Talk:Continuum hypothesis

message: I've removed this section:
:If a set S were found that disproved the continuum hypothesis, it would be impossible to make a one-to-one correspondence between S and the set of integers, because there would always be elements of set S that were "left over". Similarly, it would be impossible to make a one-to-one correspondence between S and the set of real numbers, because there would always be real numbers that were "left over".
It's a good thing to add some intuition, but I don't know that this passage helped much. CH isn't about whether such a set S can be found in any ordinary sense, but just about whether one exists. --Trovatore 21:12, 22 October 2005 (UTC)
replies: text: CH cannot be disproved from ZFC. No, I'm afraid that's not particularly close to settling the issue. --Trovatore (talk) 08:40, 17 May 2008 (UTC)
thread_title:  Investigating the continuum hypothesis 

message: The form \neg \exist m \; \aleph_0 < m < 2^{\aleph_0} makes the continuum hypothesis meaningful even if the continuum isn't well-ordered. The article states:
:The continuum hypothesis is closely related to many statements in analysis, point set topology and measure theory. As a result of its independence, many substantial conjectures in those fields have subsequently been shown to be independent as well.
Do we have any examples which would make sense without AC, and would they depend on the form of CH? Also, on the other side of the coin, my understanding is that there could be no purely arithmetic consequences, roughly because ZF + V=L proves CH anyway (and even GCH), and relativising everything to L makes no difference to (first-order) arithmetic. If correct, I believe this is worth stating. Also, the article states:
:Assuming the axiom of choice, there is a smallest cardinal number \aleph_1 greater than \aleph_0, and the continuum hypothesis is in turn equivalent to the equality 2^{\aleph_0} = \aleph_1.
The "assuming the axiom of choice" bit should really attach to the "in turn equivalent to", because even from the axioms ZF without AC, aleph_1 exists and is an immediate successor of aleph_0, right? --Unzerlegbarkeit (talk) 02:39, 1 June 2008 (UTC)
replies: text: Without AC, there could be other cardinals which are larger than \aleph_0 but incomparable with \aleph_1, for example, the cardinality of a set which is the union of natural numbers with a Dedekind finite infinite set. In this case, neither the new cardinal nor \aleph_1 can be said to be the smallest cardinal greater than \aleph_0. JRSpriggs (talk) 06:38, 1 June 2008 (UTC)
thread_title:  Forms of CH without AC 

message: I notice there aren't any references to Cantor's original writings. I'll try to dig some up, but has anyone seen any English translated letters, etc? Libertyblues (talk) 00:03, 3 August 2008 (UTC)
thread_title:  Original Sources 

message: Edited out a minor error: ZFC stands for "Zermelo-Frankel with the axiom of choice", whereas ZF is simply "Zermelo-frankel". Would be nice to see some superscript references in this article, I'm afraid I don't know how to do that yet. 83.71.3.220 (talk) 00:41, 1 October 2008 (UTC)
replies: text: See Wikipedia:Citing sources#Footnote summary for instructions on how to put in references. JRSpriggs (talk) 01:39, 1 October 2008 (UTC)
thread_title:  Need references 

message: Cantor introduced the concept of cardinality to compare the sizes of infinite sets, and he gave two proofs that the cardinality of the set of integers is strictly smaller than that of the set of real numbers. His proofs, however, give no indication of the extent to which the cardinality of the natural numbers is less than that of the real numbers.
Shouldn't natural numbers be rational numbers? Since the natural numbers is a subset of the integers, it's quite obvious that the cardinality of the natural numbers isn't greater than the integers.  —Preceding unsigned comment added by 131.215.42.208 (talk) 23:21, 20 January 2009 (UTC)
replies: text: The passage seems to be written for readers who know without having to think about it that the cardinality of the naturals is the same as that of the integers (or, possibly, who use the word integer to mean natural number, a convention used informally by some set theorists).  Since not all readers will fall into this category, we should probably change both occurrences to be the same.  The question is, should we say integers, or naturals?  Integers sounds better (the repetition of natural numbers could be a bit stilted) but natural numbers is more likely to be strictly historically accurate when talking about what Cantor strictly speaking proved. --Trovatore (talk) 23:48, 20 January 2009 (UTC)
thread_title:  Natural numbers? 

message: I was just reading a book about Cantor. It mentioned that one of the main reasons Cantor believed in the continuum hypothesis was because he managed to show every nonempty perfect set is the same size as the reals, and then showed a closed set C is either countable or has cardinality of the reals by partitioning C into a perfect set and a countable set. And then he said "In future paragraphs it will be proven that this remarkable theorem has a further validity even for linear point sets which are not closed,..." Should this be in the article? Also I personally believe his reasoning is wrong: since the reals have a countable base, the number of open sets is the same as the reals, and thus so is the number of closed sets (because they are complements of open sets). But there are "so many more" subsets, because 2^(aleph0) is negligible compared to 2^(2^(aleph0)). Right???Standard Oil (talk) 14:28, 14 August 2009 (UTC)
replies: text: Cantor never thought he had proved CH, iirc.  Yes, a quote from him (and a little more info on the origins of the problem--Cantor did a lot of work on it) would be nice. 66.127.52.47 (talk) 11:32, 24 March 2010 (UTC)
thread_title:  Cantor's fanatism 

message: Could someone clueful please check whether this edit (by me) is any good, and revert it if necessary.  I started having doubts after making it.  Thanks.  66.127.55.192 (talk) 17:57, 17 February 2010 (UTC)
replies: text: Sounds better to me. JRSpriggs (talk) 04:19, 18 February 2010 (UTC)
thread_title:  Goedel 

message: http://citeseerx.ist.psu.edu/viewdoc/summary?doi=10.1.1.57.6594
an article explaining(?) Woodin's recent work.  I don't understand it at all.  Maybe someone here can make sense of it and possibly cite it or use something from it. 66.127.55.192 (talk) 06:24, 18 February 2010 (UTC)
thread_title:  Dehornoy article 

message: Why is this hypothesis called the "continuum" hypothesis? The proposition "There is no set whose cardinality is strictly between that of the integers and that of the real numbers." is asserting the lack of any set of size (or cardinality) between integers and real numbers, which sounds far more like the a "non-continuum hypothesis." or since it is saying that there is not a single set that stands in that range (far less a continuum of size/cardinalities of sets) the theory might even be called the "not-even-close-to-continuum hypothesis" or "discontinuum hypothesis" or "discrete hypothesis." Since continuum is a word in normal English, I guess it would help to know why this theory has the name it has. I can't understand the name, let alone the math :-) --Timtak (talk) 10:09, 14 October 2010 (UTC)
replies: text: Answering my own question, looking at the article on Continuum_(set_theory), the latter seems to be saying that the fact that there are no sets between the size (cardinality) of integers and real numbers, proves that the latter forms a continuum? I.e. the continuum hypothesis, is called such, not because it says something about a continuum of sets, but because it posits a continuum of real numbers. So perhaps by way of explanation, one might write, "'There is no set whose cardinality is strictly between that of the integers and that of the real numbers.' implies that real numbers form a continuum, and hence the name of this theory." Is that it? Or indeed, since from the the Continuum_(set_theory) article it says that the cardinality of real numbers is the cardinality of the real line (is that the same as "a or any real line"?) then the continuum hypothesis is called such because it implies reality is continuous?!--Timtak (talk) 10:18, 14 October 2010 (UTC)

text: Perhaps because it could be worded "Any uncountable subset of the continuum is equinumerous with the continuum.". JRSpriggs (talk) 11:55, 14 October 2010 (UTC)

text: Where was your teacher trained?  Tkuvho (talk) 13:32, 13 January 2011 (UTC)

text: Truncating the continuum at 0 and 2\pi seems a bit unusual.  I have certainly not seen any trace of such a convention at any of the wiki pages on varieties of continua.  Also, historically mathematicians have studied various notions of the continuum but it always tends to be a linear one, see continuum (theory).  Tkuvho (talk) 17:31, 14 February 2011 (UTC)
thread_title: Why the name?

message: Is it consistent with ZFC to have 2^{\aleph_0} = some large cardinal?YohanN7 (talk) 07:48, 9 May 2011 (UTC)
replies: text: Well, not the "normal" ones, so to speak.  But there are variants that still fit neatly into the consistency-strength hierarchy for which the answer could be yes.  For example, it's consistent that the continuum is weakly inaccessible (equiconsistent with the existence of an inaccessible) or that it's real-valued measurable (equiconsistent with the existence of a measurable). --Trovatore (talk) 07:53, 9 May 2011 (UTC)
replies: text: The two cases you mention are related in a way, right? I vaguely recall reading in "Geometric Measure Theory" by Herbert Federer (a horribly difficult book for me by the way, but good) that the existence of a nontrivially measurable infinite set requires the existence of an inaccesible cardinal. I need to refresh my brief knowledge on that. By the way, the articles here on inaccessible cardinals and related definitions are kind of messy. (I can be more precise regarding that last statement if needed, given some time.)YohanN7 (talk) 09:12, 9 May 2011 (UTC)

text: Edit: In the above I meant measurable set with all subsets measurable. YohanN7 (talk) 19:03, 9 May 2011 (UTC)
replies: text: Right, I believe that requires a real-valued measurable, but I'd have to check myself on that.  Real-valued measurables are certainly all weakly inaccessible, and quite a bit more than that (e.g. weakly Mahlo).  I have heard of something called a "real-valued supercompact", so it doesn't necessarily stop at measurable.  I wonder if there's some abstract way for taking a large-cardinal property, defined say in terms of extenders, and turning all the measures into real-valued measures, and thereby getting a "real-valued" version. --Trovatore (talk) 19:08, 9 May 2011 (UTC)
replies: text: As far as the measures go, signed measures and complex-valued measures can be expressed in terms of real-valued measures (positive semidefinite ones) using the Hahn-Jordan decomposition. But you probably knew that. [I'm way out of my league here;)]YohanN7 (talk) 21:04, 9 May 2011 (UTC)
replies: text: Ah, just figured out what you were talking about here.  That's not really what I meant.  The measures given by extenders, when defining large cardinals, are ordinarily two-valued measures — the only possible values are 0 and 1.  Going to real-valued measures is a liberalization, not an extra requirement. --Trovatore (talk) 03:07, 10 May 2011 (UTC)
replies: text: Hm. I should perhaps have read real-valued measurable before I opened my mouth;). YohanN7 (talk) 09:52, 10 May 2011 (UTC)
replies: text: Indeed, we are talking about the same thing, at least we were at the outset. My reference [Federer] refers to non-measurable cardinals as Ulam Numbers. The only thing that appears to differ is the term of atomless that gives the transition from measurable to real-valued measurable (implicitely there in Federer I believe). Do you mean that you want to take an arbitrary measure and expand it into some sort of sum of real-valued measures? And, in addition, somehow express various large cardinal properties to see exacly how strong large cardinal axiom one needs to actually find a cardinal that at least might be real-valued measurable? In other words, you want to find exactly where in the supposedly linearly ordered consistency-strength hierarcy of large cardinal axioms each of the involved axioms belong? (I'm only guessing here so don't laugh too much!)  YohanN7 (talk) 10:28, 10 May 2011 (UTC)

text: Easton's theorem allows one to make 2^{\aleph_0} equal to be any uncountable regular cardinal since they all have cofinality greater than \omega \,. However, when you do that the cardinal in question would cease to be a strong limit cardinal and thus would no longer satisfy the definitions of most types of large cardinals. JRSpriggs (talk) 10:03, 9 May 2011 (UTC)
replies: text: Yes, that makes sense. But you must mean that the cardinal in question would cease to be a strongly inaccessible cardinal (if that is what we happened to start with)? Might it still be a weakly inaccessible cardinal? YohanN7 (talk) 22:28, 9 May 2011 (UTC)
replies: text: If you do it in the most obvious way (just add κ Cohen reals) then I think the forcing preserves cardinalities and cofinalities, so κ should still be both a limit cardinal and a regular cardinal, thus weakly inaccessible. --Trovatore (talk) 01:06, 10 May 2011 (UTC)
thread_title:  Large cardinal? 

message: One intuitive argument against CH is said to be Freiling's Axiom of Symmetry. I wonder where the intuition comes from. I agree that the Axiom of Symmetry (AX) sounds quite convincing on first reading, but that's a little thin for mathematical reasoning. I thought about a way to base the intuition on provable facts using smaller sets than the reals. What about this?
Modify AX the following way:
1.) We throw darts at a countable set instead of the unit interval (I). Say the natural numbers (N) for definiteness.
2.) The function f:I->{x:x is a countable subset of I} occuring in AX is exchanged for f:N->{x:x is a finite subset of N}
Leave everything else untouched in AX. Call this thing the Axiom of Countable Symmetry (ACX).
I think that ACX "sounds" just about as convincing as AX on "first reading". Lets call a counterexample of AX (or ACX) a Freiling function. ACX is false. Proof: f(n) = {m: m <= n} is a Freiling function.
I don't draw any conclusions from this. But provided I'm correct in the above [and I might very well not be, the night is not young;)], I can't see the validity of the claim that AX is an argument against CH. If anything, it might be an argument in favour of CH, but I wouldn't go as far as that. In the article Freiling's axiom of symmetry there are listed two objections against AX. My example above might just be an example of objection number two, but I can't see exactly what that number two says. At any rate one does not need CH to disprove ACX, and I can't see the need for AC either. For a countable set there exists a bijection between it and N. That bijection does provide a wellordering (with the same length as N). Or doesn't it? Correct me if I am wrong. YohanN7 (talk) 00:00, 10 May 2011 (UTC)
thread_title:  The Axiom of symmetry 

message: I added a note about an article by Solomon Feferman, though I think I may have messed up the explanation somewhat—I figured the main thing was to get the citation into the article.  Any review/fixes would be appreciated.  The EFI page (url in the Feferman citation) looks interesting in general and there's an overview by Koellner of the current state of CH, that also seems worth summarizing in the article.  I might try, but I'm not very knowledgeable, so it would be great if someone else did it. Feferman has some other slides about his "definiteness" theory that are a little more detailed than his EFI slides, if that's of any interest. 64.160.39.72 (talk) 01:57, 5 February 2012 (UTC)
thread_title: Feferman/EFI

message: Does anyone know, if;
:2^{\aleph_\alpha} = \aleph_{\alpha+1}.
What this evaluates to?
:2^{{{2}^2}^...}.
Edit: It won't let me format that how i wanted to - basically I mean what is 2 to the power of 2 to the power of 2 to the power of 2 to the power of 2 and so on. — Preceding unsigned comment added by 109.149.174.130 (talk • contribs) 16:24, 23 October 2012‎
:::Hi 109.  This is an article talk page, which is intended for discussions aiming at improving the article.  General questions like this can be asked at Wikipedia:Reference desk/Mathematics. --Trovatore (talk) 19:18, 23 October 2012 (UTC)
thread_title:  Questions 

message: Since a recent edit has implicitly raised the question of how the formulas in Continuum hypothesis#Implications of GCH for cardinal exponentiation are justified. I will provide proofs here.
First, take notice of the following facts:
* α < β implies \aleph_{\alpha} < \aleph_{\beta} \,;
* the properties of exponentiation given at Cardinal number#Cardinal exponentiation;
* GCH implies \aleph_{\beta+1} = 2^{\aleph_{\beta}} \,;
* GCH implies the axiom of choice which, in turn, implies \kappa \cdot \nu = \max (\kappa, \nu) \, and in particular \kappa \cdot \kappa = \kappa \, for infinite cardinals κ and ν; and
* AC (and thus GCH) also implies König's theorem (set theory).
Now, suppose that α ≤ β+1, then
:\aleph_{\beta+1} = 2^{\aleph_{\beta}} \leq \aleph_{\alpha}^{\aleph_{\beta}} \leq \aleph_{\beta+1}^{\aleph_{\beta}} = (2^{\aleph_{\beta}})^{\aleph_{\beta}} = 2^{(\aleph_{\beta} \cdot \aleph_{\beta})} = 2^{\aleph_{\beta}} = \aleph_{\beta+1} \,
which means
:\aleph_{\alpha}^{\aleph_{\beta}} = \aleph_{\beta+1} \,.
On the other hand, suppose that β+1 < α, then
:\aleph_{\alpha} = \aleph_{\alpha}^{1} \leq \aleph_{\alpha}^{\aleph_{\beta}} \leq (2^{\aleph_{\alpha}})^{\aleph_{\beta}} = 2^{(\aleph_{\alpha} \cdot \aleph_{\beta})} = 2^{\aleph_{\alpha}} = \aleph_{\alpha+1} \,
which means
:\aleph_{\alpha} \leq \aleph_{\alpha}^{\aleph_{\beta}} \leq \aleph_{\alpha+1} \,.
If we further suppose that \aleph_{\beta} < \operatorname{cf} (\aleph_{\alpha}) where cf is the cofinality operation, then any function from \aleph_{\beta} \, to  \aleph_{\alpha} \, must be bounded above by some \gamma < \aleph_{\alpha} \,. And γ has a cardinality \vert \gamma \vert = \aleph_{\delta} \, where δ < α. The cardinality of the set of functions so bounded by γ is
:\aleph_{\delta}^{\aleph_{\beta}} \leq 2^{(\aleph_{\delta} \cdot \aleph_{\beta})} = \aleph_{\max(\delta,\beta)+1} \leq \aleph_{\alpha} \,.
Adding these together for the \aleph_{\alpha} \, possible values of γ gives
:\aleph_{\alpha}^{\aleph_{\beta}} \leq \aleph_{\alpha} \cdot \aleph_{\alpha} = \aleph_{\alpha} \,
which means
:\aleph_{\alpha}^{\aleph_{\beta}} = \aleph_{\alpha} \,.
On the other hand, if we further suppose \operatorname{cf} (\aleph_{\alpha}) \leq \aleph_{\beta} \,, then by one of the corollaries of König's theorem we have
:\aleph_{\alpha} < \aleph_{\alpha}^{\operatorname{cf} (\aleph_{\alpha})} \,
and thus
:\aleph_{\alpha+1} \leq \aleph_{\alpha}^{\operatorname{cf} (\aleph_{\alpha})} \leq \aleph_{\alpha}^{\aleph_{\beta}} \,
which means
:\aleph_{\alpha}^{\aleph_{\beta}} = \aleph_{\alpha+1} \,.
These were what was to be proved. JRSpriggs (talk) 10:05, 13 November 2012 (UTC)
thread_title:  Proofs of certain consequences of GCH 

message: User:JRSpriggs made a short remark in a recent edit correcting (thank you!) my previous edit.  I would like to expand this remark here for two reasons: I want to help others avoid the mistake that I had made; also, perhaps we can add a short form of this remark into the article itself. 
I will write Z0 for ZF without the axiom of foundation and without the axiom of replacement. 
There are two versions of GCH: 
: GCHpower: For all sets X, every subset Y\subseteq P(X) of the power set of X either is equinumerous with P(X) or can be injected into X.
: GCHaleph: For all ordinals \alpha we have 2^{\aleph_\alpha}=\aleph_{\alpha+1}. 
* GCHpower is apparently  stronger than GCHaleph; the implication GCHpower \Rightarrow GCHaleph is already provable in Z0. 
* GCHaleph implies that
*# ... the power set of every \aleph_\alpha is well-orderable. (trivial, in Z0)
*# ...  the power set of every well-orderable set is well-orderable. (easy from 1., but needs replacement)
*# ... every set has a well-order. (The implication from 2. is non-trivial but well-known. Needs foundation. The standard proof uses induction of von Neumann's hierarchy.) 
*# ... AC holds. (easy from 3.)
*# ... every set is equinumerous with some \aleph_\alpha. (Easy from 4. Replacement used again.) 
*# ... GCHpower holds.
So the two versions are indeed equivalent over the base theory ZF. Still I think that the (apparently) "stronger" version GCHpower should be mentioned as the GCH.
When we mention Sierpinski's proof, we should clarify that he proved AC from  GCHpower.  
(In ZF, AC also follows from GCHaleph, but the interesting part is the implication from "P(well-order)=well-orderable" to AC, which has nothing to do with cardinal arithmetic.  Was Sierpinski also the author of this theorem?)
--Aleph4 (talk) 12:48, 12 September 2013 (UTC)
replies: text: You did notice this is discussed in the section above, right?—Emil J. 13:47, 12 September 2013 (UTC)
thread_title:  AC and the two versions of GCH 

message: JR's point seems to be that Goedel's contribution is listed first and Cohen's second (same as the chronological order).  But the unusual word order "disproved nor proved" really does not get this across; it just looks weird.  I suggest that if we want to say that Goedel showed it couldn't be disproved in ZFC and Cohen showed it couldn't be proved in ZFC, then that is what needs to be said.  But at that level of the lead, I don't know that we need to get that specific, and I think we should just go back to "proved nor disproved".
:Again, it should be ZFC, not ZF. --Trovatore (talk) 07:41, 9 January 2014 (UTC)
replies: text: Yes, ZFC if anything. Also, could we get rid of the annoyingly prudent "if ZF set theory is consistent" clause in the lead? Pathological cases aren't of interest to the general reader (and hardly anyone else). (How on earth has it not become standard to implicitly assume consistency in the math lingo.) YohanN7 (talk) 15:57, 9 January 2014 (UTC)
replies: text: Because it might turn out to be inconsistent. Tkuvho (talk) 16:00, 9 January 2014 (UTC)
replies: text: Yes, and in the intervening 101001000 years we all have to use awkward language in encyclopedias and elsewhere so that we don't forget. You miss the point. In the rest of the article we could be as precise as we wish, provided the sun rises tomorrow. YohanN7 (talk) 16:14, 9 January 2014 (UTC)
replies: text: You appear to be comfortable in your faith that the sun is less likely to rise tomorrow than ZFC is likely to be consistent, and you are certainly entitled to it. However issues of faith may constitute POV and shouldn't necessarily influence issues of content. Tkuvho (talk) 16:35, 9 January 2014 (UTC)
replies: text: No, I am not - provided of course that both the sun and ZFC exist. I am just expressing my surprise that mathematics hasn't rationalized its internal language in this case so that it becomes less cumbersome. In most contexts there are obvious implicit assumptions not spelled out each and every time. If a not very obvious assumption is used then it is spelled out. YohanN7 (talk) 18:14, 9 January 2014 (UTC)

text: JRSpriggs, are you going to address this?  I continue to object to "disproved nor proved".  The unusual word order gives the reader a problem to solve; the fraction of readers who come up with your intended solution, I expect to be small.  There is just no real indication that the word order corresponds to which mathematician did what.  I'm concerned that some readers will come up with an unwanted solution, such as losing the parallelism between "disproved" and "proved". --Trovatore (talk) 23:17, 9 January 2014 (UTC)
replies: text: Would "deduced" be better than "proved"? Tkuvho (talk) 10:10, 13 January 2014 (UTC)
thread_title: "disproved nor proved"

message: I may be old-fashioned, but I don't see the specific relevance of Hampkins' paper to CH; although I haven't yet read a the paper, it would seem to apply to any independent (in some sense) statement, not just CH.
Even assuming the paper is considered significant in the field.  — Arthur Rubin  (talk) 13:51, 27 January 2014 (UTC)
replies: text: Hamkins has a more general notion of a "switch", namely an axiom that can be adjusted at will to either be satisfied or violated (like the CH) by passing to a larger model. However, his discussion does tend to center on CH, and has immediate relevance to the philosophical debate around CH. Tkuvho (talk) 15:43, 27 January 2014 (UTC)
thread_title:  Multiverse approach to set theory 

message: The book:
 Cohen, P. J. (1966). Set Theory and the Continuum Hypothesis. W. A. Benjamin.
is not accessible: you can request it, pay for it and the there is no book and no refunds.
Suspicious?  — Preceding unsigned comment added by 87.218.117.85 (talk) 09:00, 30 June 2014 (UTC)
replies: text: Buyer beware. Before you pay out money, you should make sure that the book-seller you are dealing with is reputable. If the book is out-of-print, then the book-seller should have told you that rather than taking your money and stiffing you. JRSpriggs (talk) 04:49, 1 July 2014 (UTC)
replies: text: Perhaps "Cohen, Paul Joseph (1966). Set theory and the continuum hypothesis. Addison–Wesley. ." was meant? JRSpriggs (talk) 05:08, 1 July 2014 (UTC)
thread_title:  Danger in Bibliography 

message: It's not a big deal, but I think that the recent edits have been for the worse. The size of a set is, in informal terms, its cardinality. The lead can, at this point, stay informal, the term cardinality coming one sentence later. The article now links Set (mathematics), not infinite set. Worse, it speaks of sets with infinite elements. YohanN7 (talk) 13:16, 14 November 2014 (UTC)
replies: text: I agree. It violates WP:TECHNICAL. We should keep the wording in the lead as simple as possible, even at the expense of precision, and save the more precise wording for later. Size vs cardinality is an example of that. I would be in favor of reverting to the earlier wording. —David Eppstein (talk) 16:08, 14 November 2014 (UTC)
thread_title:  Recent edits 

message: Can a better example for bijection be provided? The example given, "Hence, the set {banana, apple, pear} has the same cardinality as {yellow, red, green}." lists three fruits that exist in more than one of the colors given.
Mjpollard (talk) 14:17, 6 November 2016 (UTC)
replies: text: There also exists several distinct bijections. YohanN7 (talk) 14:20, 8 November 2016 (UTC)
thread_title:  Improved example requested 

message: Is saying that |\mathbb R| = |\mathcal P (\mathbb N )| really equivalent to the continuum hypothesis? There seems to be a bijection between the set of real numbers and powerset of the natural numbers here, which would imply the continuum hypothesis is true if they were equivalent. --AkariAkaori (talk) 00:03, 10 July 2017 (UTC)
replies: text: Thanks for pointing out that error. I tried to fix it. JRSpriggs (talk) 02:33, 10 July 2017 (UTC)
thread_title:  "equivalently ... i.e. it (the cardinality of the set of real numbers) equals the cardinality of the power set of the integers" 

message: Explains Cohen's proof of independence of CH.  It looks pretty self contained yet is short enough for one lecture.  I'm going to try to read it since other places have made the proof sound very mysterious and I have never understood it.  Leaving link here for future reference and for possible use in improving the article. 2602:24A:DE47:BB20:50DE:F402:42A6:A17D (talk) 05:17, 18 August 2020 (UTC)
replies: text: The proof is not terribly mysterious.  There is a lot of machinery to build and a lot of details to check, but the basic idea is straightforward.  Something of that definitely could, and probably should, be conveyed in this article.  We would want to be careful not to let it get out of hand; this isn't the article in which to explain forcing in detail.  But we can probably say more than what the article currently says, which is Essentially, this method begins with a model of ZF in which CH holds, and constructs another model which contains more sets than the original, in a way that CH does not hold in the new model.  That single sentence also does not really stand out. --Trovatore (talk) 05:29, 18 August 2020 (UTC)
replies: text: Thanks.  Those slides seem easier to understand than the current article forcing (mathematics), maybe because the terminology is more familiar.  Maybe the forcing article could be updated a little from it.  2602:24A:DE47:BB20:50DE:F402:42A6:A17D (talk) 00:06, 22 August 2020 (UTC)
thread_title:  Lecture slides 

message: I just came across this article about a new mathematical proof implying that the hypothesis is true. But I don't understand the subject matter well enough to add it to the page in a sensible manner, perhaps someone else can take a gander?
Here's the article that might be worth including: https://www.quantamagazine.org/how-many-numbers-exist-infinity-proof-moves-math-closer-to-an-answer-20210715/
CheersCreapuscularr (talk) 13:42, 6 August 2021 (UTC)
replies: text: All this shows is an equivalence between two other axioms for set theory.  While this doesn't directly prove anything about the continuum hypothesis, the article in question goes as far as to say this "result strengthens the case against the continuum hypothesis", not for the continuum hypothesis as you have suggested.  I dont know enough about the subject matter to decide if this article is relevant to this wikipedia article though. 2607:F140:400:A016:2874:B0DF:D7F7:5B4B (talk) 00:56, 7 April 2022 (UTC)
thread_title:  Update 

message: Since this conjecture is "a conjecture which can be disproved using a counterexample", if this conjecture is in fact false, we can use this counterexample to disprove it, thus the conjecture is decidable, and if the conjecture is undecidable, then it must be true, thus, if we can prove that this conjecture is undecidable, then we can prove that this conjecture is true, which is a contradiction, like Goldbach conjecture (a counterexample is an even number ≥4 which cannot be written as sum of two primes), Fermat Last Theorem (a counterexample is an integer n≥3 and positive integers x, y, z such that xn + yn = zn), Riemann hypothesis (a counterexample is a complex number z such that Re(z) ≠ 0, Im(z) ≠ 1/2, but ζ(z) = 0), and four color theorem (a counterexample is a graph which cannot be drawn using ≤4 colors), a counterexample of continuum hypothesis is a subset S of R such that Card(S) > Card(N) and Card(S) < Card(R). 2402:7500:916:25D6:DD5D:1928:6166:9583 (talk) 09:25, 20 August 2021 (UTC)
replies: text: You are wrong. Even if such a set S exists, it cannot be exhibited and its required properties verified in a manner which is independent of the model of set theory within which one is working. JRSpriggs (talk) 16:39, 20 August 2021 (UTC)

text: Per the talk-page guidelines, we're actually not supposed to discuss this here, but it would be a very interesting topic at the mathematics reference desk.  Would you be interested in raising it there?  Remember to state it in the form of a question, like on Jeopardy. --Trovatore (talk) 17:57, 20 August 2021 (UTC)
thread_title:  This conjecture is true.