Where are the irrationals in this set?

Question

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!!!

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(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!

Answer 1

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.

Answer 2

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.