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I have what many on here would consider an elementary question, but I would very much appreciate responses that use only elementary ideas, if possible, so that I can understand them. I would also appreciate detailed rather than brief responses.

By construction, $\mathbb{Q} \subseteq \mathbb{R}$. The rationals are countably infinite, while the irrationals are uncountably infinite. This got me thinking about what properties the irrationals have that the rationals do not have that would cause such a huge difference in their cardinalities (although this question isn't specific to their cardinalities -- the rationals have 0 Lebesgue measure and the irrationals have infinite Lebesgue measure).

Both the rationals and the irrationals are dense in $\mathbb{R}$. So density does not play into cardinality. But why doesn't it? There are uncountably many irrationals, and so there are uncountably many intervals $(a,b)$ with irrational endpoints. Since each of these intervals contains a rational number, shouldn't there be uncountably many rationals? There aren't uncountably many rationals, which means there is at least one rational that is contained in uncountably many of these intervals. But which one? Are they all contained in uncountably many of these intervals? What if we only look at the subcollection of all intervals with irrational endpoints that also have infinitesimally small length (if possible)? I know I'm rambling now...

I guess my main question is: what properties do the irrationals have that the rationals don't have that lead to the irrationals being uncountable? Although, I would also like to hear thoughts on my statement above about the density of the rationals.

layman
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  • There's no such thing as 'infinitesimally small length' (in $\mathbb{R}$). You hit the nose on the head; every rational is in fact contained in uncountably many of these intervals (as might be obvious since for instance, you can just take the intervals $(q-r, q+r): r\in\mathbb{R}^+$ ). In fact, for any $\epsilon\gt0$, every rational is contained in uncountably many intervals of length $\epsilon$. – Steven Stadnicki Jul 18 '14 at 03:40
  • Why can't we talk about infinitesimally small lengths in $\mathbb{R}$? I have always felt queasy about infinitesimally small numbers, and now I am confused about why they are not applicable in this scenario. – layman Jul 18 '14 at 03:41
  • You can talk about arbitrarily small lengths, but you can't talk about infinitesimally small lengths - there are no such things as infinitesimals in $\mathbb{R}$ (or $\mathbb{Q}$, for that matter). There are nonstandard theories that do have such things as infinitesimal numbers, but they're not generally regarded as 'the' theory of the reals, for instance. – Steven Stadnicki Jul 18 '14 at 03:44
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    With, for instance, digital expansions or continued fraction representations, specifying a real requires an infinite amount of data, while specifying a rational only requires a finite amount. The only difference between rational and irrational is that rationals terminate/repeat. – anon Jul 18 '14 at 03:44
  • I understand the idea of what you are saying. Irrationals are real numbers with non-repeating, non-terminating decimal expansions, while rational numbers are either repeating or finite decimal expansions. But I'm not making a rigorous connection as to why this leads there to be uncountably many irrationals. I do understand Cantor's diagonalization argument. – layman Jul 18 '14 at 03:48
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    @blue That's an information-theoretic distinction (or, arguably, a set-theoretic one); I think the germ of the OP's question here is looking for a topological distinction between $\mathbb{Q}$ and $\mathbb{R}\setminus\mathbb{Q}$, and the 'infinite data' explanations of the irrationals don't really provide that. – Steven Stadnicki Jul 18 '14 at 03:49
  • Also, here is another question that has been bothering me. If $x$ is an irrational number, does its decimal expansion contain every possible finite string of numbers? Should I ask a separate question on this? – layman Jul 18 '14 at 03:50
  • @StevenStadnicki I consider it a metrical distinction. The OP is looking for something at the heart of why one is countable why the other isn't. If for some reason they don't find this answer satisfactory, they should say so (and preferably explain why). Of course, how they're different as topological spaces is also a valid question - I am not sure at the moment if it's good for discussing cardinality, though. – anon Jul 18 '14 at 03:55
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    @user46944 no $\sum 10^{n^2}$ is irrational since it doent repeat. – Rene Schipperus Jul 18 '14 at 03:55
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    @user46944 That question can be answered immediately without even thinking: just create a nonrepeating sequence that doesn't contain all possible finite strings! – anon Jul 18 '14 at 03:55
  • Does .123411223344111222333444... work? – layman Jul 18 '14 at 03:56
  • Yes, that's non-repeating and fails to contain every digit. – anon Jul 18 '14 at 03:57
  • @blue, Thank you for clearing that one up. :) I think I understand your answer to my original question as well, but I am looking for as many perspectives on the subject as I can get. – layman Jul 18 '14 at 03:58
  • Digits, to whatever base a culture may use, are of at most peripheral interest. – André Nicolas Jul 18 '14 at 04:05
  • One could be less artificial and just speak of (equivalence classes of) Cauchy sequences to say the same things, sans the arbitrary choice of base (and choice of using digital representations at all). – anon Jul 18 '14 at 04:15
  • The irrationals can be made into a complete metric space without isolated points, and any complete metric space without isolated points must be uncountable. –  Jul 18 '14 at 06:02

2 Answers2

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In a metric space one can speak of a sequence of approximations which grow arbitrarily precise in the limit. One can phrase this in terms that do not actually assume there is a limit - namely, through the use of Cauchy sequences. A metric space fails to be complete if one can provide such a sequence of approximations growing arbitrarily precise but fails to converge to something. In a sense, we are specifying something that is not actually there - it is sort of a ghost, it transcends the space. It exists outside, in the completion of that space. One of the ingeniously clever tricks of math was instead of seeking to fully describe something infinite or beyond description, describe how one gets there (even though one never truly gets there). By identifying the destination with the journey, one gains the ability to speak of things like real numbers, limits from analysis, and even more exotic limits that exist in algebra (like, in the category of topological rings, the $p$-adic numbers).

There are different ways of going about "specifying" a real number. In general, any Cauchy sequence (modulo null sequences - those converging to $0$) will do, but kinds of specifications that come with rules are also nice. One can use digital expansions with respect to a chosen base - this is convenient and practical, if artificial. One can use continued fraction expansions (search this on google for more information). One can use Dedekind cuts, as a theoretical tool.

In these sorts of schemes (which are all metrically based - if one wants to use algebra and minimal polynomials to describe elements, or speak of computable/definable reals etc. then things can go a different way), rational numbers require only a finite amount of data to specify, whilst an arbitrary real will in general require an infinite amount of data. (I consider the repetition of digits a finite amount of data.) Since it shows up in all these metrical "specification" schemes, and the reals are defined metrically by such specifications (the reals are the unique archimedean linearly ordered complete field - and completeness refers to these things), I think this answer gets at the heart of why $\Bbb R$ is uncountable while $\Bbb Q$ isn't. To illustrate with binary: choosing a finite sequence of flips of a coin essentially encodes a natural number in binary, of which there are countably many, but choosing an infinite sequence of flips has a sample space of $2^{\aleph_0}$ which is $>\aleph_0$ by Cantor (and this theorem is not just auxillary to the discussion: to have a fair appreciation for set theory, it needs to be absorbed into one's intuition of what it means to be uncountable).

I think it is better to think of $\Bbb Q$ vs. $\Bbb R$ (the first contained in the second) instead of $\Bbb Q$ vs. $\Bbb R\setminus\Bbb Q$; for purposes of discussing cardinality there is no reason to partition $\Bbb R$ into $\Bbb Q$ and $\Bbb R\setminus\Bbb Q$.

Getting to the points you made: the fact that rationals are dense essentially means that we can use them in our specification of real numbers, but it fails to touch on the fact that it can take an infinite number of them to do that job, involving an infinite number of choices (even after getting rid of redundancies via null sequences), and the number of choices is what ultimately controls the cardinality. Yes, every rational $x$ is contained in uncountably many intervals with irrational endpoints - namely $(x-\epsilon,x+\epsilon)$ for every irrational $\epsilon>0$ - but this doesn't mean there are uncountably many rational numbers. If you try to pinpoint why you might suspect this implication of holding in the first place - a potential correspondence between intervals and points that forces $\Bbb Q$ to be uncountable - you will fail to find any meaningful correspondence. Why we begin with that kind of intuition in the first place I'm not sure, but intuition must be honed in light of facts.

anon
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  • Given the emphasis in this answer on the reals as the completion of the rationals, I'd be curious to see you elaborate a bit more on the $p$-adic numbers in this context. – Semiclassical Jul 18 '14 at 05:08
  • @Semiclassical $\Bbb Q_p$ is also a completion of the rationals, and all of the same language I used still applies, just with a different metric. (One can motivate this metric: often we speak of congruences, and getting residues to higher powers of primes means more is known about the original integer - greater precision. By allowing sequences of approximations of arbitrary precision on this view, we get $\Bbb Z_p=\varprojlim\Bbb Z/p^n\Bbb Z$.) – anon Jul 18 '14 at 05:16
  • I actually meant that it might improve the answer further if you were able to incorporate that fact more explicitly. Right now the reader might be forgiven for thinking that the reals are the only game in town as far as completions of the rationals go. – Semiclassical Jul 18 '14 at 05:23
  • I've read your post three times. I took a set theory course, so I am familiar with describing real numbers as Dedekind cuts. It seems to me the crux of your argument, which I agree with, is that the irrationals contain an infinite amount of "data", while the rationals do not. This concept is still nebulous to me. Also, I am wondering if it's possible to find more than one defining characteristic of $\mathbb{R}$ and $\mathbb{R} \setminus \mathbb{Q}$ that leads to these differences in cardinality. Thanks for your answer. – layman Jul 18 '14 at 11:50
  • @user46944 Is the fact that the collection of infinite binary strings is uncountable, while the collection of finite strings is countable, nebulous? One can make a visual depiction (this is for $\Bbb Z_2$, but the same holds for $\Bbb R$ with only a few details changed) with trees: put down a node, and two child nodes, each having two more child nodes, and each of those having two child nodes, and so on. An integer is a node, while a $2$-adic integer is literally an infinite chain of nodes. Perhaps you should become familiar with more than just Dedekind cuts for describing real numbers. – anon Jul 18 '14 at 18:13
  • @blue I'm not familiar with $p$-adic numbers, but I do understand why the set of infinite binary strings is uncountable. It can be proved using Cantor's diagonalization argument. But this begs the question in my mind: what properties does the set of infinite binary strings possess that make its cardinality differ from the set of all finite binary strings? The answer to that question is what is fuzzy to me, especially since we are talking about sets of infinite cardinality, where intuition is thrown out of this 100mph moving truck's window. – layman Jul 18 '14 at 18:23
  • By the way, my quest to receive 20 different perspectives clearly didn't work out since not many people were interested enough to answer this question. :( – layman Jul 18 '14 at 18:25
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    Then the "nebulousness" of this concept to you is much more fundamental than just reals and rationals, it goes to Cantor itself. Like I said, you need to make it intuitive, or else you'll be stuck in that rut. Another explanation for why you haven't received 20 different perspectives other than lack of attention (you did get 7 upvotes and a star, which is above average), is that there just aren't that many. Perhaps there is another one involving topology that forces $\Bbb Q$ to be countable and $\Bbb R$ uncountable - that doesn't reference completeness, a defining property. – anon Jul 18 '14 at 18:31
  • @blue Ok, I'll do my best to make this concept intuitive. I strongly believe I have an over-exposure to finite sets (and also a handicap of trying to visualize infinite mathematical concepts in the setting of this arguably finite world) that keep me from accepting these ideas and moving on. I'll eventually be able to rationalize it to myself. Thanks for all of your input, and if you have anything more to add, I'd love to hear it. – layman Jul 18 '14 at 18:36
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$\mathbb R\setminus \mathbb Q$ has the property of being the complement of a proper subgroup of an uncountable group.

If $G$ is an uncountable group and $H$ is a subgroup with $H\neq G$, then $G\setminus H$ is uncountable. If $H$ is uncountable, let $a$ be any element of $G\setminus H$, and then $aH$ is an uncountable subset of $G\setminus H$. If $H$ is countable, then because $G$ is the union of cosets of $H$ and a countable union of countable sets is countable*, $H$ has uncountably many distinct (and therefore disjoint) cosets contained in $G\setminus H$.

*[assuming choice]

Jonas Meyer
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  • What is so special about $\mathbb{Q}$ that makes it dense in $\mathbb{R}$? $\mathbb{Q}$ is countable, like $\mathbb{N}$ and $\mathbb{Z}$, but neither of these two are dense in $\mathbb{R}$. $\mathbb{R} \setminus \mathbb{Q}$ is uncountable, but it shares the same property of density as $\mathbb{Q}$, which indicates that density is independent of cardinality (as unintuitive as that seems...) -- so why is density independent of cardinality, when it seems like it shouldn't be? This really bothers me. – layman Aug 21 '14 at 18:51
  • To answer your first question, staying somewhat in the spirit of this answer, every subgroup of $\mathbb R$ is either dense or has a smallest positive element. $\mathbb Q$ does not have a smallest positive element. On the other hand, the complement of any proper subgroup of $\mathbb R$ is dense. (If the subgroup itself is dense, just take any coset not equal to the subgroup. If the subgroup is not dense, it is nowhere dense.) No mention of cardinality is needed to verify the claims in this comment. – Jonas Meyer Aug 21 '14 at 18:55
  • I'll work on proving the claims in your comment. Hopefully it won't be too hard, and maybe it will give me some intuition about this like it seems to have given you. Thanks for your input! – layman Aug 21 '14 at 19:02
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    @user46944: I hope you enjoy doing so, but I do not make claims that this approach is particularly geared toward building intuition. That would be a bonus. It is just a perspective I find amusing. – Jonas Meyer Aug 21 '14 at 19:04