I am having trouble understanding this proof from the book "Understanding Analysis" by Stephen Abbot. The problematic part is on page 29 of the book, which is available on google books.
Assuming that $\mathbb{R}$ is countable, it can be represented as a sequence of numbers:
$\mathbb{R}_s = \{ x_1, x_2, x_3, … \}$
The idea of the proof is to use nested interval theorem to show that there exists $x \in \mathbb{R}$ that is not an element of the sequence $\mathbb{R}_s$. This would mean that no such sequence of real number exists.
The proof goes on to construct finite and nested intervals $I$ such that:
(1) $x_{n} \notin I_n$ and
(2) $I_{n+1} \subseteq I_n$.
It is assumed that $\mathbb{R}_s$ contains all real numbers. Since $x_{n} \ne I_n$, it is clear that for the first number we choose
(3) $x_{n0} \notin\cap_{n=1}^\infty I_n$
However, the proof claims that since $\mathbb{R}_s$ contains all real numbers, this leads to $\cap_{n=1}^\infty I_n = \emptyset$
This contradicts with the nested interval theorem that states the opposite: $\cap_{n=1}^\infty I_n \ne \emptyset$
Therefore no such sequence $\mathbb{R}_s$ can be constructed, which means $\mathbb{R}$ is not countable.
My questions
Q1: If all the intervals are finite, how can we even use $\cap_{n=1}^\infty I_n$?
Example: Choose $x_3$ for $x_{n0}$ and $\{x_4, x_5, …. x_{n}\}, n \in \mathbb{N}$ for $I_1$ – following (1) and (2). How can I construct infinitely many nested intervals from a finite $I_1$ following (1) and (2)? There are maximally $n-1$ such nested sub-intervals, if $I_{n+1}$ is constructed from $I_n$ such that the first or the last element of $I_n$ is chosen as $x_{n+1}$.
Q2: This part:
(3) $x_{n0} \notin\cap_{n=1}^\infty I_n$
However, the proof claims that since $\mathbb{R}_s$ contains all real
numbers, this leads to $\cap_{n=1}^\infty I_n = \emptyset$
(3) is perfectly clear, but how does that lead to $\cap_{n=1}^\infty I_n = \emptyset$? I mean, sure, (3) is valid for any $x_{n0} \in \mathbb{R}_s$, but that doesn't break the nested interval theorem. The nested interval theorem states basically that $I_n \ne \emptyset$.
Is this the point? Rules (1) and (2) will enforce a constructuion of such $I_{n}$ that is an empty set, causing the intersection of all intervals to be empty?
Best Answer
First of all, rather than $x_n\ne I_n,$ $(1)$ should read $x_n\notin I_n.$ What would it even mean for a real number to be equal to an interval of real numbers? This may be mistranscribed by you, or it may be a mistake in your text.
Second of all, there is the statement "there are maximally $n-1$ nested subintervals, if $I_{n+1}$ is constructed from $I_n$ such that the first or the last element of $I_n$ is chosen as $x_{n+1}.$" I suspect that you're misunderstanding what is meant by "finite" in the sense of intervals. It does not mean that the interval has only finitely-many elements. Rather, it means that the interval is bounded (or, equivalently, of finite length). In particular, each interval in the construction has the form $[c,d]:=\{x\in\Bbb R:c\le x\text{ and }x\le d\}$ for some $c,d\in\Bbb R$ with $c<d$. I'm not sure how your proof constructs these intervals, but here is one way we can proceed:
You should be able to see/prove that, regardless of which case we have $I_{n+1}$ is a finite, non-empty, closed subinterval of $I_n$ such that $x_{n+1}\notin I_{n+1}.$ Note, however, that there were many ways we could have chosen the subinterval! For example, in the first case, we could have chosen any (non-empty, closed) subinterval of $I_n$. In the second case, we could have chosen $c$ in many ways--for example, given any $0<t<1,$ we could let $c=tx_{n+1}+(1-t)d.$ Similarly, we could choose $d'$ in many ways in the third case. (It turns out that there are uncountably-many ways to choose an appropriate sub-interval, but that isn't important, and is tantamount to what we're trying to prove!) Moreover, the above procedure recursively constructs all of the $I_n$ for us.
Finally, there is the statement "(3) is valid for any $x_{n_0}\in\Bbb R_s,$ but that doesn't break the nested interval theorem. The nested interval theorem states basically that $I_n\ne\emptyset.$" But this is not so! Rather, the NIT states that if there is a sequence of intervals $I_n$ of real numbers such that (i) each $I_n\ne\emptyset,$ (ii) each $I_n$ is of the form $[c,d]$ for some real $c,d,$ and (iii) $I_{n+1}\subseteq I_n$ for all $n,$ then we have that $$\bigcap_{n=1}^\infty I_n:=\{x\in\Bbb R:x\in I_n\text{ for all }n\ge 1\}\ne\emptyset.\tag{$\star$}$$ Note that it is necessary that each $I_n$ is non-empty, but the conclusion is $(\star).$ Since our construction meets the criteria for NIT, we can conclude that $(\star)$ holds, and so there is some $x\in\bigcap_{n=1}^\infty I_n.$ By assumption, $x\in\Bbb R_s,$ so $x=x_{n_0}$ for some $n_0.$ But by (3), this is impossible: since $x_{n_0}\notin I_{n_0}$ by construction, then $x=x_{n_0}\notin\bigcap_{n=1}^\infty I_n$. Thus, we have a contradiction.