Examples of Axiom of Choice used in introductory-level undergradute math

Question

All of the applications of AoC I've encountered have been in upper level undergraduate or graduate math courses. Are there any basic results from courses like Calc I-III which (unbeknownst to students) rely on AoC?

Answer 1

The axiom of choice can be used to prove that the sequential definition of continuity at a point (for real functions of a real variable) is equivalent to the $\varepsilon$-$\delta$ definition. If your calculus textbook proves that sequential continuity implies epsilon-delta continuity without mentioning the axiom of choice, it's doing something like this:

. . . Then, for each $n\in\mathbb N,$ there is a real number $x_n$ such that $|x_n-x_0|\lt\frac1n$ while $|f(x_n)-f(x_0)|\ge\varepsilon.$ Thus the sequence $x_n$ converges to $x_0,$ while $f(x_n)$ does not converge to $f(x_0)$. . .

Do you see where I used the axiom of choice?

Answer 2

The Axiom of Choice can be used to prove that every vector space has a basis. That's certainly a basic result from a linear algebra course but it's definitely not unbeknownst to students.

Answer 3

A bounded, closed, convex subset of $\Bbb{R}^n$ has extreme points.
Amazingly we need (a weaker version of) axiom of choice to prove this simple fact and this has many applications. For example: To solve a linear program problem only needs to consider extreme points of its feasible region.

Answer 4

Here are two which you usually see in the first chapter of most intro to analysis books, and in many "quick set theory coverage" of the first couple of weeks:

1. Every infinite set has a countably infinite subset. Or equivalently, a set is finite if and only if every proper subset has smaller cardinality. Or equivalently, a set is infinite if and only if there is an injection which is not a surjection from the set to itself.

   You need strictly less than countable choice to prove this, even though the standard and easy proof is using a bit more. But this is certainly not provable without any use of choice. Not even if these are sets of reals.

2. The countable union of countable sets is countable. Well, even the real numbers can be a countable union of countable sets if choice fails badly enough.

If it is up to me, I'd also include a third, but I am not sure if fits your criteria:

3. A function is surjective if and only if it has a right inverse. This one is in fact equivalent to full choice, and it is often used in basic level courses like discrete mathematics and stuff like that. And it makes sense, but for infinite sets you generally still need choice.

Answer 5

A typical example that requires the axiom of choice in full is that every field has an algebraic closure. One can prove without the axiom of choice that every countable field has an algebraic closure, and also that the reals have an algebraic closure (say by using complex analysis instead), which means that for practical purposes one does not appear to need the axiom of choice. But in general you do need choice.

Answer 6

Note I posted this answer with a big delay and now see many answers. Just leaving it up now as a 'summary' answer.

The following assumes that we do not learn about the the Lebesgue integral in any undergraduate math class. Yes, the professor's talk about it, but leave it as extra credit (see comments below). Of course any undergraduate interested in quantum mechanics should examine that material; see When is Lebesgue integration useful over Riemann integration in physics? over at physics.stackexchange.com. For more, see Lebesgue theory and axiom of choice on this site.

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Do you need FULL $\mathsf {AoC}$ in undergraduate courses?
Answer (necessarily a bit subjective):

$\quad$ Multivariate calculus - no
$\quad$ Linear algebra - yes (for those infinite dim spaces), but 98% of coursework not dependent on it
$\quad$ Abstract algebra (Junior level?) - perhaps

If you are taking a real-analysis courses, you might be interested in

In mathematics, the axiom of dependent choice, denoted by $\mathsf {DC}$, is a weak form of the axiom of choice ($\mathsf {AC}$) that is still sufficient to develop most of real analysis.
It was introduced by Bernays (1942).

See wikipedia - Axiom of dependent choice

For example, you need $\mathsf {DC}$ to prove the following statement:

If $(p_n)$ is a sequence in a compact metric space $X$, then some subsequence of $(p_n)$ converges to a point of $X$.

$\quad$ (see for example Rudin's Theorem 3.6)

Most undergraduate students in advanced calculus will see the implicit use of $\mathsf {DC}_{\mathbb R}$, but you can't beat calculus for those math-intuition-insight correlations.

Answer 7

Already mentioned in the answer by @Asaf Karagila:

The Axiom of Choice is required to prove a basic fact from elementary set theory, commonly used in calculus:

Let $f : A \to B$ be a surjection. Then $f$ has a right inverse, i.e. there exists a function $g : B \to A$ such that $f \circ g = \mathrm{id}_B$.

Proof:

Since $f$ is surjective, $f^{-1}(\{y\})$ is nonempty for every $y \in B$. Hence, $\{f^{-1}(\{y\}) : y \in B\}$ is a family of nonempty pairwise disjoint sets. Using The Axiom of Choice, choose $x_y \in f^{-1}(\{y\})$, and define $g(y) = x_y$ for every $y \in B$. We have $f \circ g = \mathrm{id}_B$, by definition of $g$.

In contrast to this, the converse claim does not require Choice:

Let $f : A \to B$ be right-invertible. Then $f$ is surjective.

Proof:

Since $f$ is right-invertible, there exists $g : B \to A$ such that $f \circ g = \mathrm{id}_B$. Let $y \in B$, and consider $g(y) \in A$. We have $f(g(y)) = y$, by definition of $g$. Therefore, $f$ is surjective.

A similar fact that $f$ is injective if and only if it is left-invertible also follows without Choice.

Answer 8

I realise this is not at all at basic undergraduate level (in fact, it is rather technical), but I think it is a good addition to the results mentioned above.

The existence of a non Lebesgue measurable set is independent of ZF set theory and requires Axiom of Choice. This is a classical result due to Solovay.

However Thychonoff's Theorem, that the product of any collection od topological spaces is compact with respect to product topology is a classical undergraduate level result which is equivalent to Axiom of Choice (apologies if this is mentioned earlier).

Answer 9

Again, more upper-level undergrad, or intro grad level: "in a commutative ring, every proper ideal is contained in a maximal ideal" is equivalent to AC.