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If we have a space $(0,1)$ we can find two irrational numbers $i_1$ and $j_1$ so that $0 < i_1 < j_1 < 1$.

Then we can find two rational numbers $a_1$ and $b_1$ so that $i_1 < a_1 < b_1 < j_1$.

Then we can find two irrational numbers $i_2$ and $j_2$ so that $a_1 < i_2< j_2 < b_2$.

Then we can find two rational numbers $a_2$ and $b_2$ so that $i_2 < a_2 < b_2 < j_2$.

$\dots$

Then we can find two irrational numbers $i_n$ and $j_n$ so that $a_n < i_n < j_n < b_n$.

As $n \to \infty $ what becomes of the segments $(i_n,j_n)$ and $(a_n ,b_n)$ and the density of rational numbers vs irrational numbers?

Will the vales of $i_n,a_n,b_n,j_n$ approach $i_n = a_n = b_n = j_n = 0.5$ or do we have to say $i_n \to a_n \to b_n \to j_n \to 0.5$

I will try to flesh out the question with some actual numbers.

Let $i_1 = \frac{1}{10^{1!}} + \frac{1}{10^{2!}} + \frac{1}{10^{3!}} + \dots$

Let $j_1 = 1 - \frac{1}{10^{1!}} + \frac{1}{10^{2!}} + \frac{1}{10^{3!}} + \dots$

Let $a_n = \frac{2^n-1}{{2^{n+1}}}$

Let $b_n = \frac{2^n+1}{{2^{n+1}}}$

If $n > 1$ then let $i_n = \frac{2^n-1}{{2^{n+1}}} + \frac{1}{10^{(n+1)!}} + \frac{1}{10^{(n+2)!}} + \frac{1}{10^{(n+3)!}} + \frac{1}{10^{(n+4)!}} \dots$

If $n > 1$ then let $j_n = \frac{2^n+1}{{2^{n+1}}} - \frac{1}{10^{(n+1)!}} + \frac{1}{10^{(n+2)!}} + \frac{1}{10^{(n+3)!}} + \frac{1}{10^{(n+4)!}} \dots$

Ivan Hieno
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    So your claim is that $\frac12=\frac13=\frac1\pi$? I don't understand how from "$n\to\infty$ you deduce $a_n=b_n$ for some $n$. – Asaf Karagila Oct 16 '18 at 11:27
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    Or in other words, $\frac1n\to 0$ as $n\to\infty$. Could you point out the natural number for which $\frac1n=0$? – Asaf Karagila Oct 16 '18 at 11:55
  • So approaches and equals are not the same. If you answer this question I will give you top answer. https://math.stackexchange.com/questions/2955013/when-can-we-say-that-something-is-equal-to-rather-than-something-approaches-a-l – Ivan Hieno Oct 16 '18 at 12:16
  • @IvanHieno You have to be careful: "the sequence $(x_i){i\in\mathbb{N}}$ approaches $x$" is the same as "the limit of the sequence $(x_i){i\in\mathbb{N}}$ is exactly equal to $x$." The crucial point here is avoiding conflating a sequence and a number, and it accounts for the confusion going on. – Noah Schweber Oct 16 '18 at 12:50
  • Incidentally, this title is highly misleading: there's nothing here about Cantor, you're just asking a question about sequences. Even if this is part of an attempt to come to grips with Cantor's results, if that's not directly germane to the question it shouldn't be in the title. – Noah Schweber Oct 16 '18 at 13:01
  • However, since you have mentioned it, I've said a bit about it in my answer. – Noah Schweber Oct 16 '18 at 13:07

2 Answers2

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Not necessarily. You can pick initial $0 < c < d< 1$, and then choose your $a_n, i_n, j_n, b_n$ so that $$ a_n < i_n < c < d < j_n < b_n $$ for all $n$.

Mees de Vries
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As in your other question, the issue here is a conflation of sequences and objects.

The situation you have here is:

  • Each sequence approaches ${1\over 2}$: $$\lim_{n\rightarrow\infty}a_n=\lim_{n\rightarrow\infty}b_n=\lim_{n\rightarrow\infty}i_n=\lim_{n\rightarrow\infty}j_n.$$

  • However, no individual term is exactly ${1\over 2}$.

Incidentally, a closely-related conflation is often the key mistake made in "refutations" of Cantor's diagonal argument: the argument in question claims that the "antidiagonal real" $D$ appears in in the sequence, since arbitrarily good approximations to $D$ appear in the sequence. Here the conflation is between "a number appearing in a given sequence" and "a sequence of approximations to a number being a subsequence of a given sequence."

A lot of the confusion around arguments about infinity comes from "discontinuous behavior at infinity": behavior happens at infinite stages which doesn't occur at any finite stage. Distinguishing carefully between sequences and individual numbers is an important step towards resolving this, since once we fully realize the distinction we can recognize that there's no reason a priori to expect "finitary facts" to yield directly-analogous "infinitary facts."

Noah Schweber
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  • Noah said "no individual term is exactly ${1\over 2}$". That is true, but what is the density of the rationals vs the irrationals when our segments approaches an infinitesimal length? If there exists more irrationals than rationals, then they must exist someplace else. – Ivan Hieno Oct 16 '18 at 14:18
  • @IvanHieno "what is the density of the rationals vs the irrationals when our segments approaches an infinitesimal length? If there exists more irrationals than rationals, then they must exist someplace else." I'm not sure what that means but I think you're trying to claim that at some point we need to get an interval where there are some irrationals with no rationals between them? If so, that's false. The basic intuitions one has about how cardinality interacts with geometry are generally false so even after phrasing your claim precisely you need to prove it - even if it seems obvious. – Noah Schweber Oct 16 '18 at 14:22
  • Would you be able to explain what you mean when you say "The basic intuitions one has about how cardinality interacts with geometry are generally false". How does it differ, given that Cantor theorized about points in lines, planes and solids etc. – Ivan Hieno Oct 16 '18 at 14:40
  • @IvanHieno Cantor theorized by actually proving things; he didn't let his sense of what's "obvious" get in the way (e.g. he was surprised by the fact that $\mathbb{R}^2$ and $\mathbb{R}$ have the same cardinality). Statements like "there can't be more irrationals than rationals, since the rationals "separate out" the irrationals" are perfectly obvious at first glance, but when you sit down to try to prove them rigorously you quickly find that you can't actually seem to build the bijection you want without waving hands and saying things like "clearly" and "obviously." – Noah Schweber Oct 16 '18 at 14:44
  • @IvanHieno Regardless, the point is: the onus is on you to state precisely and prove your claims, not on us to decipher and disprove them. This is especially true when pushing back against an established theorem; there's nothing wrong with doing that, indeed it's a good thing to do to understand the results, but you need to be careful to back up your assertions and recognize the burden of proof. (After all, we have already rigorously proved Cantor's theorem!) – Noah Schweber Oct 16 '18 at 14:46
  • I want you to know that I really do value any time that you and others give me. The problem is that I just can't "see" Cantor's conclusion, like I can "see" other ideas like Euclid's Proof of the Infinitude of Primes. – Ivan Hieno Oct 16 '18 at 14:59
  • @IvanHieno That's very fair, and common. I think there are two things to do. First, try to phrase your counterarguments precisely (e.g. what exactly does " If there exists more irrationals than rationals, then they must exist someplace else" mean?). Even before trying to prove them, this is useful: e.g. for me, something that often happens is that I'll run into a kind of "slipperiness" right off the bat, where I suddenly realize I don't know exactly what I want to say. That doesn't mean I'm wrong, of course, and sometimes I can fix it after all, but it certainly indicates something worrying. – Noah Schweber Oct 16 '18 at 16:17
  • The second is to separately look step-by-step at Cantor's argument and convince yourself that each basic step works. This can often be tricky for longer proofs, but here it's not bad since there's only three steps: (1) given a map $f:\mathbb{N}\rightarrow\mathbb{R}$, we define the "antidiagonal" real $D_f$. (2) We show that $D_f$ isn't in the range of $f$. (3) We conclude that no $f:\mathbb{N}\rightarrow\mathbb{R}$ is surjective, and this is exactly what we wanted. Now, this sort of step-by-step analysis may not be enough on its own to get an intuitive understanding. – Noah Schweber Oct 16 '18 at 16:19
  • The other piece of it is to look at different ways to phrase the proof. For example, one important shift is from contradiction to direct proof, which I've done above: rather than start by assuming that $f$ was surjective and then getting a contradiction, I've started with $f$ arbitrary and (constructively!) produced a real not in its range. The proof by contradiction is fine, but eliminating the contradiction generally makes it easier to understand (and I would argue reveals the actual content and is a Good Thing). Another useful tweak is to change the codomain: (cont'd) – Noah Schweber Oct 16 '18 at 16:21
  • instead of showing that $\mathbb{R}$ is uncountable, show that $\mathcal{P}(\mathbb{N})$ (the set of sets of natural numbers) is uncountable. The reason this is an improvement is because it's easier to define the "antidiagonal set:" given $f:\mathbb{N}\rightarrow\mathcal{P}(\mathbb{N})$, we just let $S_f={n: n\not\in f(n)}$. (Note that this more generally shows that no map from a set to its powerset is ever surjective; see e.g. here). Once "$\mathcal{P}(\mathbb{N})$ is uncountable" becomes intuitively clear, (cont'd) – Noah Schweber Oct 16 '18 at 16:25
  • and then focus on understanding that $\mathcal{P}(\mathbb{N})$ and $\mathbb{R}$ have the same cardinality (and hence $\mathbb{R}$ is uncountable). This latter is quite simple, so we've isolated the hard work on understanding the following three steps: (1) Given a function $f:\mathbb{N}\rightarrow\mathcal{P}(\mathbb{N})$, let $S_f={n: n\not\in f(n)}$. (2) We know $S_f$ can't be in the range of $f$: for every $n$, either $n\not\in f(n)$ in which case $n$ is an element of $S_f$ but not $f(n)$, or $n\in f(n)$ in which case $n$ is an element of $f(n)$ but not $S_f$. Either way, $S_f\not=f(n)$. – Noah Schweber Oct 16 '18 at 16:27
  • Finally, (3) notice that what we've proved is "Any function from $\mathbb{N}$ to $\mathcal{P}(\mathbb{N})$ is non-surjective," and we can rephrase that as "There is no surjection from $\mathbb{N}$ to $\mathcal{P}(\mathbb{N})$" - that is, $\mathcal{P}(\mathbb{N})$ is uncountable. – Noah Schweber Oct 16 '18 at 16:28