Formalizing an idea

Question

All rational numbers of the unit interval [0, 1] can be covered by countably many intervals, such that the $n$-th rational is covered by an interval of measure $1/10^n$. There remain countably many complementary intervals of measure $8/9$ in total.

Does each of the complementary intervals contain only one irrational number? Then there would be only countably many which could be covered by another set of countably many intervals of measure $1/9$.

Is there at least one of the complementary intervals countaining more than one irrational number? Then there are at least two irrational numbers without a rational between them. That is mathematically impossible.

My question: Can this contradiction be formalized in ZFC?

I'm not sure about your question, but in any interval there are infinitely many irrational numbers... — Asaf Karagila, Apr 15 '12 at 10:31
I wouldn't be so sure about "countably many complementary intervals". Imagine removing just the rationals from $[0,1]$. This way you remove only countably many points. The set of irrationals that remains, contains no non-degenerate intervals, i.e. its connected components are singletons. An uncountable number of singletons. So, I'm pretty sure the construction you mention does something similar. — Dejan Govc, Apr 15 '12 at 10:59
I remember this which seems to be somewhat related to this question. — Asaf Karagila, Apr 15 '12 at 11:04
Thanks Asaf for the hint. But it is obvious that the answer given there with "indeed" and "ín fact" is not mathematics. By induction it can be proven that n intervals yield n-1 complementary intervals. This is true for every n and, therefore, for all n. We count the complementary intervals. Otherwise, if induction would not reach till the first infinity, Cantor's counting of the rationals could not cover all rationals; it woulde be invalid too - like the counting of the complementary intervals here. If bijection is denied in one example then it can also be denied in the other one. — , Apr 15 '12 at 11:22
But you only prove that by induction for finite $n$. Your claim is about an infinite collection of intervals. Induction does not prove the infinite case! — Asaf Karagila, Apr 15 '12 at 11:35
Much the same posted to MO, http://mathoverflow.net/questions/94105/why-is-counting-remaining-intervals-different-from-counting-rationals — Gerry Myerson, Apr 15 '12 at 12:48
@ Gery: Asaf hinted me to the recent discussion. The answer there was poor and unmathematical. Indeed and in fact ma be used in real life, but not in mathematics. @ Asaf: The pairing function or bijection applied for all finite cases proves, according to set theory, countability. Therefore there are countably many complementary intervals. — , Apr 15 '12 at 13:18
@BelsaZarkin, you are wrong. There is no interval left behind. Every interval contains at least one rational number, so if you remove all rational numbers (let alone a bunch of intervals containing all of them), there can be no interval left over. — Bruno Joyal, Apr 20 '12 at 06:47

score 15 · Answer 1 · edited Apr 13 '17 at 12:20

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There is a mistake in your argument, which follows from the fact that the complement of this union contains no interval.

Let $C=\bigcup I_n$ the open set which covers the rationals whose measure is at most $\frac19$. You are considering $B=[0,1]\setminus C$, which is a subset of the irrationals.

If $B$ would contain an open interval then it would contain a rational number. Since there are no rational numbers in $B$ there is no interval subset. The irrational numbers form a totally disconnected space, namely every connected component is a singleton.

Note that by DeMorgan we have $B=\bigcap [0,1]\setminus I_n$, this is an intersection of closed sets. Indeed for every $n$ we have that $[0,1]\setminus (I_1\cup\ldots\cup I_n)$ contains an interval, in which there are infinitely many rational and irrational numbers.

The limit, however, is not require to have the properties of the sequence, and we have that $B$ contains no interval.

(This thread can be useful here: Fake induction proof)

edited Apr 13 '17 at 12:20

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answered Apr 15 '12 at 12:12

Asaf Karagila

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In a totally disconnected space there must be points or intervals disconnecting it. The number of these points or intervals is countable and in bijection with the remaining intervals. The bijection is the same as that from N to Q. (For instance at every step n the configuration of intervals could be determined in principle.) If you deny the validity of this bijection for the limit, why don't you deny the validity for the limit of the bijection from N to Q? – Apr 15 '12 at 12:51
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@Belsa No. The definition of a totally disconnected space is a space in which every connected space is a singleton. E.g. a discrete space. Are you saying that every discrete space is countable? – Asaf Karagila Apr 15 '12 at 12:52
I say only that before reaching the limit, we have n intervals and n-1 complementary intervals. If something weird happens in the limit, why then is everybody convinced that the same does not happen when counting the rationals? – Apr 15 '12 at 12:58
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@Belsa: In this case of covering a subset of $[0,1]$, sure. But why does that mean that the limit has to have the same properties? Read the answer I posted on your MO question. – Asaf Karagila Apr 15 '12 at 13:00
qAsaf: "But why does that mean that the limit has to have the same properties?" My answer is this: From the possibility of pairing all elements n of N and m of M as (n, m) we can conclude that the set M is countable. That is the definition of countability. It holds in the limit and only in the limit when all n are used. – Apr 15 '12 at 13:02
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@Belsa: What is $M$? What is $N$? You have two sequences here. One has a countable limit and the other does not have a countable element. – Asaf Karagila Apr 15 '12 at 13:35
@ Asaf: N is the set of natural numbers. M is the set of left endpoints of the complementary intervals. – Apr 15 '12 at 15:06
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@Belsa, indeed $\mathbb N$ is countable, and at the stage $k$ you have only finitely many end points. However once you take the union of infinitely many intervals you may have uncountably many endpoints. To see this in an even fouler way consider the Cantor set which is closed. Its complement is open, and therefore is the union of countably many open intervals. However the boundary of this countable union is the Cantor set which is uncountable, so the union of countably many intervals may have uncountably many endpoints. – Asaf Karagila Apr 15 '12 at 15:18
I do not take any union but simply consider the bijection of naturals and left endpoints. This proves the countability of complementary intervals. Or the countability of the rationals is not provable. There is no reason to consider analogies like the Cantor set. Should we change to chat? What do we have to do in order to change? – Apr 15 '12 at 16:24
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@Belsa: Yes, you have countably many closed intervals complementing the countable number of open intervals. I really don't get the point you are trying to make. Furthermore, if the intervals are centered at rational points and their length is rational then the endpoints are also rational numbers. – Asaf Karagila Apr 15 '12 at 16:51
I really don't get the point you are trying to make. – Apr 15 '12 at 16:55
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@Belsa: This is what I take from your questions (here and on MO), you cover the rationals in the interval with countably many open intervals whose measure is small. In the complement of this cover every interval has only one irrational, which is a contradiction since there are only countably many intervals in the complement. Am I correct? – Asaf Karagila Apr 15 '12 at 16:59
"I really don't get the point you are trying to make." @Asaf: The point is simply that, by the rules of set theory,in the limit there are not more than countably many endpoints of complementary intervals, hence not more complementary intervals with total measure 8/9. My question is: How to formalize that? – Apr 15 '12 at 17:03
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@Belsa: In which limit? In the intersection of all complements? – Asaf Karagila Apr 15 '12 at 17:07
The "limit" is the enumeration of all left endpoints of complementary intervals. There is no intersection required. The cardinality of the set of complements cannot surpass the cardinality of left endpoints. – Apr 15 '12 at 17:11
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@Belsa: No, the number of complements to the intervals is indeed countable. I just don't get two things: the first is that every complementary interval is an interval and by definition contains infinitely many rationals and infinitely many irrational numbers; the second is what does ZFC have to do with all that? – Asaf Karagila Apr 15 '12 at 17:17
How should countably many endpoints make up a disconnected space with uncountably many disconnected points? 2) ZFC is the basis of all mathematics. How can this idea be formalized?

Apr 15 '12 at 17:31

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@Belsa: I am officially at loss. Your question (at the top of the page) says that the complements is made of intervals of measure of total 8/9. This is not true, the complement is the intersection of intervals whose measures approach 1. The intersection has measure of at least 8/9. Your question asks whether at least one of those intervals contain two irrational points which will lead to contradiction because we covered all the rationals. again not true because the complement does not contain any interval. The closed intervals which are complements of the intervals used in the cover... – Asaf Karagila Apr 15 '12 at 17:38

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...are indeed intervals and as such contain infinitely many rational and irrational points. The intersection of all those closed intervals, however, make a closed set which is totally disconnected as a subspace of the irrational numbers. As for the remark about ZFC, it is fine to assume ZFC as the foundation for some parts of mathematics, however insisting on "writing a claim in ZFC" is like insisting to use only assembler because all compilers essentially turn code into assembler, but also insisting to write your code in binary machine code via hex editors. – Asaf Karagila Apr 15 '12 at 17:39

"The intersection of all those closed intervals, however, make a closed set which is totally disconnected as a subspace of the irrational numbers." – Apr 15 '12 at 17:58

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@Belsa: Yes. The intersection of countably many intervals may result in a totally disconnected space, e.g. the Cantor set. – Asaf Karagila Apr 15 '12 at 18:04

"The intersection of all those closed intervals, however, make a closed set which is totally disconnected as a subspace of the irrational numbers." This is not true. The number of all complementary intervals in the n-th step is at most n+1. Therefore we have the bijections 1) between naturals k and the rationals q (k, q) - 2) the bijection between the naturals k and the left endpoints of complementary intervals (k, p_k) - and 3) the bijection between the naturals k and the number n of complementary intervals existing at step k (k, n_k) with n_k =< k + . These show all intervals – Apr 15 '12 at 18:06

all intervals at every stage being countable.(n_k is the number of complementary intervals that have formed in the k-th step. – Apr 15 '12 at 18:08

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@Belsa: I hope that you are willing to agree with me on the following: (1) Every rational number is inside the open cover. (2) Every number not in the open cover is irrational. (3) In any interval which contains more than one point there is a rational number. (4) If the set of the points not inside the open cover does not contain rational numbers then there is no interval $(a,b)$ which is a subset of that set. (5) Every countable set is of measure zero. (6) If we removed points of measure $\le\frac19$ then we are left with uncountably many points. (cont...) – Asaf Karagila Apr 15 '12 at 18:18

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(...) and (7) None of the remaining points, after removing the open cover, is rational. *The conclusion* is that the remainder is a totally disconnected space which contains no interval and every point inside this space is an irrational number. (Note that I did not talk about endpoints. I did not talk about what happens at the $k$-th stage. I talked about what happens after all the finite stages**) – Asaf Karagila Apr 15 '12 at 18:21

Sorry, this discussion has become too lengthy. So I will try to ask a new question essentially considering the majorat criterion: Does the majorant criterion hold for infinite sequences in your opinion? Is it possible that the number n_k of complementary intervals formed in the n-th step surpasses the number m_k of intervals around the rationals in the k-th step? If not, is it possible that the sequence (n_k) has a larger limit than the sequence (m_k)? – Apr 15 '12 at 18:55

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@Belsa: I do not know the term "the majorant criterion" and Google is of no help. Note that the length of each complementary interval in the $n$-th stage will be split by an interval from the open cover at some further stage. This means that the length of these complementary intervals approaches to zero (their cumulative measure is not!). Closed intervals of length zero are singletons. – Asaf Karagila Apr 15 '12 at 19:06

http://www.skipperw.com/wiki/Majorant_criterion Further the complementary intervals need not be closed. We can use closed intervals to cover the rationals. But please let us continue at http://math.stackexchange.com/questions/132191/is-the-majorant-criterion-valid-for-sequences-with-improper-limit-inifinty – Apr 15 '12 at 19:15

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@Belsa: But this not a sum of numbers. This is a union and intersection of intervals. – Asaf Karagila Apr 15 '12 at 19:16

I am calculating the maximum number of intervals into which the complementary part of measure 8/9 can be split at most. – Apr 15 '12 at 19:19

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@Belsa: Any interval can be split into infinitely many parts. Furthermore the limit of two eventually increasing sequences of natural numbers is always infinite. – Asaf Karagila Apr 15 '12 at 19:20

Any interval cannot be split in more parts than there are splitting intervals. The limit of two sequences should obey lim a =< lim b if for every k a_k < b_k, in my opinion. – Apr 15 '12 at 19:48

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@Belsa: But the open intervals are fixed! You don't "count the ways to split" the splitting is predefined by the enumeration of the rationals. Furthermore every interval in the complement of the finite open cover will be split at some point (since it contains a rational which was not yet covered) and therefore the general length of the intervals in the complements is decreasing. – Asaf Karagila Apr 15 '12 at 20:02

I do not consider the lengths of the intervals but their number. One point is sufficient. What is your opinion about the following: Let all positive rationals q_k be enumerated by the naturals k. Let m_k be the number of covered intervals within [0, 1], and let n_k be the number of uncovered intervals within [0, 1] after step k. Then for every k we have k >= m_k >= n_k. You claim lim n_k > lim m_k. Why then should not lim m_k > lim k? – Apr 16 '12 at 07:02

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@Belsa: That makes absolutely no sense whatsoever. You are confusing cardinality with sequences of real numbers, in fact sequences of natural numbers. You are counting how many intervals you have in every step. Yes these are natural numbers and the sequences are non-decreasing and eventually increasing (they cannot be constant more than finitely many steps). BOTH SEQUENCES HAVE THE SAME LIMIT, $\infty$. Note that $\infty$ says nothing about cardinality. It is not countable nor uncountable it is just... infinite. – Asaf Karagila Apr 16 '12 at 07:48

@ Asaf: If you can enumerate every element of a set, then the set is said to be countable. When counting its elements, then the initial segments of elements have natural numbers as their cardinal numbers. Therefore natural number are appropriate to be considered in this connection. Further: I can count all endpoints of all intervals $m_k$. It is impossible by mathematical reasons that there is a complementary interval $n_k$ that has any different endpoint. Therefore the set of complementary intervals is countable because the set of their endpoints is. – Apr 16 '12 at 15:07

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@Belsa: This is the last comment I am willing to write here, or anywhere, on this topic. I really hope that this one will get through to you. Functions into the cardinal numbers are not the same as functions into the real numbers. Cardinal and real numbers are different beasts altogether and how these function behave at limit points is a different story altogether. Even if the function, when restricted to finite cardinals, is into finite cardinals - which makes it a real-valued function, the limit point is a different story. I really hope you understand this. Good luck in the future. – Asaf Karagila Apr 16 '12 at 17:19

Why don't you talk to the topic? "The limit point is a different story" does not apply to the proof that there are only countably many endpoints. Countably many endpoints make countably many intervals with rationals and countably many intervals without rationals. There is no way to circumvent this fact. But I understand that you are lacking arguments. – Apr 16 '12 at 20:15

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@Belsa: Your argument is murky, unclear, and I honestly believe that I did way more than I could to try and understand it (at least judging by the comments and the question posted at the top of this page). You insist on being unclear and incoherent. I no longer wish to try and understand what is it that you were trying to achieve here. If you find a way to phrase yourself which is absolutely clear about what is it that you want, I might bother with this again. (cont) – Asaf Karagila Apr 16 '12 at 20:58

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(cont) My main problem is that you write "this thing happens at finite stage, therefore the limit has to have this property" which is a known fallacy and pitfall. You gave absolutely no indication as to why your argument really holds except a handwaved explanation that "it has to be!" and you consistently seem to think that the fact that you have two sequences of natural numbers which are increasing imply that the function defining both would behave the same way when applying to the $\aleph_0$ (which, I should probably explain, is not a real number). – Asaf Karagila Apr 16 '12 at 20:59

Clearest possible formluation of my argument: The total set T of all endpoints of the intervals $M_k$ covering all rationals is countable. The total set of all endpoints (including single points in case the interval is a single point) of all complementary intervals $N_k$ is a subset of T. A subset of a countable set is countable. The intervals $N_k$ don't overlap. Therefore every endpoint belongs to at most two intervals. The set {$N_k$} is countable – Apr 17 '12 at 05:52

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@Belsa: There are countably many disjoint intervals creating $[0,1]\setminus C$ where $C$ is the Cantor set. There are only countably many endpoints to these countably many intervals. Are you hinting that therefore there are only countably many points in the complement of $[0,1]\setminus C$? – Asaf Karagila Apr 17 '12 at 06:12

@Alsa: Please refrain from analogies. Answer to your question: Countably many endpoints imply countably many intervals. – Apr 17 '12 at 06:45

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@Belsa: Who is Alsa? I am not going to refrain from any analogies as you as you make these ridiculous claims that countably many intervals with countably many endpoints imply that there have to be only countably many degenerate closed intervals. If you wish to continue claiming this, either do it elsewhere or write to me in full details (namely give me the exact endpoints) for all your open intervals and then we can count together how many intervals exist in your complement. (Inductively naming your intervals does not hold here for the reasons you wish to ignore that were mentioned earlier.) – Asaf Karagila Apr 17 '12 at 07:07

qAsaf: Sorry for misnoming you, Asaf. I was in a hurry. Concerning intervals: the countability of all M_k and of all their endpoints is obvious without computing each one. The remaining complementary intervals, whether "degenerate" ornot, have exactly the same endpoints except some which do not separate complementary intervals. I don't know how you can believe that there should be more. Further I need not use open intervals M_k but can use closed ones. The argument is valid in both cases. – Apr 17 '12 at 07:34

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@Belsa: I can prove that there is a countable collection of disjoint open intervals, and the complement of the union contains no interval and cannot be written as a countable union of closed, nor open, intervals. What more do you want me to tell you? You insist on not listening to me, but you insist on writing replies. If you want to discuss your question please be open to critique from people that might be slightly more experienced and knowledgeable in the topics of your questions. – Asaf Karagila Apr 17 '12 at 07:51

Hi Asaf: I do not deny that you can prove what you say. But I can prove what I say: Every endpoint of a non-covered interval $N_k$ including the only single point of a degenerate interval either is an endpoint of an interval $M_k$ or the limit of a sequence of such endpoints. (The latter is necessarily always true if the $M_k$ are closed intervals.) But the elements of the sequence do not persist. ctd. ... – Apr 17 '12 at 09:07

ctd. They disappear by the overlapping of intervals $M_k$ . They cannot apply to more than one limit. We have the fundamental theorem that there are not more limits than elements of sequences. If you dislike my proof, please say what step is wrong. If your proof is also correct, then we have a contradiction. – Apr 17 '12 at 09:07

@Belsa: Suppose that $a$ is an irrational number such that every rational $q$ has the property that the interval covering $q$ is much shorter than $|a-q|$. Will $a$ be in the cover? No, of course not. Will it be an end point of any of the open intervals? Again, no. If it were an endpoint it would have been relatively close to $q_n$ for some $n$, but that is a contradiction. To see that such $a$ exists simply consider the sets of real numbers which are "too far" from $q_k$ for $k>n$ (where too far means $|x-q_k|>10^{k+10}$ for example). (cont...) – Asaf Karagila Apr 17 '12 at 21:29

(...) Each of these sets contain compact sets which form a decreasing chain of closed sets whose intersection is non-empty and the point of intersection is irrational since any rational is some $q_n$ which is of distance $0$ from $q_n$. I have just procured a point which is not an endpoint of any interval. Note that this irrational number is the limit of rational numbers which can get arbitrarily close to it, but never "too close". – Asaf Karagila Apr 17 '12 at 21:30

Hi, Asaf: There are limits of Cauchy sequences with no doubt. Your limit is uncovered, hence there must be an endpoint between itself and the covered intervals. What do you conclude from your construction? Looking from outside (i. e. the set of uncovered intervals including the elements of Cantor-dust, which we will call intervals too, shows more doors than looking from inside (the set of covered intervals)? Impossible. Or: The irrational numbers are countable because the same trick may be applied to Cantors list? ctd. – Apr 18 '12 at 12:16

ctd. ("Trick" means here: the limit is never a rational number but Cantor argument excludes only rational numbers.). This contradiction is just what I wish to pojnt to. Concerning countabiliy of real numbers: Have you ever seen a Cauchy-sequence defining a limit unless the Cauchy sequence has had a finite definition? Do you know how many finite definitions can be given in a language that can be applied to conversation and identification? – Apr 18 '12 at 12:16

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@Belsa: You are officially entering my crackpot list now. Your arguments are those of a man who would not listen, and when presented a mathematical argument answers in naive ideas that indicate lack of basic understanding. Up until now I was hoping you will turn up to be an undergrad student, but it appears that you are not. I hope that you will be the exception of the crackpot rule and will actually try and learn mathematics and not just "refute it" and "invent your own". God speed. – Asaf Karagila Apr 18 '12 at 12:23

Construct a Cauchy sequence. Find no endpoint of an $M_k$ that can be mapped on the limit. Believe, counterfactually, that there are uncountably many limits, Cantor dust, formed by countably many intervals. That's a mathematical argument in your opinion??? Have you ever seen the limit of an infinite sequence being defined by its terms? Would you believe to know the limit after having seen 10000 or 100000000 terms? No. Limits are defined by finite definitions. And there are not more than countably many. – Apr 18 '12 at 14:44

Hi, Asaf, you said: "I have just procured a point which is not an endpoint of any interval." And I repeat, you can prove your claim, I can prove mine: There are countable many endpoints. If your point is not in a covered interval $M_k$, then it must be separated from the points within all covered intervals. By what? There is but one possibility: By an endpoint, because other possibilities are not available. There is no way to circumvent that conclusion. – Apr 18 '12 at 20:34

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@BelsaZarkin, I heartily second Asaf's last claims. I read through this entertaining discussion and I everything Asaf said was correct, whereas most of what you said either meant nothing, or was blatantly false. I would recommend being a little more modest, and actually trying to understand what Asaf had to say. There is not a single interval contained in the complement of your open set. – Bruno Joyal Apr 20 '12 at 07:04

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Fairly soon, there will be an uncountable infinity of comments on this answer, and then we'll be able to test all the theories directly. – Gerry Myerson Apr 20 '12 at 13:07

@Gerry: That depends on the enumeration of the rationals and the choice of intervals! :-D – Asaf Karagila Apr 20 '12 at 14:52

There is a mistake in your argument, which follows from the fact that there is no union at all! I use only intervals and their borders. There is no necessity to union any of the intervals. We let them just as they are constructed. So we have aleph_0 left borders and aleph_0 right borders, in total 2*aleph_0 = aleph_0 borders. They define aleph_0 intervals. And aleph_0 of these so defined intervals contain 8/9 of the measure and uncountably many irrational numbers. That's a contradiction. – May 18 '12 at 17:25

No matter what the definition of a totally disconnected space is: We can prove that between every two irrational numbers there is a rational one. And therefore between every two irrational numbers that are not covered by any of the intervals of measure 1/9 there is such an interval. – May 18 '12 at 17:25

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@Belsa: Are you claiming that there is a set which is not the union of its singletons? That's a big idea. Furthermore, yes - we can prove that between two irrational numbers there is some rational number. This is exactly why if you have covered all the rational numbers but something of measure $\frac19$ the remaining set cannot contain any rational numbers, and therefore contains no open interval. – Asaf Karagila May 18 '12 at 18:20

@Asaf: 1. Of course, if R was uncountable, then it could not be the set of its singletons. It is impossible to distinguish uncountably many elements. But only distinct elements can belong to a set. Otherwise extensionality was meaningless. – May 18 '12 at 19:18

@Asaf: 2. Without any unioning or other magical tricks I construct aleph_0 intervals, not more and not less. Some of them contain all rational numbers. Others do not. I don't ask how they look like. But if there were uncountably many irrationals, then there was a contradiction. – May 18 '12 at 19:19

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@Belsa: So you admit that your ideas are not ideas in ZFC, because in ZFC a set is equal to the union of all its singletons. Why would you ask to "formalize" something which is inherently contradictory to ZFC "in ZFC"? This very much like I would ask you to formalize $-1$ in the natural numbers under the axioms of Peano. – Asaf Karagila May 18 '12 at 19:50

@Asaf: Please don't deviate from the subject: Without any unioning or other magical tricks I construct aleph_0 intervals, not more and not less. Some of them contain all rational numbers. Others do not. I don't ask how they look like. But if there were uncountably many irrationals, then there was a contradiction. This conclusion (no two irrationals without a rational in between) is within ZFC. – May 18 '12 at 19:55

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@Belsa: Do you know the axioms of ZFC? These axioms prove that every set is the union of the singletons of its elements. If you reject this fact you reject ZFC. Period. I do not deviate from the subject, this is what you originally asked for. – Asaf Karagila May 18 '12 at 20:01

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I randomly came across this answer, and I just wanted to say that Asaf seems like the most patient person I have seen in this website. It is simply incredible how you kept going until the OP could understand (even when the OP was clearly being difficult). – Prism Jan 07 '19 at 22:04

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@Prism: The OP was a well-known crank who appears on the site from time to time and posts these kind of questions in order to argue with people that Cantor was wrong and modern mathematics is mostly mistaken when it comes to the infinite. I was not patient, I was young and naive. – Asaf Karagila Jan 07 '19 at 22:16

@Asaf: Ah okay, thanks for clarifying! – Prism Jan 08 '19 at 00:15

Formalizing an idea

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