Talk:Dedekindinfinite set
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Hello, I'm rather new to Wikipedia, so forgive me if this isn't the place for what I'm about to write:
I thought it might be a good idea to add here an interesting implication of the existence of infinite, though not Dedekind infinite, sets: the existence of 2 such sets, A and B, with A's cardinality being strictly smaller than B's, and the existence of a function f: A>B which is surjective. This being rather counterintuitive to the "normal" idea of cardinality, I think this is the main reason why the existence of such sets (infinite but not Dedekindinfinite), only true without AC, can be taken as an argument for AC, as mentioned in the article. Alex Jan 10 2005
 Actually, it would not really be an argument in favor of AC, it would be an argument in favor of an extremely weaker statement than AC, namely something weaker than even countable choice. And there are lots of convincing "reasons" from classical analysis why one would want countable choice.
inconsistency
In the section "Dedekindinfinite sets in ZF", it is written that
 Every Dedekindinfinite set A also satisfies the following condition: [...] This is sometimes written as "A is dually Dedekindinfinite".
And a few sentences below:
 It is not provable in ZF that every Dedekind infinity implies dual Dedekind infinity.
The second statement, apart from being formulated in a rather strange way, directly contradicts the first statement. — Emil J. 10:43, 27 April 2009 (UTC)
The first statement is the correct one: we can define a surjective and not injective function on A by taking a suitable "shift" on a countably infinite subset of A, and identity elsewhere. — Emil J. 14:07, 27 April 2009 (UTC)
Boundary?
I'm definitely not an expert but the italics doesn't make any sense to me:
With the general acceptance of the axiom of choice among the mathematical community, these issues relating to infinite and Dedekindinfinite sets have become less central to most mathematicians. However, the study of Dedekindinfinite sets played an important role in the attempt to clarify the boundary between the finite and the infinite, and also an important role in the history of the AC.
What does this mean? From my understanding of this article Dedekind finite means there does not exist a function in the model to make it Dedekind infinite. These models are definitely not "maximal", in that certain functions that "should" exist, don't exist, right? Hence infinite sets are always Dedekind infinite (intuitively of course, because for one the negation of AC implies that certain sets that "should" exist, don't exist). Besides that, in a twelve year old's sense, I honestly don't think there's something less infinity and greater than all the natural numbers. My reasoning of this is just like the proof of how there are no infinitesimal in complete Archimedean fields (I'm aware of nonstandard analysis but that's a bit too difficult for me) Breath of the Dying (talk) 04:14, 10 September 2009 (UTC)
Ring?
Heldergeovane (talk · contribs) added "A ring is Dedekindfinite if xy = 1 implies yx = 1." to the lead, and gave a reference. I will revert this edit for the following reasons: (1) this is an article about set theory, not algebra, so focusing on rings (especially in the lead) is inappropriate; and (2) the statement appears to be false as it stands (perhaps it could be corrected, I do not know). A simple counterexample is the real numbers, a ring, indeed a field. Clearly this is an infinite set includes the natural numbers and is thus not Dedekindfinite; and clearly it satisfies xy=1 implies yx=1 because its multiplication is commutative. JRSpriggs (talk) 03:13, 30 June 2010 (UTC)
 What the refs appear to be doing is this: Formulate Dedekind finiteness arrowtheoretically in the category of sets, then apply that same definition in the category of rings, and (rather surprisingly!) this is what you get. Seems to be wellreferenced enough, and closely enough related (even if in a strange way), that it should be mentioned. I agree that care needs to be taken to explain that this is not the same notion, but an analogous notion in a different category (e.g. the reals are not Dedekind finite as a set, but apparently they are as a ring). Trovatore (talk) 03:58, 30 June 2010 (UTC)

 Then perhaps you could restore his references and put a section at the bottom about this variant notion of "Dedekindfinite". JRSpriggs (talk) 06:27, 30 June 2010 (UTC)
 I apologize for the inconvenience. I found this article because Dedekind finite redirects here and I was studing this concept in the context of rings. Unfortunatelly, I wasn't able to write a better explanation of it, though I think it's worth to mention it in this article... Helder (talk) 17:14, 30 June 2010 (UTC)
 Then perhaps you could restore his references and put a section at the bottom about this variant notion of "Dedekindfinite". JRSpriggs (talk) 06:27, 30 June 2010 (UTC)



 I tried to integrate the definition in a less confusing way. Perhaps you can all have a look to see if I was successful and make sure I didn't get anything wrong. Hans Adler 07:54, 1 July 2010 (UTC)
 Thanks to EmilJ for fixing an embarrassing mistake in my formulation. Hans Adler 23:16, 12 July 2010 (UTC)


Inadequate proof
The proof in section Dedekindinfinite set#Proof of equivalence to infinity, assuming axiom of countable choice fails to show how the elements of the finite sets united into U are ordered. However, a real proof is available (from me) at Talk:Finite set#Countable choice is enough. The key difference is using sets of injections (or other structures) from n+1 to X rather than just subsets of cardinality n. JRSpriggs (talk) 21:02, 16 September 2012 (UTC)
 If the proof is bad then it should definitely be removed. This is the type of proof we are supposed to avoid including in articles anyway. Ideally, one of us would have a text to reference it in. Do you happen to have one, JRS? Rschwieb (talk) 00:26, 17 September 2012 (UTC)
The lead
From the lead:
 The axioms of Zermelo–Fraenkel set theory without the axiom of choice are not strong enough to prove that every set that is Dedekindfinite is finite in the sense of having a finite number of elements.
This is because it isn't true in ZF (not only not provable). I understand what is meant, but if this sentence is to be kept, it needs to be explicitly established that ZFC is assumed and that the statement is true in ZFC. YohanN7 (talk) 20:43, 6 February 2014 (UTC) Should add as well that I think the sentence should stay. YohanN7 (talk) 21:17, 6 February 2014 (UTC)
 What do you mean, "it isn't true in ZF"? Truth is relative, if anything, to models, not to theories. Theories only prove or disprove; they don't decide truth. Trovatore (talk) 21:57, 6 February 2014 (UTC)
 I mean with my post that there is a model of ZF set theory in which there are infinite Dedekindfinite sets (hence the "not true in ZF"). Thus the statement can hardly be provable in ZF. You know I'm not a set theorist. I understand that my formulation hurts your ears (or eyes) a bit, but I'm simply not used to the lingo. YohanN7 (talk) 22:32, 6 February 2014 (UTC)
 Oh. Well, that's actually equivalent to "not provable in ZF", because of Gödel's completeness theorem. My view is that the "intended model" of ZF is the same as the intended model of ZFC. ZF is just weaker, not strong enough to be able to prove everything that ZFC can. Trovatore (talk) 22:40, 6 February 2014 (UTC)
 Ok, but I'll have to digest "Well, that's actually equivalent to "not provable in ZF", because of Gödel's completeness theorem". But do you agree that the lead needs to establish its context?
 B t w, I didn't come here to critically read the article, I came for some info (and found it). YohanN7 (talk) 23:14, 6 February 2014 (UTC)
 Oh. Well, that's actually equivalent to "not provable in ZF", because of Gödel's completeness theorem. My view is that the "intended model" of ZF is the same as the intended model of ZFC. ZF is just weaker, not strong enough to be able to prove everything that ZFC can. Trovatore (talk) 22:40, 6 February 2014 (UTC)
 I mean with my post that there is a model of ZF set theory in which there are infinite Dedekindfinite sets (hence the "not true in ZF"). Thus the statement can hardly be provable in ZF. You know I'm not a set theorist. I understand that my formulation hurts your ears (or eyes) a bit, but I'm simply not used to the lingo. YohanN7 (talk) 22:32, 6 February 2014 (UTC)
 I have tried to add some context around that sentence (that I added myself to the lead some time ago). Is it better now? —Tobias Bergemann (talk) 09:50, 8 February 2014 (UTC)
 Hmm, not bad, except for my standard reservations about how "naive set theory" is treated. Trovatore (talk) 09:53, 8 February 2014 (UTC)
 Too critical? The term "naive set theory" itself is a bit problematic, I think, even if it is wellestablished (by Paul Halmos?). Of course, mathematicians before Zermelo were not "naive" or "careless" or "negligent". However, they were not always very explicit with their use of arbitrary choices (or denumerable choices, or dependent choices, or...). And Cantor himself for a long time did not believe that a definition of finite sets was even necessary.
 Of course, feel free to change that treatment in any way you see fit. —Tobias Bergemann (talk) 10:48, 8 February 2014 (UTC)
 Hmm, not bad, except for my standard reservations about how "naive set theory" is treated. Trovatore (talk) 09:53, 8 February 2014 (UTC)



 My actual concern is not about the term naive being "critical" exactly. It's that it feeds into this narrative that informal Cantorian set theory was responsible for the paradoxes. I don't think that's true, at least not quite the way people think it is. But it's a complicated discussion probably out of place here.
 Anyway, I don't have a better suggestion at the moment, just a sort of inward cringe when I see that narrative being even indirectly reinforced. Trovatore (talk) 21:55, 8 February 2014 (UTC)


(Co)Hopfian
Dedekind finite sets are the coHopfian objects in Set. Dually Dedekind finite sets are the Hopfian objects in Set. GeoffreyT2000 (talk) 14:56, 16 February 2015 (UTC)