Axiom of Choice: Can someone explain the fallacy in this reasoning?

Question

Not a "set theory" guru (apologies if my terms are imprecise), but I have heard that it is an elementary result that the set of rational numbers has a measure of zero - intuitively meaning that the set of rationals is "less dense" than the set of real numbers (or equivalently than the set of irrational numbers).

However, I read this article which seems to provide a rather simple proof showing that the rationals have a cardinality at least as big as the irrationals, which would contradict the above paragraph. So at least one of the results should be wrong? However I cannot find the fallacy in the reasoning in the article - the author places the blame on the axiom of choice (bit of a whipping boy of set theory I think).

Any ideas? or is the author's argument sound?

Answer 1

The author's argument is, of course, not solid.

His main argument proceeds as follows: let $(\mathbb{X},\preceq)$ be a well ordering of the positive irrational numbers.

He then attempts to construct, through transfinite induction, an embedding of $\mathbb{X}$ into a subset of $\mathbb{Q}\cap[0,\infty)$ as follows: let $\zeta\in\mathbb{X}$.

• Assuming that all elements $\xi\prec \zeta$ have been mapped into $\mathbb{Q}$, there exists $q\in\mathbb{Q}$ such that either $q=0$ or $q$ is the image of some $\xi\prec\zeta$. Let $q'$ be any rational such that $q\lt q'\lt \xi$, and $q\neq 0$ and $q$ is not the image of any $\xi\in\mathbb{X}$ with $\xi\prec \zeta$. Define the image of $\zeta$ to be $q'$.

He claims that $q'$ will always exist "because there is always a rational between any two real numbers". But this assertion is empty: the existence of such a rational does not imply the existence of a rational that has not yet been chosen. The author never actually establishes the existence of such a $q'$.

Added. Note also that your first paragraph is not very apt. While (Lebesgue) measure is very weakly connected to cardinality (in that any set that is countable must have Lebesgue measure $0$), the connection is not very good. The Cantor set has measure $0$, but has the same cardinality as the entire real line; and the Cantor set has empty interior, so it is very much "not dense".

Answer 2

Have you never seen a simple proof that the cardinality of the rationals is less than that of the irrationals? It's easy to find such proofs in texts or on the web, they are quite satisfactory in every way. There is no need to look at the article to which you refer. It must be wrong, and it is the author's job, not yours or mine, to find out where the mistake is.

Answer 3

This is a work of someone who clearly does not understand the idea behind the axiom of choice and ordinal arithmetics.

For to address the minor issue your first paragraph has, the rational has measure zero because singletons have measure zero, and the measure is countably additive, that is a union of countably many disjoint sets is the sum of the measure.

Now to address the "paper".

Firstly he is very mistaken in the part where he claims that cuts are characterized by their end points. How can you characterize the rational numbers smaller than $\pi$? It has no end point. The thing is that those are subsets, as the excerpt from Levy's text say. And Cantor's theorem says that $|P(X)|>X$ (a theorem whose proof requires no choice whatsoever), so there are many more subsets (cuts) than end points, and possibly many of those are unrealized cuts/gaps (i.e. irrational numbers).

This shows that it is very possible to have many many more cuts than end points within the actual ordering.

As for the bijection, the writer has no grasp about the idea of infinitary processes. Consider the following:

Take an enumeration of $\mathbb{Q}=\{q_n\mid n\in\omega\}$. Now take an enumeration of the irrationals such that the first $\omega$ many are those of the form $\{q_n+\pi\mid n\in\omega\}$ and then enumerate the rest of the irrationals as you wish. The bijection described in the article will exhaust exactly after $\omega$ steps while you still have at least $2^{\aleph_0}$ many steps to go with the irrationals.

In order to work with infinite well-ordered sets one needs to say what happens at limit points, i.e. steps which has no preceding ones, this is not handled in the construction and he just says "continue until $\mathbb X$ is exhausted".

This is similar to proving that every ordinal number is finite:

Start with $0$, finite. Then assume $n$ is finite therefore $n+1$ is finite. Continue until you exhaust the class of ordinal numbers. Therefore all ordinals are finite.

This is wrong because eventually we exhaust natural numbers and we find ourselves at the realm of infinite ordinals, as we did not specify what is going to happen at the limit stages, this induction can (and will) fail at the first limit point - $\omega$.