Confusion about Lemma 13.2 in Munkres' topology (property which implies that a collection is a basis for a topology)

Question

Lemma 13.2 and its proof confuse me.

$X$ is a topological space and $C$ is a collection of open sets of $X$ satisfying a property. A specific topology is not mentioned in the lemma. In the proof, he assumes the properties of topologies are properties of $C$. Is $C$ a topology? Is this something that should be proven, or was it implicit in the language of the lemma?

He says he will often omit mentioning the topology $T$. But he defined open just in terms of the basis, and so in my mind, I don't associate an open set with any particular topology. Or in this case, a collection of open sets doesn't translate to topology. Is it supposed to?

Answer 1

As you say, the topology is usually omitted, i.e. left implicit. The statement

Let $X$ be a topological space.

means

Let $X$ be a set, together with a chosen topology $\mathcal{T}$ on $X$.

the $\mathcal{T}$ being implicit. The collection $\mathcal{C}$ is exactly what he says in the statement of the theorem, namely,

$\mathcal{C}$ 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 $x\in C\subset U$.

In other words,

$\mathcal{C}$ is a subset of $\mathcal{T}$ such that, for each $U\in\mathcal{T}$ and each $x\in U$, there is an element $C\in\mathcal{C}$ such that $x\in C$ and $C\subset U$.

The topology on $X$ was already chosen; he did not define the topology on $X$ in terms of any basis. The goal here is to prove that, if $\mathcal{C}$ is a collection of subsets of $X$ satisfying the above criterion, then $\mathcal{C}$ is a basis for the topology $\mathcal{T}$ on $X$.

Because elements of $\mathcal{C}$ are open subsets of $X$ (as defined by the topology $\mathcal{T}$), or in other words $\mathcal{C}\subset\mathcal{T}$, the elements of $\mathcal{C}$ will have the properties of open sets: for example, an intersection of elements of $\mathcal{C}$ will be an open set, because the intersection of any two open sets is again an open set. It need not be the case that an intersection of elements of $\mathcal{C}$ will again be an element of $\mathcal{C}$.

Answer 2

Lemma 13.2 on p.80 states: Let $X$ be a topological space. Suppose that $C$ 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 $K$ of $C$ such that $x \in K \subset U$. Then $C$ is a basis for the topology of $X$.

$C$ is not a topology, but a basis for a topology. A basis generates a topology by taking all unions of all elements of $C$.

I think your confusion is that "\'a priori", you don't know which sets are open. He assumes that you have a topology (so you DO know which sets are open), and then proceeds to show that the conditions on $C$ he provides are sufficient to guarantee that $C$ is a basis and generates the same topology as the assumed one.