Let's define a sequence of families of sets inside $[0, 1]$ interval:

$$ F_n= \left\{ \left(\frac{j-1}{2^n},\frac{j}{2^n}\right] \mid \forall j \in \{ 1, \cdots, 2^n\} \wedge n \in \mathbb{N} \right\} $$

Now, we know that $[0, 1] = \{0\} \cup(0, 1]$. Now we are goint to define a new set $S_n$ as follow:

$$ S_n = \bigcup_{X \in F_n} X $$

For me it's clear that $S_n = (0, 1]$ as long as $n$ is a natural fixed number, and so, it contains all irrational numbers in $(0, 1]$. The question is: does $S_\infty$ contain the irrationals? It is obvious that this set is not empty and contains rationals numbers. Now the questions:

  1. Does $S_\infty$ loses the irrationals?
  2. Are all rationals contained in the set $S_\infty$?
  3. Is $S_\infty = (0,1]$??

Many thanks in advance!!!

(RE)$^2$EDIT: Lets define $S_\infty$. Let's start defining $F_\infty$:

$$ F_\infty = \{ x \in (0, 1] \; | \; x \in X_j \wedge X_j \in F_n \; \text{with} \; j \leq n \; \forall n, j \in \mathbb{N} \} $$

This implies $F_\infty$ would become a infinite (but countable?) partition of $(0,1]$ in disjoint subsets containing just a point. So, $S_\infty$ would just be:

$$ S_n = \bigcup_{X \in F_\infty} X $$

Thanks @Zev Chonoles and @Asaf Karagila for the useful comments!

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    $\begingroup$ How do you define $\lim$ for sets? $\endgroup$ – Asaf Karagila Aug 4 '16 at 21:38
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    $\begingroup$ Well, then, take a deep breath, and now figure out the answer on your own. Then, to dispel any confusion leftovers, recall that a lot of things in mathematics are not continuous. And there is no reason to expect that a limit of a constant sequence will be any different. $\endgroup$ – Asaf Karagila Aug 4 '16 at 21:48
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    $\begingroup$ You still haven't gotten around the fact that $S_n = (0,1]$ for each $n$. The intersection of a set with itself, no matter whether you intersect the set with itself finitely many times or infinitely many times, is still the same set. So if $S_\infty$ is the intersection of all the sets $\{S_1 = (0,1], S_2 = (0,1], ...\}$ then $S_\infty = (0,1]$. $\endgroup$ – Lee Mosher Aug 4 '16 at 21:59
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    $\begingroup$ Your new definition fails you as well. For $n\neq m$ we have that $F_n\cap F_m=\varnothing$. So the intersection is a subset of $F_1\cap F_2$ which is pretty much empty, so it has to be empty as well. $\endgroup$ – Asaf Karagila Aug 4 '16 at 22:10
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    $\begingroup$ Have you read my answer? $\endgroup$ – Asaf Karagila Aug 4 '16 at 22:19

You have defined a set $S_n$ for each natural number $n$; however, you have not defined a set $S_\infty$ since $\infty$ is not a natural number.

It is an understandable and common mistake to assume that doing some operation, or defining some object, for each natural number $n$ necessarily results in some "limiting" operation or object for when "$n=\infty$". However this is usually either not possible at all, or it is quite subtle to do.

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    $\begingroup$ It is even more common to automatically assume that all our definitions are continuous unless proven otherwise. Namely, if you defined something, that the limit will be the limit of the definitions. We see this in the case of "the power set of the natural numbers is the union of power sets of its initial segments", or in other similar definitions like that which "challenge" basic set theoretic truths. $\endgroup$ – Asaf Karagila Aug 4 '16 at 21:33
  • $\begingroup$ @Zev Chonoles Thanks for the comment! I hope my edit solves your point : ) $\endgroup$ – Carlos Toscano-Ochoa Aug 4 '16 at 21:39

Recall that if $A=B$ then $A\cap B=A=B$. Just as well, if $A_n=A$ for all $n$, then $\bigcap\limits_{n\in\Bbb N}A_n=A$.

This shouldn't be surprising, if $x_n=x$ for all $n$, then $\lim x_n=x$. We are not surprised by this at all.

What is supposedly surprising here, is that the intuition we have is inherently discontinuous, and this is a result of infinity not being intuitive. We expect that when we reach infinity, we make "the leap" across the chasm which lies between the finite and infinite. But a limit means only reaching to the edge of the chasm, not actually leaping across. Continuity is what helps us to cross-over when things are "nice enough". But not everything is continuous.

Okay, I digress a bit into metaphors.

Let's get back to your question. You expect that the set $S_\infty$ will be a set where we already refined all the intervals, and this is most certainly not the intersection of all the $S_n$'s. This is something finer than that. It seems that what you expect is that $S_\infty$ will be the set whose members lie in an intersection "along a branch" of the tree of intervals. Namely, it seems that you expect that $$x\in S_\infty\iff\exists f\colon\Bbb N\to\Bbb N\text{ with }f(n)<2^n,\text{ such that }\forall n,\ x\in\left(\frac{f(n)}{2^n},\frac{f(n)+1}{2^n}\right].$$

Sure, we can define it this way, but just note that this is not the limit of the sequence of sets $S_n$, and very much not the intersection of that sequence of sets.

Note that this is not too different from the common mistake the $\bigcup\Bbb Q^n$ and $\Bbb{Q^N}$ are two distinct sets, and exactly one of them is uncountable.

  • $\begingroup$ I'm in right now $\endgroup$ – Carlos Toscano-Ochoa Aug 4 '16 at 22:20
  • $\begingroup$ Dear Asaf, my problem arises from the fact that $\bigcap [a, a+\frac{1}{n}) = \{a\}$. And I've thought in this problem seeing the infinite intersection of the "branches path" of nested intervals, and accepting that you will end up always in a number as defined in the closed frontier of the interval; and because I've defined such numbers as rationals, I've assumed that at the end you would end up with a collection of rationals (2-adics, I think). Can you understand my point? $\endgroup$ – Carlos Toscano-Ochoa Aug 4 '16 at 22:29
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    $\begingroup$ But you are not intersecting $[a,a+\frac1n)$. $\endgroup$ – Asaf Karagila Aug 4 '16 at 22:31
  • $\begingroup$ Choose $j=2^n$ and you will get the following nested intervals: $\left(\frac{2^n-1}{2^n},\frac{2^n}{2^n}\right] = \left(1-\frac{1}{2^n},1 \right]$, which is the same case as exposed. This is what is in my mind and the main motivation for this problem. $\endgroup$ – Carlos Toscano-Ochoa Aug 4 '16 at 22:50
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    $\begingroup$ You have uncountably many possible functions. $\endgroup$ – Asaf Karagila Aug 4 '16 at 23:06

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