# Comments on From Vector Spaces to Periodic Functions

## Hubryan said:

Linear algebra, all powerful.

## Eschborn said:

This is a nice example for why the axiom of choice is controversial. The proof from this blog post only works because of it. However, the result seems counter-intuitive. Let us have a look at whether this is only our intuition being off or whether functions with these properties are just "weird".

When speaking of periodic functions the first that comes to mind is the \( \sin \) function. Can something similar to such a function be used to construct the functions \( g \) and \( h \) from the blog post? If we define them similar to \( \sin \) as periodic and continuous, then the answer is no. Every periodic and continuous function has a maximum value. The sum of two functions with a maximum value has a maximum value. However, the identity is unbounded. The sum can therefore not be the identity.

Okay, so \( \sin \)-like functions are out. What about \( \tan \)-like functions? Take \( \tan(x) \) and \( -\tan(2x) \) for example. Those are not continuous everywhere. Well, we can also rule out many functions like this. Suppose \( g \) and \( h \) are two periodic functions with period lengths \( a \) and \( b. \) If \( \frac{a}{b} \) is rational, then the sum of \( g \) and \( h \) cannot be the identity function. Suppose that \( \frac{a}{b} \) was rational, then \( ab \) is a common period of both functions, i.e., \( g(x) = g(x + ab) \) and \( h(x) = h(x + ab). \) It follows that \( (g + h)(x) = (g + h)(x + ab). \) As \( a \) and \( b \) are non-zero, it follows that \( g + h \) cannot be the identity function. Functions of the form \( a \tan(bx) \) for rational \( b \) thus do not work. Any two periodic functions where you can reason about their sum by just looking at what happens for individual periods are thus ruled out.

My conclusion is that the two periodic functions from the blog post must look "weird".

A common theme with axiom of choice proofs is also that they show the existence of something but give no hint at how this something actually looks like. The proof in this blog post is a good example of this. We know that the functions "exist" but have absolutely no clue about how they look like. It is also an open question whether describing them in any other way than "the function from this proof" is even possible.

## Andi said:

Jerry Bona said, "The axiom of choice is obviously true, the well-ordering principle obviously false, and who can tell about Zorn's lemma?"

## MathematicsPadawan said:

The axiom of choice is allowed.

Why does this need to be stated explicitly? Why does the axiom of choice not enjoy the privilege of being a first-class axiom like other axioms? Why do many proofs need to state the assumption of this axiom explicitly before using it?

## GenerationP said:

MathematicsPadawan, Because it gets you applications like this (or the Banach-Tarski paradox), which make you wonder how much it deserves its place. To some extent this is more due to Tertium Non Datur than to the axiom of choice, but my impression is ZF doesn't know enough about infinite sets to get to the really counter-intuitive stuff.

## QSort said:

MathematicsPadawan, Proofs (at least when they are written "the usual" way) are human-to-human communication, not human-to-machine encoding. A proof depending on the axiom of choice is a more interesting fact than it depending on, say, the axiom schema of specification, so it makes sense that the author might want to say "hey, pay attention to that".

## Hansvm said:

GenerationP,

There are counter-intuitive results in ZF without choice too:

- Collection of non-empty sets with empty cartesian product.
- Infinite set without a countable infinite subset.
- There is a pair of sets such that neither is equinumerous with a subset of the other.

More or less, our intuition about seemingly obvious ideas is completely thrown off without choice. Banach-Tarski at least has the property that it probably doesn't directly apply to the real world (if you can split an object into probably physically impossible sets and then rejoin them correctly then you can double the volume of the physical object) and so doesn't really violate our intuition — the premise doesn't apply in the real world, so no conclusion really matters. It's like claiming that every element of the empty set is a leprechaun with a pot of gold — it's true, but it doesn't matter in any meaningful sense.

## Im3w1l said:

Hansvm, If I understand correctly, your examples
are *undecidable* rather than true in ZF. That's why we can
add the axiom of choice without getting contradictions.

## 6gvONxR4sf7o said:

I seem to be missing something. I'm reading this as looking for functions \( f \) and \( g \) from \( \mathbb{R} \) to \( \mathbb{R}, \) such that \( f(x) + g(x) = 1 \) and \( f(x + p) = f(x). \) If I were reading that correctly, then for any periodic \( f(x), \) \( g(x) = 1 - f(x) \) would do, meaning any old periodic function would do. What am I missing?

Am I also reading it right that the author is saying the axiom of choice is equivalent to the statement that every vector space has a finite basis? I don't get how that allows infinite dimensional vector spaces. If not, and it's just that every vector space has a basis that maybe infinite, then what's the justification of the Hamel basis being finite?

I feel like I'm missing a lot between the lines here.

## Shakow said:

6gvONxR4sf7o said:

I'm reading this as looking for functions \( f \) and \( g \) from \( \mathbb{R} \) to \( \mathbb{R}, \) such that \( f(x) + g(x) = 1 \) … What am I missing?

6gvONxR4sf7o, Identity function means that \( f(x) = x, \) not \( f(x) = 1. \)

## 6gvONxR4sf7o said:

Shakow, Doh. Thanks.

## Hansvm said:

Im3w1l, Well, sure. Without choice there exist models of ZF where those properties hold. Adding choice precludes those models — i.e., it restricts our axioms in a way which prevents certain kinds of unexpected results.

## LopsidedBrain said:

MathematicsPadawan, It is partly historical notoriety. It allows one to prove certain statements that seem counter-intuitive. So proofs that employ it can often be viewed with some skepticism.

Moreover, this axiom is *independent* of the other axioms in
ZFC. It is in fact possible to have entirely self-consistent
"worlds" of mathematics, ones where the axiom of choice is true, and
ones where it is false.

More details and examples of alternate axioms are in the Wikipedia article: Axiom of choice.

If it seems weird that mathematics can give you contradictory results, remember that the difference only shows up when you deal with some form of infinity (e.g., when performing an operation on an infinitely large set). For any usage of mathematics in the real world, the truth or falsity of this axiom won't give you contradictory results.

## Kevin Milner said:

6gvONxR4sf7o said:

Am I also reading it right that the author is saying the axiom of choice is equivalent to the statement that every vector space has a finite basis?

6gvONxR4sf7o, Not quite. The axiom of choice is equivalent to saying every vector space has a Hamel basis, which is to say every element can be represented as a finite combination of elements of the Hamel basis. It doesn't imply that the Hamel basis is finite itself.

## Phillip B said:

Very nice solution! It is quite interesting that this does not work for \( e^x. \) I wonder if there is any general way to tell for any function.