Implicit uses of Countable or Dependent Choice

Question

What are instances of implicit reliance on countable or dependent choice in classic books? Two examples are

Introduction to Commutative Algebra by M.F. Atiyah and I.G. MacDonald

where it is claimed, in remark 1 after corollary 1.5, that a certain argument does not require the Axiom of Choice but in fact the argument does use Dependent Choice, as pointed out in this question (now deleted), and on page 42 in

P. Halmos, Measure Theory, Graduate Texts in Mathematics 18, Springer-Verlag, New York, 1974 (reprint of the edition published by Van Nostrand, New York, 1950)

where countable additivity of the Lebesgue measure is proved without mentioning Countable Choice (which is needed).

Note that I am not interested in opinions as to whether the authors were really aware of the implicit use and did not mention it, or were unaware of it; only the factual instances of such arguments not mentioning reliance on countable or dependent choice explicitly.

Answer 1

Rudin's Principles of Mathematical Analysis (and most every book on Mathematical Analysis) in the proof that $\lim_{x\to p}f(x)=q$ is equivalent to "$\lim_{n\to\infty}f(p_n)=q$ for every sequence $\langle p_n:n\in\mathbb{N}\rangle$ such that $p_n\neq p$ for all $n$ and $\lim_{n\to\infty}p_n=p$" (and hence that $\varepsilon$-$\delta$-continuity is equivalent to sequential continuity).

Given the $\varepsilon>0$ without a suitable $\delta$ we get "Taking $\delta_n=1/n$ ($n=1,2,3,\ldots$), we thus find a sequence satisfying $\lim_{n\to\infty}p_n=p$ for which $\lim_{n\to\infty}f(p_n)=q$ is false."

Rudin uses the same argument to show that a limit point of a set is the limit of a sequence in that set.

Addendum To answer Gro-Tsen's question in the comment below: part of that magic can be found in Exercise 3.2.(b) of Engelking's General Topology, the original source is Note de tératopologie. II (Rev. Sci. 77, 180-181 (1939)) by Bourbaki and Dieudonné. The formulation that Herrlich uses is: if $D$ is dense in $X$ and $f:X\to Y$ is a map, where $Y$ is regular then $f$ is continuous iff it is continuous on the subspace $D\cup\{x\}$ for every $x\in X$.

The second part of the magic is that Herrlich only works in $\mathbb{R}$. First you fix an enumeration of $\mathbb{Q}$ and show, for every $x$ that if $f\mathbin\upharpoonright(\mathbb{Q}\cup\{x\})$ is sequentially continuous at $x$ then it is is $\varepsilon$-$\delta$-continuous at $x$ by letting $p_n$ be the first rational after $p_{n-1}$ with $|x-p_n|<1/n$ and $|f(x)-f(p_n)|\ge\varepsilon$.

That argument shows that the equivalence holds for separable metrizable domains.

For larger metrizable spaces you'll need a well-orderable dense set to make this argument go through.

Answer 2

Halmos is not alone in proving the countable additivity of Lebesgue measure without explicitly mentioning (countable) choice. I have the third edition of H. L. Royden's Real Analysis in front of me. In Chapter 3, Section 2 ("Outer Measure"), the proof of Proposition 2 says, "the collection … is countable, being the union of a countable number of countable collections." The axiom of choice is not mentioned. (Later on in the same chapter, when proving the existence of a non-measurable set, Royden does mention the axiom of choice explicitly.)

While I haven't checked, I expect that it is the norm rather than the exception for textbooks to mention the axiom of choice explicitly when proving the existence of a non-measurable set, but to omit mentioning countable choice explicitly, even though the latter is essential to the standard development of measure theory.

Answer 3

The axiom of dependent choice is sometimes implicitly used in the context of chain conditions in algebra.

If $(X,\le)$ is a partially ordered set, then we say that it is well-founded if every nonempty subset of $X$ has a minimal element with respect to $\le$. (This is different to asking that $(X,\le)$ is well-ordered. A partially ordered set is well-ordered if and only if it is well-founded and totally ordered.)

Theorem (ZF+dependent choice). Let $(X,\le)$ be a partially ordered set. Then, $(X,\le)$ is well-founded if and only if there does not exist a descending chain $a_0>a_1>a_2>\cdots$ of elements of $X$.

Proof. $(\Rightarrow)$ This direction doesn't require any choice whatsoever. Suppose a descending chain $a_0>a_1>a_2>\cdots$ does exist. Then, $\{a_0,a_1,a_2,\dots\}$ does not have a minimal element.

$(\Leftarrow)$ If $(X,\le)$ is not well-founded, then there exists a nonempty subset $E$ of $X$ which does not have a minimal element. That is, for all $x\in E$ there is a $y\in E$ such that $x>y$. By applying the axiom of dependent choice to the relation $>$, we see that a descending chain of elements of $E$ exists. QED.

Taking $X$ to be the set of ideals of a commutative ring $R$, and $\le$ to be the relation $\supseteq$, we derive the following corollary.

Corollary. The following definitions of a Noetherian commutative ring $R$ are equivalent in the presence of dependent choice: (i) any nonempty collection $\mathcal A$ of ideals of $R$ has a maximal element with respect to $\subseteq$; (ii) there does not exist a chain $\mathfrak a_1\subsetneq\mathfrak a_2\subsetneq\mathfrak a_3\subsetneq \cdots$ of ideals of $R$.

(Note that the chain $\mathfrak a_1\subsetneq\mathfrak a_2\subsetneq\mathfrak a_3\subsetneq\cdots$ is descending from the point of view of the relation $\supseteq$, even though it is usually called ascending in commutative algebra.)

Actually, both the theorem above and the corollary also appear in Atiyah-Macdonald (Chapter 6 on Chain Conditions), giving us another place where dependent choice is implicitly used in that book.