Confusions about axiom of choice and the “deception” in Munkres' “Topology”

Question

In Munkres' "Topology", page 59, Axiom of choice, it states: Given a collection $\mathcal{A}$ of disjoint nonempty sets, there exists a set $C$ consisting of exactly one element from each element of $\mathcal{A}$; that is, a set $C$ such that $C$ is contained in the union of the elements of $\mathcal{A}$, and for each $A\in\mathcal{A}$, the set $C\cap A$ contains a single element.

In Munkres' “Topology”, page 61, it states: Now we must confess that in an earlier section of this book there is a proof in which we constructed a certain function by making an infinite number of arbitrary choices. And we slipped that proof in without even mentioning the choice axiom. Our apologies for the deception. We leave it to you to ferret out which proof it was!

Aha, I think I may ferret out that proof. It is lemma 21.2!

Lemma 21.2 (The sequence lemma). Let $X$ be a topological space; let $A\subset X$. If there is a sequence of points of $A$ converging to $x$ then $x\in\bar{A}$; the converse holds if $X$ is metrizable.

In Munkres' “Topology”, page 130, Lemma 21.2 (The sequence lemma), it states: Conversely, suppose that $X$ is metrizable and $x\in\bar{A}$. Let $d$ be a metric for the topology of $X$. For each positive integer $n$, take the neighborhood $B_{d}(x,1/n)$ of radius $1/n$ of $x$, and choose $x_{n}$ to be a point of its intersection with $A$. We assert that the sequence $x_{n}$ converges to $x$: Any open set $U$ containing $x$ contains an $\epsilon$ -ball $B_{d}(x,\epsilon)$ centered at $x$; if we choose $N$ so that $1/N<\epsilon$ , then $U$ contains $x_{i}$ for all $i\geq N$.

Here are my questions:

(i) In page 61, Munkres require me to ferret out “the proof”. Is it exactly the proof I mention above?

(ii) The proof above does make an infinite number of choices, but it may not satisfy the requirements of “axiom of choice” in page 59. Since in axiom of choice, the sets in the collection $\mathcal{A}$ must be disjoint. But all the sets $B_{d}(x,1/n)\cap A$, $n=1,2,\cdots$ are not disjoint. It makes an infinite number of choices to get the sequence $(x_{n})$ but it is not an example of “axiom of choice”. So what is it? How can we name it?

Answer 1

He is not using the axiom directly, but a corollary of the axiom that is equivalent.

Specifically:

If there is function $S:I\to \mathcal P(X)\setminus \{\emptyset\}$ there is a function $x:I\mapsto X$ such that $x(i)\in S(i).$

We usually use subscript notation $S_i$ to indicate that $S$ is an "indexed collection of sets," and not $S(i).$

$I$ is called the index set, and we use it because we sometimes need the $S_i$ to be allowed to be equal.

In your case, the ball of radius $1/n$ will have index $n,$ and the ball of index $n$ and the ball of index $n+1$ might be equal.

You prove this corollary by taking the set $Y=X\times I$ and define $A_i=\{(u,i)\mid u\in S_i\}\subseteq Y$ and then the $A_i$ are disjoint and non-empty.

Answer 2

Aha, I think I may ferret out that proof. It is lemma 21.2!

I do not think that Munkres means Lemma 21.2. Yes, its proof uses the axiom of choice, but Munkres explicitly states

Now we must confess that in an earlier section of this book there is a proof in which we constructed a certain function by making an infinite number of arbitrary choices.

Lemma 21.2 does not belong to an earlier section of the book.

Munkres certainly means the proof of

Theorem 7.5. A countable union of countable sets is countable.

In case $J = \mathbb Z_+$ one chooses for each $n \in \mathbb Z_+$, i.e. for infinitely many $n$, a surjective function $f_n : \mathbb Z_+ \to A_n$.

The existence of such a function $f_n$ is due to the fact that $A_n$ is countable. However, unless $A_n$ is a one-element set, there are infinitely many such functions and we have to make a choice to select one.

Before we explain how the axiom of choice is used here, let us rephrase the axiom of choice as follows:

Given a collection $\mathcal A$ of disjoint nonempty sets, there exists a function $\phi : \mathcal A \to \bigcup_{A \in \mathcal A} A$ such that $\phi(A) \in A$ for all $A \in \mathcal A$.

In fact, if we have a set $C$ as in Munkres's formulation, then $\phi(A)$ is the unique element in $C \cap A$. Conversely, if we have a function $\phi$ as above, then $C = \{\phi(A) \mid A \in \mathcal A\}$.

The commonly used variant of the axiom of choice, however, is this:

Given a family $(A_i)_{i \in J}$ of (not necessarily disjoint) nonempty sets $A_i$ indexed by a set $J$, there exists a function $\phi : J \to \bigcup_{i\in J} A_i$ such that $\phi(i) \in A_i$ for all $i \in J$.

This variant clearly covers Munkres's version because we can identify the collection $\mathcal A$ with the family $(A)_{A \in \mathcal A}$ ("self-indexing" of a set of sets).

To see that Munkres's version implies the above general variant, consider the set $\mathcal A = \{A_i \times \{i\} \mid i \in J\}$ which consists of disjoint nonempty subsets of $(\bigcup_{i \in J} A_i) \times J$. There exists a function $\phi : \mathcal A \to \bigcup_{A \in \mathcal A} A$ such that $\phi(A) \in A$ for all $A \in \mathcal A$. Now let $p : (\bigcup_{i \in J} A_i) \times J \to \bigcup_{i \in J} A_i$ denote the projection forgetting the $J$-coordinate. Then $\phi': J \to \bigcup_{i \in J} A_i, \phi'(i) = p(\phi(A_i \times \{i\}))$, has the property $\phi'(i) \in A_i$ for all $i \in J$.

In the context of Theorem 7.5 we take $J = \mathbb Z_+$ and $B_n$ = set of surjective functions $\mathbb Z_+ \to A_n$ and apply the axiom of choice to the family $(B_n)_{n \in \mathbb Z_+}$.