differently.
A: I'm answering my own question because I have a clear answer:
When defining a "basis", we start out with a set $X$.  We don't define a topology on it before we define $\mathcal{B}$ axiomatically.
A set of subsets of $X$, $\mathcal{B}$, is said to be a base to the set $X$ if the following two conditions hold:
(i) $\forall x \in X$, $\exists B \in \mathcal{B}$ such that $x \in B$
(ii) Given $B_{1}, B_{2} \in \mathcal{B}$, if $\exists x \in B_{1} \cap B_{2}$, then $\exists B_{3} \in \mathcal{B}$ such that $x \in B_{3} \subseteq B_{1} \cap B_{2}$
Now that we have defined what it means for a set $\mathcal{B}$ of subsets of $X$ to be a base to the set $X$, we can define $\mathcal{T}_{\mathcal{B}}$, the topology generated by $\mathcal{B}$, as the set of all unions of elements in $\mathcal{B}$.  That is, a set $U$ is open in $\mathcal{T}_{\mathcal{B}}$ if it is a union of elements of $\mathcal{B}$.  It is easy to prove that this is a topology.
Now, if we start out with a topology $\mathcal{T}$ on a set $X$, and we say $\mathcal{B}$ is a basis for the topology $\mathcal{T}$, this is defined as $\mathcal{T}$ actually being the topology $\mathcal{T}_{\mathcal{B}}$, the set of all unions of elements in $\mathcal{B}$.
If $(X, \mathcal{T})$ is a topological space, it is possible to have a set $\mathcal{B}$ of subsets of $X$ satisfy the two properties that make it a base to the set $X$ without it being a basis for a given topology.  But if $\mathcal{B} \subseteq \mathcal{T}$, then we are assured $\mathcal{T}_{\mathcal{B}} \subseteq \mathcal{T}$.  We don't have equality, though, unless we are also given that $\mathcal{B}$ is a basis for $\mathcal{T}$.
Here is an example of a topological space in which a base to the set $X$ is contained in the topology of $X$ but is not a basis for the topology.  Let $(X, \mathcal{T}) = (\mathbb{R}, \mathcal{T}_{\text{indisc}})$ where $\mathcal{T}_{\text{indisc}}$ is the indiscrete topology (i.e., $\mathcal{T}_{\text{indisc}} = \{ X , \emptyset \}$).  Then since every topology acts as a basis for itself, $\mathcal{T}_{\text{indisc}}$ is a basis for itself, and it is also a base to the set $X$.  However, this base to the set $X$ is contained in $\mathcal{T}_{\text{disc}}$, the discrete topology, but it is not a basis for the discrete topology.
So, the main point here is that when we define a base to the set $X$, it is independent of any topology on $X$.  But should the elements of the base be in a topology, then the topology generated by the base is a subset of the original topology.  Furthermore, if we say the base is a basis for the original topology, by definition that means the original topology is equal to the topology generated by the base.
