The abstract for a paper on arxiv.org (http://arxiv.org/pdf/math/0408089v3.pdf) reads (with my emphasis):

"Transfinite set theory including the axiom of choice supplies the following basic theorems: (1) Mappings between infinite sets can always be completed, such that at least one of the sets is exhausted. (2) The real numbers can be well ordered. (3) The relative positions of real numbers which are enumerated by natural numbers can always be determined, in particular the maximum real number below a given limit. (4) Any two different real numbers are separated by at least one rational number. These theorems are applied to map the irrational numbers into the rational numbers, showing that the set of all irrational numbers is countable."

It concludes: "[W]e can say that there are no different infinities. If the axiom of choice is abolished, then well-ordering of the continuum and of larger sets is impossible, and there is no chance of attributing a cardinal number to those sets. If the axiom of choice is maintained then the continuum can be proved countable, also contradicting transfinite set theory."

I was just getting comfortable with $\omega$, $\omega+1$, $\omega 2$, $\omega^{2}$,$\omega^{\omega}$,$\epsilon_{0}$,$\Gamma_{0}$,$\Omega$, and even $\Omega_{\Omega}$. Is there really only one $\infty$?

P.S. I am still relatively new here: if it's inappropriate to discuss the literature here, or I didn't do it right, please let me know and I'll do better next time.


closed as off-topic by Andrés E. Caicedo, Will Jagy, user147263, Did, Camilo Arosemena-Serrato Jul 10 '15 at 0:27

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  • "This question is not about mathematics, within the scope defined in the help center." – Andrés E. Caicedo, Will Jagy, Community, Did, Camilo Arosemena-Serrato
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    $\begingroup$ What Mückenheim claims is wrong. Whether it originates from a true misunderstanding of (canonical) set theory, or is an intentional misrepresentation is anyone's guess. There may be contradictions in set theory, but what Mückenheim writes does not expose any. $\endgroup$ – Daniel Fischer Jul 9 '15 at 13:47
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    $\begingroup$ I think your post is fine. It seems like he doesn't really know what he's talking about. Particularly, he seems to think that if you have a map from one set to another, that the cardinalities of the sets somehow are related. This is not even close to being true. You need injectivity to conclude something like that. $\endgroup$ – Cameron Williams Jul 9 '15 at 13:48
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    $\begingroup$ Also do not downvote this post simply because you disagree with the author of the paper on arXiv. This is a perfectly acceptable post. Downvoting like this is rampant on other sites (reddit) and really ruins the integrity of the site. $\endgroup$ – Cameron Williams Jul 9 '15 at 13:49
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    $\begingroup$ See meta.math.stackexchange.com/q/2290/442 and meta.math.stackexchange.com/q/16736/442 for discussion of whether posts like this one should be allowed. $\endgroup$ – GEdgar Jul 9 '15 at 14:21
  • $\begingroup$ Michael, you should perhaps edit your post to be more in line with the guidelines on how to ask questions about preprints. $\endgroup$ – Asaf Karagila Jul 10 '15 at 4:40

This author is notorious for his claims about set theory, and mainstream mathematics does not take them seriously.

For a detailed criticism of his style of argument, see this review (in German) by Franz Lemmermeyer of a book he wrote.

Regarding the specific text you link to, arguments like this:

The rational numbers are countable whereas the irrational numbers are uncountable. It is argued that they make up intervals of the continuum separated by rational numbers which are only points. The set of intervals formed by these points necessarily is countable too. In order to support the idea of uncountably many irrational numbers, we should be able to find at least one interval containing uncountably many of them.

should give you a pretty clear idea of whether you want to take this seriously or not.

  • $\begingroup$ Maybe I'm a bit too sleepy, but I don't see anything exactly wrong with the quoted statement - unless he means disjoint intervals, as I'm not aware of an order-preserving enumeration of the rationals. Is the issue that he's ignoring that the rationals are dense in the reals? $\endgroup$ – G. Bach Jul 9 '15 at 19:11
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    $\begingroup$ @G.Bach: Yes, he means disjoint intervals. Yes, he really is that clueless. (He infests the Usenet group de.sci.mathematik.) $\endgroup$ – Brian M. Scott Jul 9 '15 at 20:10
  • $\begingroup$ @G.Bach: by my reading, he appears to be stating that the proof of uncountability of rationals requires believing that there are intervals in the real line with rational endpoints but no rationals in the interior. Of course, nobody has ever claimed that, and the usual proof does not use any such assumption. I'm sure the author would argue that my reading is flawed, but to me that just goes to show that this kind of writing is too vague to count as mathematics. $\endgroup$ – Schemer Jul 9 '15 at 21:48
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    $\begingroup$ The author would argue that $\Bbb R$ does not equal to $\bigcup_{r\in\Bbb R}\{r\}$. So it really doesn't matter what the author argues. $\endgroup$ – Asaf Karagila Jul 9 '15 at 22:44

I was bored, so I read the paper in order to figure out what's wrong with it. The paper purports to create an injection from $\mathbb{X}_+$ (the positive irrationals) to $\mathbb{Q}_+$ (the positive rationals) by repeatedly removing one element from each set. The glaring flaw I found was this (emphasis preserved):

If the set $\mathbb{Q}_+$ were exhausted prematurely and no $q_n$ remained available to map $ξ_n$ on it, this proof would fail. We would leave the countable domain and could no longer make use of Cantor's "Lagenbeziehung" to select the largest rational number $q \in Q$ with $q < ξ_n$. But that cannot occur because there is always a rational number between two real numbers.

This argument doesn't seem to be justified, and it is indeed invalid. Supposedly, $\mathbb{Q}_+$ will not be exhausted before $\mathbb{X}_+$, and so every element of $\mathbb{X}_+$ will successfully be mapped onto an element of $\mathbb{Q}_+$. In fact, $\mathbb{Q}_+$ is exhausted first, and so the proof does fail.

This paper, in a footnote, also purports to refute Cantor's diagonal argument. Unsurprisingly, this refutation doesn't make sense, either. Here it is (emphasis preserved):

By the way, here lies the reason why Cantor's diagonal argument must fail, nevertheless. Every number smaller than $ω$ is a finite number, and it is surpassed by other finite numbers [Cantor, p. 406]. Hence, if the list contains "every number smaller than $ω$", then there are other numbers, not contained in the list. The list is not complete. Further, for finite numbers the miracle of the discrepancy between ordinal and cardinal number is void of its witchcraft. Hence the cardinal number of the lines of the list is finite, however large the finite ordinals may be.

The last sentence does not follow.

  • $\begingroup$ Muckenheim seems to be confused.Perhaps he doesn't understand that an enumeration of the rationals as q(0),q(1),q(2)... etc cannot preserve the arithmetic order. And his argument on Cantor's diagonal method is incoprehensible to me.I think it just gibberish. Pathologica Mathematica. $\endgroup$ – DanielWainfleet Aug 2 '15 at 2:40

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