This is the statement of Lemma 13.2 in Munkres' Topology Chapter 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 $\in$C$\subset$U. Then $\mathcal{C}$ is a basis for the topology of X.
The English description confuses me. I would like to express the collection $\mathcal{C}$ in set-builder notation. Is what I have written below what the author intends?
Letting $\mathcal{T}$ denote the topology on X:
$$\mathcal{C}=\{V\in\mathcal{T}:\forall U\in\mathcal{T}\forall x\in U\exists C\in\mathcal{C}(x\in C \subset U)\}$$
Is my understanding correct? If not, please explain to me where I may be mistaken.
Best Answer
The trouble with your suggestion is that this Lemma is not about some well-defined collection $\mathcal C$, it is instead about a certain property of a subcollection $\mathcal C \subset \mathcal T$, i.e. a certain property of a collection $\mathcal C$ of open subsets of $X$.
Without yet stating this property but introducing the notation $\mathcal P(\mathcal C)$ to represent it, what the lemma says is that for any collection $\mathcal C$ of open subsets of $X$, if the property $\mathcal P(\mathcal C)$ is true then $\mathcal C$ is a basis for the topology $\mathcal T$.
The property $\mathcal P(\mathcal C)$ is sort-of what you've written as part of your set builder notation, if you threw away the set builder part. Here's the full statement of the relevant property: $$\mathcal P(\mathcal C)\,\, := \,\, \forall U\in\mathcal{T}, \forall x\in U, \exists C\in\mathcal{C} \, (x\in C \subset U) $$
One thing to keep in mind: there can easily be many different sub collections $\mathcal C \subset \mathcal T$ which satisfy the property $\mathcal P(\mathcal C)$, so as I said $\mathcal C$ is not determined by this property.
Just as an example, if $X = \mathbb R^2$, and if $\mathcal T$ is the usual Euclidean topology on $X$, then $\mathcal C = \{\text{all open balls}\}$ satisfies the property $\mathcal P(\mathcal C)$, and $\mathcal C = \{\text{all open squares}\}$ also satisfies it, and many, many other collections $\mathcal C \subset \mathcal T$ satisfy it. As a consequence, by applying the lemma one concludes that there are many many different bases for the Euclidean topology $\mathcal T$ on $\mathbb R^2$.