Does a basis exist independent of the topology? While proving Lemma 13.2

Let X be a topological space. Suppose that $\mathcal{C}$ is a collection of open
sets of X such that for each open set U of X and each x in U, there is an element C
of $\mathcal{C}$ such that x ∈ C ⊂ U. Then $\mathcal{C}$ is a basis for the topology of X.

of Munkres', we first proved that $\mathcal{C}$ is a basis. Basis of what topology? The given topology on X? If yes, then why did we go on to prove that the topology generated by $\mathcal{C}$ is the same as the given topology on X when essentially proving the first part gives us what is said in the lemma?
 A: Munkres' notation is perhaps a bit confusing.
Given a set $X$ and a collection $\mathscr B$ of subsets of $X$, we define the collection $\mathscr T (\mathscr B)$ of subsets of $X$ generated by $\mathscr B$ via

$U \in \mathscr T (\mathscr B)$ if and only if for each $x \in U$ there exists $B \in \mathscr B$ such that $x \in B \subset U$.

Clearly $\mathscr B \subset \mathscr T (\mathscr B)$. When is  $\mathscr T (\mathscr B)$ a topology on $X$?
In the definition at the beginning of §13 Munkres introduces the concept of a basis for a topology on a set $X$. This is a collection $\mathscr B$ of subsets of $X$ such that

*

*For each $x \in X$, there is at least one basis element $B$ containing $x$.

*If $x$ belongs to the intersection of two basis elements $B_1$ and $B_2$, then there is a basis element $B_3$ containing $x$ such that $B_3 \subset B_1 \cap B_2$.

Note that $X$ is just a set, it does not yet have a topology. Thus basis for a topology on $X$ does not mean that we are given a topology on $X$, but Munkres shows that $\mathscr T (\mathscr B)$ is a topology on $X$. Thus $\mathscr T (\mathscr B)$ is the topology generated by the basis $\mathscr B$.
It is moreover easy to see that the above conditions 1. and 2. are not only sufficient, but also necessary for $\mathscr T (\mathscr B)$ being a topology.
In Lemma 13.1 Munkres gives an alternative characterization of the topology $\mathscr T (\mathscr B)$; it is the set of all unions of elements of $\mathscr B$.
This shows in particular that $\mathscr T(\mathscr B)$ is the coarsest topology on $X$ containing  $\mathscr B$ since each topology $\mathscr T$ contains all unions of members of $\mathscr T$.
However, Munkres' formulation of Lemma 13.1 may be misleading because it states "let $\mathscr B$ be a basis for a topology $\mathscr T$ on $X$". This seems to indicate that we are given a topology $\mathscr T$ on $X$, but this is not the case. Munkres should have better said

Let $X$ be a set; let $\mathscr B$ be a basis for a topology on $X$. Then the topology $\mathscr T (\mathscr B)$ generated by $\mathscr B$ equals the collection of all unions of elements of $\mathscr B$.

After that Munkres writes

We have described in two different ways how to go from a basis to the topology it generates. Sometimes we need to go in the reverse direction, from a topology to a basis generating it. Here is one way of obtaining a basis for a given topology; we shall use it frequently.
$\phantom{} $
Lemma 13.2. Let $X$ be a topological space. Suppose that $\mathscr C$ is a collection of open sets of $X$ such that for each open set $U$ of $X$ and each $x$ in $U$, there is an element $C$ of $\mathscr C$ such that $x \in C \subset U$. Then $\mathscr C$ is a basis for the topology of $X$.

In this lemma "the topology of $X$" is of course the given topology on the topological space $X$.
Thus I would say that Munkres actually has two different concepts of "basis":

*

*A basis $\mathscr B$ for a topology on a set $X$. In this case no topology is given, but $\mathscr B$ generates one.


*A basis $\mathscr C$ for a topological space $X$. In this case a topology $\mathscr T$ is given on the set $X$ and $\mathscr C$ is required to be a subset of $\mathscr T$ satisfying the following condition:

For each $U \in \mathscr T$ and each $x \in U$ there exists an element $C \in \mathscr C$ such that $x \in C \subset U$.

In Lemma 13.2 Munkres shows that if $\mathscr C$  is a basis for a topological space $X$ with given topology $\mathscr T$, then $\mathscr C$ is a basis for a topology on $X$ and $\mathscr T (\mathscr C) = \mathscr T$.
Note that the condition characterizing a basis for a topological space seems to be weaker than the two conditions characterizing a basis for a topology on a set; but the essential point is the requirement $\mathscr C \subset \mathscr T$.
In fact, a basis of a topological space allows to represent each open set as a union of elements of the basis.
A: Let be $(X,\tau_x)$the topological space. Let $\mathcal B$={$C\subseteq X$,open} with the property that for every open set $U\subseteq X $ and for every $x\in U$, $ \exists C \in \mathcal C$ such that $x\in C_x\subset U$. To prove $\mathcal C$ is a base we take a generic open set $A\subseteq X$ and we show we can write $A$ as an union of element of $\mathcal C$. Since A is open $\forall x \in A, \exists C_x$ such that $x\in C_x \subset A$. Just take $A=\bigcup_{x\in A}C_x=A$. You can do it for every $A\in \tau_x$ so $\mathcal C$ is a base. $\square$
A: The definition of topological basis comes before the definition of topology.
Definition: Let $X$ be a set and $\mathcal B\subset 2^X$ be a subset of the power set of $X$. Then, $\mathcal B$ is a topological basis if i) $\mathcal B$ is a covering of $X$. ii) For all $U,W\in\mathcal B$ and $x\in U\cap V$, there exists a $W\in\mathcal B$ such that $x\in W\subset U\cap B.$
Let $X$ be a topological space. Suppose that $\mathcal{C}$ is a collection of open
sets of $X$ such that for each open set $U$ of $X$ and each $x$ in $U$, there is an element $C$ of $\mathcal C$ such that $x ∈ C ⊂ U$. We will show that $\mathcal C$ is a topologicl basis: It is clear that $\mathcal C$ covers $X$. So, (i) is satisfied. Let $U, V\in\mathcal C$ and $x\in U\cap V$. Obviously $U$ and $V$ are open sets so $U\cap V$ is open set. So, by definition of $\mathcal C$, there is an open set $W\in\mathcal C$ such that $x\in W\subset U\cap V$. (ii) is satisfies and $\mathcal C$ is a basis. $\mathcal C$ clearly generates the given topology, since any open set of $X$ can be written as a union of some elements of $\mathcal C$. So $\mathcal C$ is a basis of the given topology.
