Talk:Cardinality of the continuum

Recent reverts about cardinality of real numbers with non-unique binary representation

I have not checked this completely, I'm just arguing off the top of my head. As far as I know, the only ambiguity arising as expressing numbers as binary expansions is that 0.1=1.0 and 0.01=0.10 . At least, I think this is the case, analogously to 0.9=1.0 in decimal. If this is true that the only real numbers without unique representation are the ones terminating in 1, isn't it obvious only countably many numbers suffer this problem? It would be equivalent to counting the finite sequence of digits that come before the infinite repeating tail, and there are only countably many such finite sequences. Addendum: Rereading the reverts, maybe I have misunderstood what the problem really is. Maybe the mistake, meriting removal, is at a juncture other than this. If so, nevermind! Rschwieb (talk) 18:45, 2 February 2012 (UTC)

I quite agree with your interpretation. I don't understand the opposing view at all. — Arthur Rubin (talk) 18:52, 2 February 2012 (UTC)

I think the problem is at another juncture; Isn't it obvious that the interleaving function is well defined when only one of the inputs does not terminate in 1? Interleaving a sequence ending in 1 with one that does not would result in one that does not. Rschwieb (talk) 18:53, 2 February 2012 (UTC)

That's worse, I'm afraid. If x and y both have nonunique expansions (x₀, x₁; y₀, y₁ respectively, then all 4 possibilities are different reals:

$\left(x_{0},y_{0}\right),\left(x_{0},y_{1}\right),\left(x_{1},y_{0}\right),\left(x_{1},y_{1}\right)$

although the first and last also have different representations as reals. If only one is non-unique, then there are two possible results, both unique. — Arthur Rubin (talk) 20:09, 2 February 2012 (UTC)

Oh, it is supposed to have domain and range with real numbers. In that case, then the map works, it's just that the statement about the countability of nonuniquely expressable reals is irrelevant. That particular blurb seems to have appeared here, contributed by a user who has a history of adding not-quite-correct and questionably relevant statements to this article. Rschwieb (talk) 20:35, 2 February 2012 (UTC)

As the person that made the original revert perhaps i should clarify. The problem is that the interleaving function is not a surjection. Consider the reverse of the interleaving function with (in decimal) $z_{1}=0.25{\overline {89}}$ and $z_{2}=0.26{\overline {80}}$, both of these will map to the same $(0.2{\overline {8}},0.6)$. Since it's not a bijection it doesn't say that the cardinality of $\mathbb {R}$ and $\mathbb {R} ^{2}$ is the same. The placement in the article makes it seem as if this function was somehow an argument for their cardinalities being equal, which is why it doesn't belong there.The statement that it is only a countable number of points that should be removed is false because the function is not surjective for an uncountable number of points in the unit square. For all $(x,y)$ where y is non-unique this problem exists (for example $(0.2{\overline {8}},0.6)$ and that set is uncountable.

I'll concede the point if someone can tell me where $z_{1}=0.00{\overline {10}}\in X$ and $z_{2}=0.1{\overline {0}}\in X$ map to in $X\times X$?

Nevermind, you're right the map obviously isn't onto X! I was so distracted by the task of removing the irrelevant analysis of where the interleaving map was defined and didn't stop to question that statement. Any more surgery on the domain and range of this function is will make the interleaving example useless, so perhaps it is best to leave it out. Rschwieb (talk) 14:34, 4 February 2012 (UTC)

Proofs

(Per request on my talk page) I have added proofs for some of the examples listed because I believe that it's helpful for the reader's understanding. The list of examples itself is justified for the same reason. DavidLeighEllis (talk) 01:43, 26 April 2014 (UTC)

Well, what I really wanted to know was, how far are you planning to take this? Are you planning on adding more examples/more proofs? How many, how detailed? Do you have a grand vision, and could you let us in on it? --Trovatore (talk) 01:48, 26 April 2014 (UTC)

I'm pretty much done with this article, actually. DavidLeighEllis (talk) 01:50, 26 April 2014 (UTC)

OK, I don't want to discourage you. I think the additions are probably an improvement on balance. I was just a little concerned about where this might end up. --Trovatore (talk) 01:58, 26 April 2014 (UTC)

Cardinality

Examples of cardinality 196.249.99.211 (talk) 17:58, 22 December 2021 (UTC)

Continuum Hypothesis

The second source clearly stated it is undecidable that ${\mathfrak {c}}=2^{\aleph _{0}}$. However, it is only briefly mentioned once at the end of the introduction that most of this section is dependent on the continuum hypothesis. The "intuitive" property section depends on Cardinal arithmetic, which appears to be dependent on GCH. For now, I proposed to merge the last paragraph of the introduction into the first paragraph and suggest experts in this field to add more clarifications (on assumptions in addition to ZF set theory). --Yangwenbo99 (talk) 11:12, 14 February 2023 (UTC)

As far as I can see, nothing in the article depends on CH. You seem to have a misimpression about what CH actually is, which might explain the confusion.

The continuum hypothesis is not ${\mathfrak {c}}=2^{\aleph _{0}}$, which is just a theorem of ZFC. The continuum hypothesis is ${\mathfrak {c}}=\aleph _{1}$, which is a different claim. --Trovatore (talk) 17:24, 14 February 2023 (UTC)

When the cardinality of the reals is first mentioned in the article the statement is simply that the cardinality of the continuum is

${\mathfrak {c}}=2^{\aleph _{0}}>\aleph _{0},$

with an epitaph that "this was proved by Cantor". It would be more correct to say that Cantor showed that the cardinality of the reals is greater than the cardinality of the naturals and the former equals the power set of the latter follows from the continuum hypothesis added to ZFC set theory. As Trovatore correctly points out, the above identity is a theorem of ZFC+CH. However simply stating it's veracity before any mention of ZFC+CH renders this article into the land of not mathematics. Izmirlig (talk) 12:41, 25 June 2024 (UTC)

Note that

${\mathfrak {c}}=2^{\aleph _{0}}>\aleph _{0}$

is provable in ZF, it doesn't need CH or even choice. CH is the nonexistence of an x such that

$\aleph _{0}<x<{\mathfrak {c}}.$

CRGreathouse (t | c) 15:59, 25 June 2024 (UTC)

I really don't understand why Izmirlig's error is so prevalent, but let's say this clearly: The fact that the cardinality of the reals is the same as the cardinality of the powerset of the naturals is a theorem of ZFC. It does not depend in any way on the continuum hypothesis. This is not subtle, this is not a matter of opinion; this is as straightforward as it gets. --Trovatore (talk) 19:53, 25 June 2024 (UTC)

Reverted contrib

Hi @Engr. Smitty,

Why was this contribution reverted? Oneequalsequalsone (talk) 00:18, 17 October 2023 (UTC)

I'm not the person you pinged, but I don't think it's true, is it? I'm pretty sure, for any $\kappa$, you have $\{0,1\}^{\kappa }$ is a compact Hausdorff space without isolated points, but clearly we can make it have cardinality bigger than the continuum. --Trovatore (talk) 00:41, 17 October 2023 (UTC)

Thanks! I'm fairly new to this space (no pun intended) and trying to piece things together Oneequalsequalsone (talk) 13:27, 17 October 2023 (UTC)