Exercise 3.4.9

Let $O_n={\bigcup <!--\; FUNCTION\; APPLICATION\; -->}_{i=1}^n{V}_{{𝜖}_i}(r_i)$, $F_n=O_n^c$, and $G_n=F_n\cap [1,3)$. Clearly $G_n\supseteq G_{n+1}$ and $G_n$ is compact, so if we can show $G_n$ is always nonempty, we can use the Nested Compact Set Property to show $F={\bigcap <!--\; FUNCTION\; APPLICATION\; -->}_{n=1}^{\infty }{F}_n\supseteq {\bigcap }_{n=1}^{\infty }{G}_n$ is nonempty.

Consider the set $M=\{m∕2^{n-1}:m\in N\}\cup [1,3)$, a set of evenly spaced numbers with $1∕2^{n-1}$ between each number. Some arithmetic shows that there are $2^n$ elements in $M$. Since the length of $O_i$ is $1∕2^{i-1}$, there can only be at most $2^{n-i}$ elements of $M$ in $O_i$, and therefore $2^n-1$ elements of $M$ in $O$. Since $G_n\supseteq M\setminus O$, $G_n$ is nonempty, completing the proof.

First, we prove the following lemma - for any $L\in R$ and a positive sequence $({\zeta }_n)\to 0$, it is possible to construct a sequence $(x_n)\in Q$ such that $[L,L+2{\zeta }_1)\setminus {\bigcup <!--\; FUNCTION\; APPLICATION\; -->}_{i=1}^{\infty }{V}_{{\zeta }_i}(x_i)=\{L\}$. (We’ll use this lemma for the sequences isolating $\sqrt{2}$.) To do so, let $V_{{\zeta }_n}(x_n)=({\alpha }_n,{\beta }_n)$. Choose $L<{\alpha }_n<L+2{\zeta }_{n+1}$.

We show by induction that $[L,L+2{\zeta }_1)\setminus {\bigcup <!--\; FUNCTION\; APPLICATION\; -->}_{i=1}^n{V}_{{\zeta }_i}(x_i)\subseteq [L,L+2{\zeta }_{n+1})$. The base case $n=1$ is trivial. For the inductive case: assume $[L,L+2{\zeta }_1)\setminus {\bigcup <!--\; FUNCTION\; APPLICATION\; -->}_{i=1}^n{V}_{{\zeta }_i}(x_i)\subseteq [L,L+2{\zeta }_{n+1})$. Then

Since ${\alpha }_{n+1}>L$, ${\beta }_{n+1}>L+2{\zeta }_{n+1}$. Recall also that ${\alpha }_{n+1}<L+2{\zeta }_{n+2}$, and therefore $[L,L+2{\zeta }_{n+1})\setminus ({\alpha }_{n+1},{\beta }_{n+1})\subseteq [L,L+2{\zeta }_{n+2})$, completing the inductive step.

Now, for any $l\in [L,L+2{\zeta }_1)>L$, since $({\zeta }_n)\to 0$ there must be some ${\zeta }_j$ so that $L+2{\zeta }_j\le l$ and $j>1$, and therefore

Also, since ${\alpha }_n>L\forall <!--\; FUNCTION\; APPLICATION\; -->n$, $L\in [L,L+2{\zeta }_1)\setminus {\bigcup <!--\; FUNCTION\; APPLICATION\; -->}_{i=1}^{\infty }{V}_{{\zeta }_i}(x_i)$. This completes the proof of the lemma.

Returning to the original problem of making $F$ not perfect, we will construct a sequence $(r_n)$ which isolates $\sqrt{2}$. Start with any enumeration of the rational numbers $q_n$. Our lemma above means we can assign $r_{3n+1}$ and ${𝜖}_{3n+1}$ for $n\ge 0$ to ensuring $(\sqrt{2},\sqrt{2}+2{𝜖}_1)\in O$ but $\sqrt{2}$ itself not in $O$. Similarly, a slight modification to the lemma lets us assign $r_{3n+2}$ and ${𝜖}_{3n+2}$ for $n\ge 0$ to ensuring $(\sqrt{2}-2{𝜖}_2,\sqrt{2})\in O$ while leaving $\sqrt{2}$ out. Finally, we assign $r_{3n+3},n\ge 0$ to enumerating through the elements of $q_n$, skipping over any elements that will be present in $r_{3n+1}$ and $r_{3n+2}$, and deferring any elements that would cause $\sqrt{2}\in V(q)$ until $𝜖$ becomes small enough that this is no longer the case.

In this manner, $\sqrt{2}$ has been surrounded, and $F=O^c$ will have $\sqrt{2}$ as an isolated point, and thus $F$ is not perfect.

We can also construct $(r_n)$ so that $F$ is perfect. (Note: To simplify the notation a bit let $V_i=V_{{𝜖}_i}(r_i)$.) Define $R_n={\bigcup <!--\; FUNCTION\; APPLICATION\; -->}_{i=1}^n{V}_i$, with $R_0=\varnothing$. We will rely on the lemma that $F$ is perfect if for all $i\in N$, either $V_i\cap R_{i-1}=\varnothing$ or $V_i\subseteq R_{i-1}$. Informally, if $V_i$ isn’t redundant (in that it covers new numbers), then it is disjoint from all previous $V_i$. To prove this lemma, consider any element $x\in F$ and let $𝜖>0$ be arbitrary. Consider the interval $(x,x+𝜖)$, and ignore any $V_p$ if $V_p\subseteq R_{p-1}$. Since none of $V_i$ overlap partially, this interval cannot be covered completely by some union of several $V_i$ since any union would have gaps. Moreover, $(x,x+𝜖)$ cannot be covered entirely by a single $V_{{𝜖}_j}$, since then $V_j$ would be centered on $r_j=x+{𝜖}_j$ which is irrational. Thus, there must be some other element $y\ne x,y\in F$ so that $y\in (x,x+𝜖)$ and therefore $x$ is not an isolated point and $F$ must be perfect.

To construct an $(r_n)$ which satisfies this condition, we start with an arbitrary $(q_n)$. Define $R_n={\bigcup <!--\; FUNCTION\; APPLICATION\; -->}_{i=1}^n{r}_i$. For each element of $(q_n)$, we add $q_i$ to $(r_n)$ only if either $V_{{𝜖}_n}(q_i)\subseteq R_n$ or $V_{{𝜖}_n}(q_i)\cap R_n=\varnothing$. Otherwise, we procrastinate on adding $q_i$ by appending any rational number $s>\max <!--\; FUNCTION\; APPLICATION\; -->(R_n)+2{𝜖}_n$. Clearly, for any $q_i$ there will eventually be ${𝜖}_n$ small enough that $q_i$ can be added to $(r_n)$ without violating our restrictions, and we don’t need to worry about $s$’s being added since they’re far enough away from everything that they can’t affect the restrictions.