Understanding lemma 13.2 in Topology by Munkres : Why do I need to show that it generates the same topology? Munkres states the following lemma in his Topology:
Lemma 13.2: Let $X$ be a topological space and suppose $K$ 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 $K$ such that $x\in C\subset U$. Then $K$ is a basis for the topology of X. 
I have shown that 1) Given $x \in X$ there is some open set in $K$ containing $x$.(2)Given $x\in C_1,C_2 \in K$, there is an open set $C \in K$ such that $x\in K \subset C_1 \cap C_2$.
Question: Shouldn't this be enough to prove that $K$ is a basis for the topological space $X$? After all, Munkres defines a basis as a collection of subsets of $X$ satisfying properties (1) and (2). Why do I need to show that that the topology generated by $K$ is the same as the given one?
 A: Read carefully the paragraph following the definition of basis on page$78$:

If $\mathcal{B}$ satisfies these two conditions, then we define the topology $\mathcal{T}$ generated by $\mathcal{B}$ as follows: A subset $U$ of $X$ is said to be ope in $X$ (that is, to be an element of $\mathcal{T}$) if for each $x\in U$, there is a basis element $B\in\mathcal{B}$ such that $x\in B$ and $B\subset U$. Note that each basis element is itself an element of $\mathcal{T}$.

In plain English, the whole definition and the paragraph says: Ok now, given the set $X$, let's define a topology on $X$ by a new trick: we specify which sets are open in $X$ indirectly in terms of basis. Once we have a basis on $X$ first, then we can define a topology $\mathcal{T}$ as follows: A subset $U$ of $X$ is said to be open in $X$ if……blah blah…...
So the definition has not predefined any topology on $X$. 
(The language of the definition is indeed prone to cause confusions. I remember I had struggle with it for a while when I was reading that as well) It just presents you a new trick to define topologies on $X$. 
Now, in practice, often we already know some topology on $X$, and we would like to know:

Is there any basis that generates the topology we have?

So we want to go into the other direction. We would like to know that our old familiar topology  indeed comes from this "new trick". Once we have verified that $\mathcal{C}$ is a basis, it automatically generates a topology $\mathcal{T}'$. But is it the product we want? Is it the same as our old $\mathcal{T}$? This is a nontrivial fact to be checked.
