I am reading "Topology 2nd Edition" by James R. Munkres.
I am new to the axiom of choice.
Unfortunately, Munkres declared on pp. 60-61 as follows:
You will find few further specific uses of a choice function in this book; we shall introduce a choice function only when the proof would become confusing without it. But there will be many proofs in which we make an infinite number of arbitrary choices, and in each such case we will actually be using the choice axiom implicitly.
Munkres used the axiom of choice in his proof of Lemma 13.1.:
Is the axiom of choice used in the proof that every open set is the union of basis elements (Mukres, Lemma 13.1)
Did he use $c\circ f$ in his proof? (Is $c\circ f(x) = B_x$?)
Let $f:U\to2^{\mathcal{B}}$ be a function such that $U\ni x \mapsto \{B\in\mathcal{B}\mid x\in B\subset U\}\in 2^{\mathcal{B}}$.
Then, by Lemma 9.2, there exists a function $$c:f(U)\to\bigcup_{\mathcal{D}\in f(U)} \mathcal{D}$$ such that $c(\mathcal{G})$ is an element of $\mathcal{G}$, for each $\mathcal{G}\in f(U)$.
Lemma 13.1. Let $X$ be a set; let $\mathcal{B}$ be a basis for a topology $\mathcal{T}$ on $X$. Then $\mathcal{T}$ equals the collection of all unions of elements of $\mathcal{B}$.
Proof. Given a collection of elements of $\mathcal{B}$, they are also elements of $\mathcal{T}$. Because $\mathcal{T}$ is a topology, their union is in $\mathcal{T}$. Conversely, given $U\in\mathcal{T}$, choose for each $x\in U$ an element $B_x$ of $\mathcal{B}$ such that $x\in B_x\subset U$. Then $U=\bigcup_{x\in U} B_x$, so $U$ equals a union of elements of $\mathcal{B}$.
Lemma 9.2 (Existence of a choice function). Given a collection $\mathcal{B}$ of nonempty sets (not necessarily disjoint), there exists a function $$c:\mathcal{B}\to\bigcup_{B\in\mathcal{B}} B$$ such that $c(B)$ is an element of $B$, for each $B\in\mathcal{B}$.
