Question about a basis for a topology vs the topology generated by a basis?

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.

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.