# Why those who dislike the fact that the reals are uncountable always attack Cantor's diagonal proof?

For reasons that I never understood, the fact that there are several sizes of infinity and, in particular, the fact that the reals are uncountable is unacceptable for certain persons. My question is: why is it that those who try to refute that assertion always (or so it seems to me) point their guns at Cantor's diagonal proof? After all:

1. Cantor himself proved (before creating the diagonal proof) that the reals are uncountable by another method (based upon the fact that, in $\mathbb R$, every bounded monotonic sequence converges).
2. There are other proofs that the reals are uncountable.

A few years ago, Wilfrid Hodges, a logician, wrote an interesting article about nearly the same question, called An Editor Recalls Some Hopeless Papers, but his article was about the validity (or lack thereof) of certain “refutations” of Cantor's diagonal argument. But my question is: why don't they try to refute the other arguments? Is it just because the diagonal argument is so well-known? Or is there some other reason?

• I don't have any sources for this, but I suspect you have things somewhat backwards. Most of these people don't start thinking the reals are countable and then attack the proof; they don't understand the diagonal proof that is shown to them, and they attack it. – Mark S. Aug 19 '17 at 14:55
• @MarkS. That's an interestaing possibility indeed. – José Carlos Santos Aug 19 '17 at 14:57
• I think that part of the answer is indeed because it is the best known one ... and many people (including myself!) believe it was the first one, i.e they believe it is the one that, historically, upset the apple cart. But the nature of the proof has a lot to do with it as well ... we can conceptually relate to it (that is, there is not a lot of complex math involved), yet also easily misunderstood (because of the proof by contradiction style), and it is one that is easily misapplied to 'show' that the set of all natural numbers is not countable either. So yes, it is an easy and obvious target. – Bram28 Aug 19 '17 at 15:01
• @JoeJohnson126 I don't see where AC comes in when for every given $f\colon A\to P(A)$, we explicitly construct $B\subseteq A$ such that $f(x)\ne B$ for all $x\in A$. – Hagen von Eitzen Aug 19 '17 at 15:02
• @JoséCarlosSantos Aha! So you can put me in that group of people that mistakenly believe that it was the first one :) And I am sure I am not the only one ... so as long as many believe it was the first one, it's still a target – Bram28 Aug 19 '17 at 15:03