Talk:Cardinality of the continuum/Archive 1

Archive 1

Archive 2

alef-omega

The article says that c can be taken equal to alef_{omega1}. But surely ω1 = ω, and we know that c cannot be alef_ω. This can't be right. -lethe ^{talk +} 16:06, 31 January 2006 (UTC)

Oh, I know. Surely it is meant the first uncountable ω_1, rather than the countable ω1. -lethe ^{talk +} 16:06, 31 January 2006 (UTC)

Oh, now that I edit it, I see that ω_1 is indeed indicated in the tex, it just doesn't display very clearly. So, nevermind. -lethe ^{talk +} 16:06, 31 January 2006 (UTC)

Is the continuum really two to the aleph-null?

I was surprised to find that this page didn't exist so I thought I better start it. Admittedly, set theory is not my area of expertise so I hope others will improve it. I would like to see a nice clean proof of the oft stated fact that $c=2^{\aleph _{0}}$. I can't think of one at the moment (nor can I find one). -- Fropuff 06:42, 2005 Mar 6 (UTC)

Okay, I found a proof that I like and have included it in the article. If anyone knows of a better proof, feel free to change it. I am curious to know whether or not there exists an explicit bijection between the real numbers (or the unit interval) and power set of the natural numbers which isn't horribly convoluted. I've certainly never seen one (convoluted or otherwise). -- Fropuff 20:50, 2005 Mar 7 (UTC)

For a slightly less convoluted map than the one in the article, note that each real number in [0,1] is represented by a sequence of bits by the usual binary expansion. so 2^N ≥ c, and the other inequality as in the article. This is perhaps more concrete, and more similar to the second injection, so I like it. Maybe I'll swap it in. -lethe ^{talk +} 16:06, 31 January 2006 (UTC)

I did think about that one, but you'd have to add a line explaining that the cardinality of unit interval is equal to c; otherwise it works. It's also probably important to emphasize that the above map is only an injection and not a bijection. -- Fropuff 17:08, 31 January 2006 (UTC)

The proof of Schroeder-Berstein does give you an explicit bijection, if you follow it through. The note in the Cantor–Bernstein–Schroeder theorem article claiming that the proof is "not constructive" is a bit misleading; "not intuitionistically valid" would be more precise. --Trovatore 06:06, 3 October 2005 (UTC)

There are beth-one transcendental numbers?

I'd like to see a proof that there are ${\mathfrak {c}}$ transcendental numbers, because I never have before. OneWeirdDude (talk) 00:50, 7 January 2008 (UTC)

OK, here's a bare-bones sketch; holler if you need more details. There are only $\aleph _{0}$ algebraic reals. Suppose the cardinality of the set of all transcendental reals were $\kappa <2^{\aleph _{0}}$. Then the cardinality of the set of all reals would be $\aleph _{0}+\kappa$, which equals $\kappa \,\!$, which is less than $2^{\aleph _{0}}$. Contradiction.

If you want an argument that gives an actual example of a bijection between the transcendental reals and the reals, that's going to be a bit more involved, I'm afraid, but I'm confident it wouldn't be too hard. --Trovatore (talk) 01:02, 7 January 2008 (UTC)

The sum or difference of two algebraic real numbers is algebraic. Thus the sum or difference of an algebraic number and a transcendental number is transcendental. Similarly, the product of a nonzero algebraic number and a transcendental number is transcendental. e is known to be transcendental. For every n in the natural numbers and α in the algebraic numbers, map ((n+1)·e)+α to (n·e)+α. Map every other transcendental number to itself. This mapping is a bijection from the transcendental numbers to the reals. JRSpriggs (talk) 03:04, 8 January 2008 (UTC)

Nicely done! I had a much more brute-force solution in mind. --Trovatore (talk) 22:17, 8 January 2008 (UTC)

What is "Intuitive Argument?"

we are using, so for simplicity, let us consider a binary real number. Each position in its decimal expansion may hold either a 0 or a 1, so the number ${\mathfrak {c}}$ of all possible ways to fill those positions must be $2^{\aleph _{0}}$. Therefore, ${\mathfrak {c}}=2^{\aleph _{0}}$.

A prove by analogy: Each position in radix-3 number may hold either a 0 or a 1 or 2. The number ${\mathfrak {c}}$ of all possible ways to fill those positions must be $3^{\aleph _{0}}$. Therefore, ${\mathfrak {c}}=3^{\aleph _{0}}$. :) --Javalenok (talk) 14:08, 26 May 2008 (UTC)

The conclusion is correct. You have "proved intuitively" that $2^{\aleph _{0}}=3^{\aleph _{0}}$.  --Lambiam 23:51, 26 May 2008 (UTC)

In fact, if $2\leq \kappa \leq {\mathfrak {c}}$, then $\kappa ^{\aleph _{0}}={\mathfrak {c}}$. JRSpriggs (talk) 01:26, 27 May 2008 (UTC)

Uncountability questioned

I think the logic used in all the proofs is clearly wrong, and I therefore find it difficult to accept the conclusion that the real numbers are uncountable.

Let me illustrate the faultiness of the style of logic used. Consider that set S of real numbers which can be defined in English. Require only that each definition be a finite string of characters from a finite alphabet. It is then trivially easy to represent each definition as an integer. Therefore S is clearly countable. It does not matter that each definable real has an infinite number of possible definitions, nor that most integers do not correspond to valid definitions.

Take it for granted that each of our definable reals has a decimal expansion. Now define a new real number between 0 and 1 such that the Nth digit after the decimal point in its decimal expansion has the value:

  4 if the integer N does not correspond to a valid definition
  5 if the Nth digit in the decimal expansion of the real number defined by N is not 5
  6 if the Nth digit in the decimal expansion of the real number defined by N is 5. 

Clearly this new number differs from each of the numbers previously defined, and so is not in S. But it is also a real number defined in English, and therefore should by definition be in S.

This is clearly nonsense. What has gone wrong? The answer seems to be that we have conceived a set S of “all” definable real numbers and then on the basis of that set we have defined a real number which is clearly not part of that “all”. The conclusion seems necessarily to be that such logic is invalid and unacceptable.

Is there any supposed proof of the uncountability of the reals which does not depend on such logic?

W J Eckerslyke (talk) 17:32, 15 January 2009 (UTC)

That only produces a paradox if you deny the law of the excluded middle or the general principle of infinite sets as objects. The general "Cantor" diagonal argument produces:

• The set of all real numbers is uncountable. (There is no function from N onto R.)
• The set of all "definable" real numbers does not have a "definable" sequencing. (When you specify exactly what you mean by "defined in English", you'll find that this definition is faulty.)
• Considering the set of all "constructive" real numbers, there is no "constructive" counting of them.
• Considering the set of all "provably" definable real numbers (i.e., there is a proof in Peano arithmetic that the definition defines a real), there is no "proof" that there is a function from N onto that set.

—Preceding unsigned comment added by Arthur Rubin (talk • contribs) 18:02, 15 January 2009

In other words, the problem is with the imprecision of the English language, not with the Cantor diagonalization argument. JRSpriggs (talk) 08:13, 16 January 2009 (UTC)

See also Proof of impossibility#Richard's paradox. Boris Tsirelson (talk) 12:11, 16 January 2009 (UTC)

Thanks greatly for the pointer to Richard's paradox, which I seem unwittingly to have re-invented (some 104 years late). I agree that Richard's paradox is a fallacy, and that the logic it uses is invalid. But I am still inclined to think that my analysis of that invalidity is correct, and that exactly the same form of invalidity appears in such things as Cantor's diagonal argument. Can you explain the difference? W J Eckerslyke (talk) 16:41, 16 January 2009 (UTC)

You may also look at Skolem's paradox and Talk:Skolem's paradox; especially, think about the opinion of Xzungg. Do you share his opinion? Also here: Wikipedia talk:WikiProject Mathematics/Archive 43#Skolem's paradox —Preceding unsigned comment added by 79.183.255.190 (talk) 21:00, 16 January 2009 (UTC)

Thanks for the pointers to Skolem's paradox and related discussions. Some immediate responses: (1) It is nice to see a bit of passion in mathematics. Even so, it is difficult to agree with somebody like Xzungg who expresses his or her views so intemperately and voluminously. (2) Skolem's paradox, as far as I understand it, does seem to present a genuine anomaly. (3) I don't fully understand why “mathematicians no longer consider Skolem's result paradoxical”, and to that extent I have some sympathy with Xzungg. (4) However, I am not myself too worried by this, because (as you will have noted) I don't really believe in uncountable sets. (5) I remain to be convinced that a consistent set theory can prove the existence of uncountable sets without using invalid logic of the kind exposed by Richard's paradox. (6) Indeed, I suspect that a set of axioms is never consistent or inconsistent in itself, but only in conjunction with a given set of deduction rules. If you allow invalid deduction rules you will get invalid results. (7) If Xzungg is saying that set theories (including “naive” and ZFC in particular) are seriously flawed then I have to agree. If he or she is saying that there could be no consistent set theory then I have to disagree. W J Eckerslyke (talk) 17:32, 18 January 2009 (UTC)

"a consistent set theory can prove the existence of uncountable sets without using invalid logic of the kind exposed by Richard's paradox" - Really? That could be interesting. I would be glad to have more details about this. Maybe not here but better on your (or mine) personal discussion page. But how does it coexist with "I don't really believe in uncountable sets"?

Oops, it seems I forgot to sign the previous paragraph. It is me, Boris Tsirelson (talk) 06:58, 19 January 2009 (UTC)

I have taken your advice and continued my doodles in my Discussion page. I shall be very grateful if you can find time to offer further assistance in my explorations. W J Eckerslyke (talk) 11:31, 26 January 2009 (UTC)

OK, I'll comment there. Boris Tsirelson (talk) 14:57, 26 January 2009 (UTC)

Do we discuss it on the formal (axiomatic) level, or not? Formally, the proof is correct in Zermelo–Fraenkel set theory. Informally... well, here is my personal opinion. Mathematics makes a number of rather risky assumptions. First, existence of an infinite set. Further, existence of a set of all subsets of (a given set, say) of the set of natural numbers, basically the same as existence of the set of all real numbers. "This is often given the (potentially confusing) term "Platonism": the objects that the theory refers to are treated as if they were independent of what the mathematician can actually and explicitly define." (Page 148 in: The Princeton Companion to Mathematics", Princeton 2008.) If you do not accept these principles, then of course you are outside the usual mathematics (which does not mean you are wrong, but...) Boris Tsirelson (talk) 20:01, 17 January 2009 (UTC)

Existence of infinite sets: no problem. But mathematics based on the existence of undefinable entities, including "all" subsets of natural numbers, seems at best questionable, and arguably useless nonsense. Unless, of course, treated as part of theology, which is where study of the ineffable properly belongs. Therefore I am clearly "outside the usual mathematics". So I expect to desist from further comment. Thanks to all for comments made. W J Eckerslyke (talk) 17:32, 18 January 2009 (UTC)

"Useless nonsense"? Think again. Evidently, useful! Now, can nonsense be useful? The choice axiom (beyond the countable dependent choice) is more debatable. However, the continuum (yes, including "all" subsets of natural numbers) is really useful. I can understand your indignation, but here is a challenge for you: WHY is it useful? Boris Tsirelson (talk) 23:03, 18 January 2009 (UTC)

I have failed your challenge. Please tell me (elsewhere) why it is useful. In the meantime, can you offer me a pointer to a "proof" in ZFC of the existence of uncountable sets? W J Eckerslyke (talk) 11:31, 26 January 2009 (UTC)

Here is a famous quote from Paul Cohen ("Set theory and the continuum hypothesis", 1966, page 151): "This point of view regards C (the continuum - B.Ts.) as an incredibly rich set given to us by a bold new axiom, which can never be approached by any piecemeal process of construction." Boris Tsirelson (talk) 21:38, 17 January 2009 (UTC)

2^R notation

Hi JR and Potatoswatter,

maybe we should have the discussion here instead of in edit summaries.

Here's my take: The notation $2^{R}$ for the powerset of the reals is occasionally used, I think. No doubt you can come up with an attestation somewhere. But in my experience it's not really very common, and I'm not sure what it adds in this article. --Trovatore (talk) 02:11, 13 July 2009 (UTC)

For sets (rather than cardinal numbers) A and B, the notation "A^B" refers to the set of functions from B to A. That is,

$A^{B}=\{f|f:B\rightarrow A\}\,.$

Remember that 2 = {0, 1}. So for example, 2^{{a, b, c}} = { {<a,0>, <b,0>, <c,0>}, {<a,0>, <b,0>, <c,1>}, {<a,0>, <b,1>, <c,0>}, {<a,0>, <b,1>, <c,1>}, {<a,1>, <b,0>, <c,0>}, {<a,1>, <b,0>, <c,1>}, {<a,1>, <b,1>, <c,0>}, {<a,1>, <b,1>, <c,1>} }. Since this is the set of indicator functions of the subsets of {a, b, c}, one may (by abuse of the language, e.g. the "crown" is the king) equate it to the powerset. JRSpriggs (talk) 03:11, 13 July 2009 (UTC)

Well, it's this last step, identifying a set with its characteristic function, that I don't really think is very standard. I don't see what's gained by insisting on this confusing notation here. --Trovatore (talk) 03:35, 13 July 2009 (UTC)

There's no reason to use notation besides what is standard, concise, and correct, when it is all three of those.

2^R looks to me like something a teacher wrote by accident. The "absolute value" sign was simply omitted from the unambiguous 2^|R|.

2 DOES NOT EQUAL {0,1}. A scalar is not a set. Just so we get that out of the way.

A^B sounds cool but do you have a reference? Nobody is likely to recognize that. Anyway, computing the cardinality of the result is non-trivial, so in any case we would specify {f: A -> B }. It's not too long and it's readable. Potatoswatter (talk) 04:48, 13 July 2009 (UTC)

Actually identifying 2 with {0,1} is standard. The notation A^B for the set of all functions from B to A is also fairly standard, though risky given that it can mean so many other things. Really it's only the identification of a subset with its characteristic function that I think goes a bit over the line here. That plus I just don't see much advantage in writing 2^R here. --Trovatore (talk) 04:53, 13 July 2009 (UTC)

Oh, I now see that this has been updated to speak of their "indicator functions". This is another slight peeve of mine. The standard term is characteristic function. In my experience the term indicator function is not used in set theory. At all.

I understand that the probabilists have a different usage of characteristic function, but I don't really see why that should matter here. --Trovatore (talk) 04:57, 13 July 2009 (UTC)

You may be right, but Characteristic function is just a disambiguation page which leads to Indicator function. JRSpriggs (talk) 05:01, 13 July 2009 (UTC)

[2] (note the brackets, it's an equivalence class) = {0,1}. A set cannot equal a scalar, and that's a fact.

A^B for {f|f: A->B} is, for the purposes of this article, the same as P( A x B ), the powerset of their Cartesian product (ignoring the function test because it's late for me). We should note that |{f|f:A->B}| = | P( A x B ) | or whatever, but don't get silly with equals signs for things that aren't equal, and don't use obscure notation AT ALL. Will you provide a citation rather than insisting that A^B is "standard"? Potatoswatter (talk) 05:13, 13 July 2009 (UTC)

See any graduate set theory text for these things. Jech, for example, or Kunen. Probably Halmos or Enderton would also work fine.

While no one (or at least not many people) are claiming that {0,1} is what the natural number 2 really is, in any deep philosophical sense, it is completely standard in set theory to code the number by this set. Officially, all objects in set theory are sets. Therefore the natural numbers must also be sets. Which ones? Well, there's a standard coding. See von Neumann ordinal for details. --Trovatore (talk) 09:34, 13 July 2009 (UTC)

Unfortunately I only have an undergraduate abstract algebra text, but my professor was pretty adamant about disallowing this kind of notation. Now that I look it up, [2] ≠ {0,1} but rather [2] = {0,n,2n,…}, an element in some Z/nZ for n>2. I would have certainly lost points for saying either 2 = {0,1} or [2] = {0,1}, rightly so in either case.

Some notations are more standard than others. Exponentiation says nothing about using a set as an exponent. Google didn't turn one up for me when I looked. Perhaps it's standard once certain concepts are introduced, but being so casual is confusing, and unnecessary when explicitly saying {f | f: A -> B } or Z/2Z isn't too verbose anyway.

For another example, lambda calculus can be seen as the foundation of all computer programming, and it's useless to attempt much theory without it, but unless you invoke those magic words, you'll unnecessarily confuse many people when you say f( a, b ) = g( b ) for some g. That's not the way to write an encyclopedia article. Potatoswatter (talk) 12:35, 13 July 2009 (UTC)

Just to add another voice here:

• A standard theoretic definition of 2 is $\{0,1\}$. See Natural number#Constructions_based_on_set_theory for additional details above what the von Neumann ordinal page says.
• It is also standard to write the set of functions from X to Y as Y^X, and likewise it's standard to write ℘(X) as 2^X. See Power set#Representing_subsets_as_functions for more details. Even if you don't feel 2^X is the strict definition of the power set, it is certainly isomorphic to it in every important way. That being said, it's totally fine to make a distinction between the power set and the set of indicator functions; it's just not that useful of a distinction, and it's very common to use 2^X as a replacement for the power set (again, because it's isomorphic in every useful way).
• Notice that the cardinality of the power set |℘(X)|=2^|X| as expected. Hence, replacing ^X by ^|X| is tantamount to replacing a set with its cardinality.

I've used and read this notation for years in everything from academic texts to lectures to research papers. It's an elegant construction that's well understood by the field. It's construction has been done with care; it helps, it doesn't hurt. —TedPavlic (talk/contrib/@) 12:12, 13 July 2009 (UTC)

In my experience, it's really not common to write 2^A for the powerset of A. This could be a subcultural difference, I suppose. But could you point me to, say, a paper in J. Symb. Logic that does this? --Trovatore (talk) 19:01, 13 July 2009 (UTC)

"Isomorphic in every way, and elegant for papers" is still different from identical, and that's why the powerset is still denoted by P(A). It's useless to write an encyclopedia article that assumes unnecessary knowledge, particularly when it must be used to draw inferences. A good academic paper will still say denoting the powerset by 2^A, even if many academics don't like to bother with such triviality. Potatoswatter (talk) 12:35, 13 July 2009 (UTC)

After re-reading, I augmented my comments before you responded to them, and your response clobbered my augmentation. :) I've put it back. It's definitely fine to make a distinction between the two, but it's valuable to leave the indicator notation as an example (because someone looking through references (perhaps for verification purposes) will see that notation often). —TedPavlic (talk/contrib/@) 12:48, 13 July 2009 (UTC)

Sorry, we edited at the same time and I got two consecutive edit conflict pages, which was confusing. (My connection was slow and it took a while to load each one.) Yeah, clarity through redundancy is prolly a good idea — although it's supposed to be a quick list of examples, we've spent all this time arguing anyway. What really needs to be fixed, though, is clearer linking from exponentiation#over sets to exponential object and/or function space. I'm not really sure I'm qualified to write that properly. Potatoswatter (talk) 17:09, 13 July 2009 (UTC)

Instead of arguing what is "standard" for whom, we'd better explain in the article, in short, both viewpoints (since they both clearly have proponents and opponents). I could add my own opinion, but this would not help, just because "both have proponents and opponents". Boris Tsirelson (talk) 06:30, 14 July 2009 (UTC)

I think the language, at this very moment, does that. See recent changes. —TedPavlic (talk/contrib/@) 13:01, 14 July 2009 (UTC)

My (justly) reverted edits

I got out of bed at the realization that what I'd written was wrong, so I'm not complaining at the reversion. But, the intuitive argument isn't too great as is. What do we do with infinite strings of 1s? Anyone got something better?--Leon (talk) 21:43, 6 October 2009 (UTC)

If we're trying to demonstrate that the cardinality of the reals (or, specifically, [0,1) ) is $2^{\aleph _{0}}$, the simplest approaches I can think of are:

1. 1. $2^{\aleph _{0}}\leq {\mathbb {R} }$, by your approach as $\{0,2\}^{\mathbb {N} }$ as ternary or perhaps base 4.
   2. ${\mathbb {R} }\leq 2^{\aleph _{0}}$ by binary representations, or...
2. ${\mathbb {R} }=2^{\aleph _{0}}-\aleph _{0}$ (the set of binary terminating representations is countable), and $2^{\aleph _{0}}-\aleph _{0}=2^{\aleph _{0}}$ by a complicated cardinal number argument, used in the proof that GCH implies AC.

— Arthur Rubin (talk) 01:27, 7 October 2009 (UTC)

I'm out for a bit, but I might try putting that (and more) in with a few references later on. I know this sounds childish, but would you mind having a glance over once I've made the changes? I'm not a mathematician, and despite having had this account for a while I've barely edited.--Leon (talk) 10:22, 7 October 2009 (UTC)

To repeat a proof which I gave at Talk:Power set#Huhh??? — The powerset of the natural numbers can be mapped bijectively onto the Cantor set which is a subset of the real numbers. The real numbers can be mapped one-to-one into the interval (0,1) of the reals which can be mapped one-to-one into the powerset of the natural numbers by interpreting each such real as an ω-sequence of binary digits (choosing the sequence which ends in all zeros where there is an ambiguity) and regarding these functions as the indicator functions of subsets of the natural numbers (n∈S <-> 1 is the coefficient of 2^-(n+1) in the binary expansion of the real). Using the Cantor–Bernstein–Schroeder theorem, one gets that the powerset of the natural numbers can be placed into a one-to-one correspondence with the real numbers. Obviously this argument does not use either the continuum hypothesis or the axiom of choice. JRSpriggs (talk) 10:13, 8 October 2009 (UTC)

Pronounciation

How is this symbol named/pronounced/articulated: ${\mathfrak {c}}$. The beth and aleph numbers are written out, but this symbol is just put there. If you would read some of the formula's out loud (e.g. while writing it on a blackboard), what would one say? Or what is the analogy of "Aleph-null denotes the cardinality of the natural numbers" -> "... denotes the cardinality of the contiuum". 129.125.6.1 (talk) 14:29, 20 November 2009 (UTC)

It's a Fraktur letter C and I pronounce it the same: "see". — Carl (CBM · talk) 14:43, 20 November 2009 (UTC)

axiom of choice

i think the section regarding the sets with the cardinality of continuum should be subidivided so as to point out precisely what sets have cardinality of continuum assuming the axiom of choice.--81.168.169.234 (talk) 17:43, 6 February 2010 (UTC)

I think you're talking about the "Cardinal equalities" subsection; is that right? Unless I'm missing something, none of the assertions in that section depends on the axiom of choice. There are certainly claims you could put in that section that depend on AC, such as that the set of all countable sets of reals has the cardinality of the continuum, but I don't see anything currently there that depends on it.

The section on the continuum hypothesis would need to be reworded a little bit if AC were not assumed in the background (which, of course, it is; this is standard). But this is problematic, because it's not really agreed what statement should be called the continuum hypothesis if AC fails. --Trovatore (talk) 19:37, 6 February 2010 (UTC)

Intuitive argument

In my opinion, the section Intuitive argument was not clear. I tried to expand it a little.

My idea is this:

• a real number can be decomposed into an integer part and a decimal part, and both these parts are natural numbers.
• Thus, ROUGHLY, the set of real numbers has the same "number of elements" as the set of all possible pairs of natural numbers (not exactly the same: notice that to the right of the decimal point, a 5 becomes equivalent to a 50, or 500, ...).
• Thus, it is a subset of the power set of $\mathbb {N}$.
• Thus, there are $\aleph _{0}$ natural numbers, and $2^{\aleph _{0}}$ real numbers.

Notice that the last point is somewhat a leap of faith, but at least this argument makes clear that the cardinality of the real numbers is higher than that of the natural numbers.

I am not sure I was able to make this crystal clear, so I was wondering if someboty else were willing to help me to improve the readability of this section. Please keep it as simple as possible. Notice that I tried to avoid technical expressions such as "Cartesian product", "power set", or "subset of the powerset", as they may not be known by beginners. Perhaps, we don't even need to explain the concept of "decimal expansion"...

Paolo.dL (talk) 13:27, 21 May 2011 (UTC)

The first point of your idea is wrong, the third does not follow from the second, and the fourth point does not follow from the first three. The number of pairs of integers is just $\aleph _{0}$, the same as the number of integers. The number of fractional parts of real numbers -- that is, the cardinality of the interval [0, 1) -- is not $\aleph _{0}$ but $2^{\aleph _{0}}$.

But it would be helpful if you could explain what about the section you find unclear, so that perhaps it could be improved.

CRGreathouse (t | c) 04:56, 22 May 2011 (UTC)

The "intuitive argument" explains the concept of decimal expansion. Nothing else. The only sentence about the cardinality of the continuum refers to a concept that the reader is not supposed to know (power set), and does not give any insight about the relationship with that concept. The question is: "why the real numbers are $2^{\aleph _{0}}$"? The answer provided in the section is "because they are equivalent to the power set of the natural numbers". Not only the reason why they are equivalent is not explained, but also the reader is supposed to accept that the power set of $\mathbb {N}$ has cardinality $2^{\aleph _{0}}$". This anwser is not intuitive. It is useless. It creates two new questions, which are as difficult to answer as the first one. Paolo.dL (talk) 15:19, 23 May 2011 (UTC)

More exactly, the section says (without explaining the reason) that the cardinality of the interval [0, 1) is $2^{\aleph _{0}}$. Thus, a subset of the real numbers has the same cardinality as the whole set. This is totally counter-intuitive (see Space filling curve). In other words, instead of explaining the reason why the cardinality of $\mathbb {R}$ is $2^{\aleph _{0}}$, it forces the reader to accept that another set, which intuitively is "infinitely smaller" than $\mathbb {R}$, actually has the same cardinality as $\mathbb {R}$. So, the "intuitive argument" is not only useless, but also counter-intuitive. Paolo.dL (talk) 16:21, 23 May 2011 (UTC)

OK, I rewrote the section to use an argument which has the merit of being correct and simple (and relying on the decimal notation system), while perhaps suffering from the demerit of being less than 100% convincing. How do you like it? JRSpriggs (talk) 02:42, 24 May 2011 (UTC)

Intuition leads in the wrong direction. The truth is counterintuitive

The argument you inserted is not useless, but far from intuitive, and in my opinion is better explained in the following section (properties). Possibly, there's no intuitive way to explain this concept. Let me explain with an example. I will give you an argument which is wrong, but perfectly intuitive (i.e. not obviously wrong):

1. The natural numbers may have $\aleph _{0}$ digits. So, $\mathbb {N}$ can be described roughly as the set of all possible ways to fill these infinitely many digit positions (except that 1 = 0001 = 0000001 = 0000000000000000001, etc.).
2. The interval [0, 1) can be also defined roughly as the set of all possible ways to fill infinitely many digit positions (except that .1 = .1000 = .1000000 = .100000000000000, etc.).

It seems that intuition may easily lead the reader to a wrong conclusion. Should we conclude that the truth is totally counterintuitive, and hence the proof cannot be made intuitive? If we accept this conclusion, we have to remove the "intuitive argument". If we reject this conclusion, we must find an intuitive way to prove that the previous intuitive statement is wrong.

Paolo.dL (talk) 12:27, 24 May 2011 (UTC)

A natural number cannot have ℵ₀ digits because ℵ₀ is not a counting number: ℵ₀ ∉ ℕ. The number of digits in a natural number is a finite, natural number, unlike the digits following a decimal point. A natural number with infinite digits would be infinite, thus not natural. Perhaps there is a choice place to insert the word "finite" to highlight the difference.

Really, a number where every digit is preceded by a more-significant digit is a somewhat useless notation. What that tells you is that, for some numeric base, the remainder of the number with a power of the base is given. For example, …342 = n * 100 + 42 for n ∈ ℕ. However, this notation gives us no way to express that N = n * 12345 + 4321 for n ∈ ℕ. The idea would more likely be expressed with such notation alone, forgoing the notion of an infinite string of digits. Potatoswatter (talk) 21:27, 24 May 2011 (UTC)

Potatoswatter, your first paragraph is interesting, but highly counterintuitive. I fail to understand your point in the second paragraph.

Up to now, we only have gathered in this talk page and in the article a collection of statements that are "true", simple, but highly counter-intuitive, and not easy to prove. They are not what I was looking for, but I think they are useful. Since they are simple enough, at least they arouse curiosity and avoid misunderstandings (which are very likely in this case, as intuition may mislead readers). The only problem is that they are all but intuitive. Thus, the section title ("intuitive argument") is misleading. We should replace it with something like "Overview", or "Properties of natural and real numbers", or "Countable versus uncountable infinite". Paolo.dL (talk) 12:53, 25 May 2011 (UTC)

Here's another bunch of counterintuitive, but simple and curiosity-arousing statements (from Unit interval#Cardinality):

• The unit interval is a subset of the real numbers $\mathbb {R}$. However, it has the same cardinality as the whole set: the cardinality of the continuum. Also, it has the same cardinality as the unit square, or unit cube, or unit hypercube (see Space filling curve). Moreover, and even more surprisingly, it has the same cardinality as an n-dimensional Euclidean space $\mathbb {R} ^{n}$.
• Namely, the number of points in a line segment of length 1 is strictly larger than the number of natural numbers, and hence it is "uncountable". However, it is equal to the number of points in an infinitely long line, a square of area 1, a cube or hypercube of volume 1, or even an n-dimensional Euclidean space of infinite area or volume.

Paolo.dL (talk) 13:19, 25 May 2011 (UTC)

An argument which included a false statement, however intuitive, would necessarily be invalid (unless part of a reductio ad absurdum) and misleading. Thus it would be worse than useless. JRSpriggs (talk) 18:43, 25 May 2011 (UTC)

Of course. I guess you totally missed my point. I made a clear distinction between false but intuitive (see my example above), and true but counterintuitive. In short, I have been repeatedly suggesting that in this case intuition leads beginners in the wrong direction, so a true and intuitive argument might not exist. Hence, we are forced to use what we do have: some counterintuitive, yet true, simple and curiosity-arousing statements. Accordingly, I suggested to change the section title (see previous posting). By the way, in my opinion your latest edit belongs in the next section, not because it is (unavoidably!) counterintuitive, but because your formal terminology-notation, and you cardinal arithmetics is definitely too difficult to understand for beginners. I think we need something informal in the first section. Paolo.dL (talk) 21:04, 25 May 2011 (UTC)

Intuitiveness is subjective, so can you clarify what part you find unclear? To me it is quite intuitive that an infinitely long integer may be infinitely large — and that sums up my first paragraph. The second paragraph is just a critical evaluation of the notion of treating an infinite string of digits as an integer, as you suggested.

Lists of wp:trivia without context are doubly useless in math articles, so I don't want to be involved in any "collection of statements" that doesn't go anywhere. Potatoswatter (talk) 04:09, 26 May 2011 (UTC)

Would you say that your first paragraph is trivia that doesn't go anywhere? I would not. The comparison between $\mathbb {N}$ and the unit interval is crucial in this section. Thus, your first paragraph is, in my opinion, the most useful piece of information that we have collected in this discussion.

It is however extremely (and unavoidably) counterintuitive the notion that an infinitely large integer has a finite number of digits. As usual, the "intuitive" version of this statement is false: an infinitely large integer has infinitely many digits (e.g. 1 followed by infinitely many zeros). Why not? I am sure you agree that this false statement is much easier to believe for beginners than the true one. By chance, do you have a simple way to prove your statement? (Your text seems to be a circular argument, not a valid proof.)

I agree that "intuitiveness is subjective". It depends on what you already know. If you have studied set theory, your intuition is likely to lead in the right direction. Cantor may have guessed the final result before proving it with his diagonal argument. In this talk page, I was always referring to the intuition of beginners. As I wrote above, the section which is now called "intuitive statement" should be an informal overview, easy to understand for beginners.

Paolo.dL (talk) 12:54, 26 May 2011 (UTC)

No, an integer with a finite number of digits must itself be finite. "Infinitely large integer" is impossible — every integer has a finite value. And "1 followed by infinitely many zeroes" does not describe a number at all in the usual positional notation.

It is impossible to write an article which agrees with the intuition of every beginner, yet is correct. A tacit assumption in any "intuitive argument" is that the readers cup is empty to intuitively grasp the logic *as presented*, rather than use it as a guide to interpret or patch-up a *preconceived* argument. I get the impression you are being a little too creative, and the article is not the source of your confusion. Potatoswatter (talk) 13:31, 26 May 2011 (UTC)

I never suggested "to write an article which agrees with the intuition of every beginner"! Let me repeat, for the fourth or fifth time, that I suggest to use simple counterintuitive statements. Another example? These sentences of yours are highly counterintuitive, but they are simple and might be quite useful in the article (especially if they were better explained):

• "1 followed by infinitely many zeroes" does not describe a number at all in the usual positional notation.
• [There are infinitely many integers, yet] every integer has a finite value.

This is not explained in the article. It is not even stated! However, these statements are clear enough to arouse the curiosity of beginners. Without simple statements like these, the "intuitive argument" is not only counterintuitive (despite its name), but also useless to beginners. I have been trying to explain to you all how difficult it is to accept that there are "more elements" in the unit interval than in $\mathbb {N}$, although both sets are infinite. In other words, the fact that some infinite sets can be "more infinite" than others. In the mind of beginners, the words "more" and "infinite" are not compatible with each other. One way to introduce this notion is to say that each integer must have a finite number of digits, (as opposed to a real number which "has $\aleph _{0}$ digits in its expansion"). Is my point clear enough now? I hope so.

Does anybody feel the need of an informal overview, simple enough to be understood by beginners? If not, I'll stop contributing to this discussion. Paolo.dL (talk) 15:40, 26 May 2011 (UTC)

The first bullet here is irrelevant to counting any infinite set. The second is relevant to the cardinality of the natural numbers, not to the reals. See countable set for that topic.

Anyway, the "intuitive argument" section doesn't address this issue at all. It only proves that |ℝ| = 2^ℵ₀, which is useless to the lay reader as it doesn't prove that ℵ₀ ≠ 2^ℵ₀. For that, see Cantor's diagonal argument. These articles might address your questions outside the immediate scope of this article. On the basis that it only goes halfway by itself, perhaps the "intuitive" section should be merged into the following section. Potatoswatter (talk) 02:16, 27 May 2011 (UTC)

Yes, I do. The lead is incomprehensible to the lay reader. -- cheers, Michael C. Price ^talk 15:44, 26 May 2011 (UTC)

I agree the lead starts off with a lot of jargon, and won't help anyone who already knows the definitions of the terms it uses. Likely the first and second paragraphs should be switched. Other than that, are there particular questions you expect it to answer? Potatoswatter (talk) 02:20, 27 May 2011 (UTC)

The first paragraph in the lead is very similar to the usual list of alternative names/symbols for the discussed topic ("also called...", "AKA...", "also denoted...", etc.), which is typically in the first sentence of Wikipedia articles. It does not matter that the readers do not know what aleph 1 is, as far as they are told it is another way to denote the cardinality of the continuum, and as far as an internal link is provided, where they can find further information. The second paragraph is very well written. Well focused, not too technical, understandable by beginners. The problem is in the section "intuitive argument", and I am very surprised that Potatoswatter still cannot see how many logical steps are taken for granted there, and how much the number of digits in integers is relevant to the topic (nalemy, it is one of the most important missing steps). I'll explain later. Paolo.dL (talk) 09:47, 27 May 2011 (UTC)

Proposed lead

(continuing the above)

The lead prose isn't very good. I think that perhaps mathematicians expect to see statement of "knowns" followed by the statement of "more interesting things." In contrast, good informative English prose puts an interesting fact at the beginning of each paragraph, to let the reader know what they can expect to find. One thing that makes the lead difficult is that both initial sentences are excessively long. Here is a proposed revision. I rearranged a little, tried to remove parenthetical phrases (which always impede readability) and unnecessarily verbose phrases like "sometimes called," and replaced some <math> tags with Unicode to improve page rendering.

The cardinality of the continuum or power of the continuum is the size of the infinite set of real numbers ℝ. It is denoted by |ℝ|, ℶ₁ (beth-one), or ${\mathfrak {c}}$ (a lowercase fraktur script c).

ℶ₁ is a transfinite cardinal number. It is greater than ℶ₀, also known as ℵ₀ (aleph-naught), the smallest transfinite, which represents the size of the set of natural numbers. These quantities were first distinguished by Georg Cantor in his 1871 uncountability proof, part of his greater study of different infinities.

There may be no cardinal numbers between |ℕ| = ℶ₀ and |ℝ| = ℶ₁, if the continuum hypothesis holds. In that case, ℶ₁ is furthermore identical to ℵ₁.

Things I removed:

• "In mathematics": this only slows the reader. It is apparent that we are talking about math.
• That the continuum is the set of real numbers. The link to Continuum (set theory) is removed, but that article only seems to restate a few things from real number. Anyway, this seems obvious and/or unnecessary, and can probably be re-added without crufting it up too much.

As for "Potatoswatter still cannot see how many logical steps are taken for granted there"… I never said it was a good proof, but we are not critiquing mathematical rigor here. The next thing I would criticize, after failure to put the conclusion in context, is that there is too much algebra. The change of base is unnecessarily cryptic. Of course, taking transfinite arithmetic for granted pretty much reduces it all to nonsense.

Michael, does this look better? Potatoswatter (talk) 04:18, 28 May 2011 (UTC)

Looks better. Might be worth mentioning the power set and/or |ℝ| = 2^|ℕ|. -- cheers, Michael C. Price ^talk 11:40, 28 May 2011 (UTC)

Overall, it is a good job. Here are my suggestions for improvement:

• Wikipedia articles in math commonly start with "In mathematics", "in linear algebra", "in set theory", "in geometry", etc. Here, "in set theory" seems appropriate and useful to me, as it provides a context, and a link from which a beginner can start studying the topic.
• The sentence "|ℝ| is greater than |ℕ|, which represents the size of the set of natural numbers" must be highlighted, as it is the most interesting piece of information in the lead. It really catches the attention of beginners, as the notion of "different infinites" is highly counterintuitive, almost unbelievable for those who don't know set theory. SO, it should be at the beginning of a paragraph. Your text does not fail to provide this information, but the sentence is mixed with less important details. For instance "the smallest transfinite" is a detail that should not be in the middle of this sentence.
• I suggest to use |ℝ| and |ℕ| as the first symbols given for both the sizes of R and N. They are easy to remember, and |ℝ| is the first symbol given for the cardinality of the continuum.

So, here's how I would write the introduction (I showed in bold typeface the parts I edited; I also rearranged the second paragraph):

In set theory, the cardinality of the continuum or power of the continuum is the size of the infinite set of real numbers ℝ. It is a transfinite cardinal number, denoted by |ℝ|, ℶ₁ (beth-one), or ${\mathfrak {c}}$ (a lowercase fraktur script c).

The cardinality of the continuum is greater than the size of the set of natural numbers, denoted as |ℕ|, ℶ₀ (beth-zero), or ℵ₀ (aleph-naught). These quantities were first distinguished by Georg Cantor in his 1871 uncountability proof, part of his greater study of different infinities. Cantor showed that |ℕ| is the smallest transfinite cardinal number, and |ℝ| coincides with the size of the power set of ℕ, thus |ℝ| = 2^|ℕ|. As a consequence, |ℝ| > |ℕ|.

If the continuum hypothesis holds, there are no cardinal numbers between |ℕ| = ℶ₀ and |ℝ| = ℶ₁. In that case, |ℝ| is furthermore identical to ℵ₁ (aleph-one).

Paolo.dL (talk) 12:26, 29 May 2011 (UTC)

I like it. -- cheers, Michael C. Price ^talk 16:57, 29 May 2011 (UTC)

I also suggest to use the <math> format for ℝ, ℕ, ℶ and ℵ (see current format in the lead), because in Internet Explorer 9, with zoom 100%, the Unicode format suggested above is too small to show the correct shape of these letters (particularly aleph and beth). The <math> format is also used in the articles Aleph number and Beth number. Paolo.dL (talk) 19:31, 29 May 2011 (UTC)

Pretty productive changes mostly. I'd trim it back a little and try to reduce the average sentence length.

• Maybe better to stick with one name for the cardinality of the naturals. ℶ₀ needs less introduction because the first sentence contains ℶ₁, and ℵ₀ is very famous, but I think |ℕ| has the best of both worlds — analogy to the previously-mentioned concepts and very common usage.
• Separate notation from basic definition in the first paragraph. This improves flow, and segues into the math.
• I'm suggesting to begin the second paragraph with a symbol, simply because the proper name is so verbose.
• Break up sentences in the second paragraph. Group math at the beginning and bio at the end. Spruce up bio.
• Remove "size of the" and magnitude signs, and replace notion of causality with mere difference of notation, in the new sentence of the second paragraph.
• Just a note, "As a consequence, |ℝ| > |ℕ|" is a simple restatement of the paragraph thesis. I think it's worthwhile here, but there's an element of "repetition in lieu of proof."
• Avoid starting the third sentence with a clause about the continuum hypothesis. Beginners may be interested to know that there is perhaps no x for |ℕ| < x < |ℝ|, or that the status of this inequality is unknown. That motivates moving on to the hypothesis. Perhaps my original text was problematically ambiguous; I fixed that by breaking up the sentence.

I'm OK with going back to TeX formatting. The problem on Firefox is that math renders larger than the rest of the text, but a problem with smallness trumps that.

In set theory, the cardinality of the continuum or power of the continuum is the size of the infinite set of real numbers ℝ. It is denoted by |ℝ|, ℶ₁ (beth-one), or ${\mathfrak {c}}$ (a lowercase fraktur script c). It is a transfinite cardinal number.

|ℝ| is greater than the size of the set of natural numbers, denoted as |ℕ|, which is the smallest transfinite cardinal number. ℝ coincides with the power set of ℕ, denoted ℝ = 2^ℕ. As a consequence, |ℝ| > |ℕ|. This was proven by Georg Cantor in his 1871 uncountability proof, part of his groundbreaking study of different infinities.

There may be no cardinal numbers between |ℕ| and |ℝ|. This is true if the continuum hypothesis holds. In that case, |ℝ| is furthermore identical to ℵ₁ (aleph-one), the second-smallest transfinite cardinal number.

Potatoswatter (talk) 07:16, 30 May 2011 (UTC)

• ℝ = 2^ℕ is clearly wrong. I wish it were true: it would make our job simpler. As Michael C. Price suggested, we need to state the relationship between |ℝ| and |ℕ|, i.e. |ℝ| = 2^|ℕ|. In this context, this is an important piece of information.
• I would rather not remove the alternative symbols for |ℕ|. They are useful as readers can compare them with the corresponding symbols for |ℝ|, and understand the subscripts for aleph and beth. (I know you don't like parentheses, but they can be removed without removing the symbols.)
• Good job in the third paragraph (thank you for correctly guessing and respecting my rationale).

Paolo.dL (talk) 08:22, 30 May 2011 (UTC)

ℝ = 2^ℕ does seem incorrect. Perhaps |ℝ| = 2^|ℕ|, or ℝ = ℘(ℕ). (Symbol being copied from power set.) These changes seem more mathematically accurate, although, as an expert mathematician, I can't really say whether it's more understandable. — Arthur Rubin (talk) 09:19, 30 May 2011 (UTC)

The reals and the powerset of the natural numbers are not the same set, and the article makes no such claim. However, they are equinumerous as the article proves. JRSpriggs (talk) 11:29, 30 May 2011 (UTC)

Arthur, when you wrote ℝ = ℘(ℕ) what did you mean? As far as I understand, ℘(ℕ) is just another symbol for the powerset of ℕ (denoted 2^ℕ in previous contributions). In this case, ℝ = ℘(ℕ) can't be correct, as by definition the power set of ℕ is different from ℝ (they only have "a few" elements in common). JRSpriggs, we are discussing a proposal by Potatoswatter for a new lead (see comment posted at 07:16, 30 May 2011). Paolo.dL (talk) 11:41, 30 May 2011 (UTC)

|ℝ| = 2^|ℕ| is fine with me. -- cheers, Michael C. Price ^talk 13:01, 30 May 2011 (UTC)

TeX versus Unicode symbols for $\mathbb {R} ,\mathbb {N} ,\aleph ,\beth$

On a computer with Windows XP and Internet Explorer 8 (standard settings; UTF-8 encoding), both updated with Microsoft Update, I cannot correctly visualize the Unicode symbols for aleph, beth, R and N used by Potatoswatter. I see a square instead. The same is true for most of the "Math and logic" symbols provided in the Editing page. (By the way, I can see most "Symbols", and all the Latin, Hebraic, Arabic, Greek and Cyrillic characters). Paolo.dL (talk) 14:48, 30 May 2011 (UTC)

That sounds like a general problem that needs addressing elsewhere. -- cheers, Michael C. Price ^talk 15:54, 30 May 2011 (UTC)

I think this is relevant to this discussion. I don't think this is a problem that can be solved by Wikipedia. Millions of people throughout the world use English Wikipedia with IE 8 in Windows XP (not compatible with IE 9). They might solve the problem by downloading some plugin or another browser, but it seems plausible that most of them would not want or be skilled enough to do it. So, if we use Unicode font for aleph, beth, R and N in the introduction, they simply would not be able to read it. Paolo.dL (talk) 19:09, 30 May 2011 (UTC)

Honestly I'm against blackboard bold N and R. Let's just use bold. --Trovatore (talk) 19:22, 30 May 2011 (UTC)

I guess you mean $\mathbb {R}$ and $\mathbb {N}$. I prefer this format (although it is too large), as bold capital letters are also used for matrices. E.g., R is typically a rotation matrix with elements in $\mathbb {R}$. But if somebody else agrees with you, I will abide.

We also need consistency in font format, at least within this article.

Paolo.dL (talk) 19:36, 30 May 2011 (UTC)

Almost all mathematical symbols are overloaded; not too much we can do about that. My usual feeling is, blackboard bold is for the blackboard. That's not to say I've never used it, but I don't see why we need it for N and R. --Trovatore (talk) 20:00, 30 May 2011 (UTC)

Yes, clearly $\mathbb {R} =2^{\mathbb {N} }$ would require very inconveniently defining several operators. Sorry about that, I'm definitely not a mathematician. And, I definitely prefer the TeX, blackboard-bold notation to boldface. It is a bit oversized, but it's universally understood and unambiguous. "For blackboards" does not reflect reality; these symbols have been common in typeset books for a long time, not to mention other Wikipedia articles. Many readers will fail to see the relation between N and $\mathbb {N}$. TeX-to-HTML (in the Wikipedia preferences under "Appearance") will one day fix the size and portability problems, although it doesn't yet. (Hopefully before IE 8 becomes unpopular.) Potatoswatter (talk) 06:40, 31 May 2011 (UTC)

Some books use them; some don't. Some books also use the (useful abbreviation but hideous when abbreviation is unnecessary) iff; consensus at the math project is clearly against the latter. I see the bbb symbols as a bit of a similar issue (though I certainly don't feel as strongly about them as about iff). --Trovatore (talk) 09:57, 31 May 2011 (UTC)

New paragraph

I also propose to add this paragraph:

• The term continuus refers to an ordered collection of elements such that, between any two of them, no matter how close they are to each other, there's always an infinite number of other elements. For instance, there are infinitely many real numbers between 0 and 1, or 0 and 0.01. Interestingly, it can be shown that they are as many as those in the whole set $\mathbb {R}$. In other words, the interval [0,1], or in general [a,b], has the same size as $\mathbb {R}$. This is also true for many other continuous sets, such as any n-dimensional Euclidean space $\mathbb {R} ^{n}$ (see Space filling curve). Thus:

$|[a,b]|=|\mathbb {R} |=|\mathbb {R} ^{n}|=\beth _{1}.$

Paolo.dL (talk) 12:56, 31 May 2011 (UTC)

I'm opposed. Assuming that you mean distinct elements (else no total order is continuous), this describes the rational numbers which are strictly smaller than the reals. So it muddies the waters needlessly. CRGreathouse (t | c)

No, I absolutely did't mean distinct, nor rational. I meant continuous, which clearly implies some notion of "order" of elements, such that you can say, for any element in the set, whether it is smaller or greater than another, or (for points in $\mathbb {R} ^{n}$) less/more distant than the other from an origin. I probably misused the expression "ordered set". I wanted to use the word "sequence", but then I realized that "sequence" is only related to discrete sets. It is clear, however, that we need the property that I described above to tell a continuous set from a discrete set, or a set with "gaps", because otherwise it would be impossible to identify which elements of a set are "between" two other elements of the same set. Paolo.dL (talk) 17:50, 31 May 2011 (UTC)

You misunderstand. You left the word "distinct" out of your definition of "continuous". Using your definition as written, the real numbers under < are not continuous because there are only finitely many real numbers between 1 and 1. If we change the definition to

The term continuus refers to an ordered set such that, between any two distinct elements there are infinitely many elements between them.

then the rational numbers are continuous under <.

CRGreathouse (t | c) 19:11, 31 May 2011 (UTC)

It's also not standard terminology. The standard term for this is dense (specifically, order-dense). As CRG says, this does not exclude countable sets like the rationals, so it seems like it would be confusing to bring it up in this context. --Trovatore (talk) 19:42, 31 May 2011 (UTC)

I see. Thank you both for explaining. The most important part of my "new paragraph", however, is the idea that a subset of R can have the same cardinality as the whole set, and even more surprisingly, |R| = |Rⁿ|. In my opinion, this should be mentioned in the introduction, for two reasons: (1) there's a section in the article which lists several sets with cardinality $\beth _{1}$, and (2) the idea that some "infinites" are identical while they definitely and deceptively appear to be different is even more fascinating than the idea that there are "different infinites". This is the first purpose of my text. A secondary purpose is to show that all of this sets which surprisingly share with R the same cardinality appear "continuous", or more properly "dense"..., and this explains the term "continuum" in the expression "cardinality of the continuum" (although there are dense sets which have a strictly smaller or greater cardinality).

I have a doubt, however: is the power set of N a dense set? I wrote that all the sets with cardinality $\beth _{1}$ are dense. Is this correct?

Paolo.dL (talk) 20:40, 31 May 2011 (UTC)

If by your purposes you mean making the article accessible to more readers, then I'm with you there. But it's not clear to me how much this article should serve as an introduction to cardinal arithmetic. I'm not opposed, on the whole, but would prefer to see what other experienced editors think.

I still disagree with the particulars you bring up, though. Consider the real numbers and the (total) order

0 ≺ r for all r in R, 1 ≺ r for all r in R \ {0}, a ≺ b if a, b in R \ {0, 1} and a < b

under which the interval (0, 1) is empty and yet the set is uncountable.

CRGreathouse (t | c) 20:47, 31 May 2011 (UTC)

To address your addition: as my example shows, you can't look at a set alone to determine that, you need its ordering or topology as well. CRGreathouse (t | c) 20:48, 31 May 2011 (UTC)

Thank you. I appreciate your detailed explanations. The total order described in your example is extremely complex (I don't even know exactly the meaning of the symbol ≺, but I trust you that it makes (0,1) empty). I tend to think that it is safe to assume that those who are able to imagine such an order are also able to understand my text. It is easy to guess that the expression "between two of them" in the first sentence of my "new paragraph" refers implicitly to two distinct elements of a set totally ordered under an elementary relation such as <. However, as you pointed out previously, my sentence is misleading for another reason. So, let's forget my "secondary purpose" (explaining the term "continuum"). A link to Continuum (set theory) somewhere in the intro will suffice.

My "main purpose" was another. For the two above-listed reasons, I proposed to state in the lead that the real numbers between 0 and 1 are as many as those in R or Rⁿ, i.e.

$|[a,b]|=|\mathbb {R} |=|\mathbb {R} ^{n}|=\beth _{1}.$

(Notice that I purposedly avoided technical terms such as interval and equinumerous in this sentence, to make it easier to understand for beginners). This statement is not referred to a generic "totally ordered set", but to R, which is typically well-ordered under <. In this case, we can safely assume that whoever is both interested to read this article and able to understand the first sentence of the lead will also be able to guess that "between 0 and 1" means 0<x<1. If you agree with me about this, I will try and re-write my "new paragraph" accordingly.

Paolo.dL (talk) 15:45, 1 June 2011 (UTC)