# Constructing perfect set without rationals by removing open neighborhood around rationals

Let $\{r_n: n\in \Bbb N \}$ be all rationals and $\epsilon_n = \frac 1 {2^n}$. Then is $$F = \bigcap_{n=1}^\infty \left ( \Bbb R\setminus B(\epsilon_n; r_n) \right )$$ perfect? If not how can it be modified so that ($F$) it can be made perfect?

This is last part of problem 3.4.9 from Understanding Analysis - Abbott. I am hard time proving or disproving it.

Does there exists an enumeration of $\Bbb Q$ which leaves an isolated point under given conditions?

My guess is if I take sub-sequence $n_k$ s.t. $r_{n_{k+1}}$ is boundary point (always on right side) of $B(r_{n_k}; 1/2^{n_k})$. Suppose I begin with rational $x$ whose index is under some enumeration is $n_1$, then I must end up at some irrational number $r$.\

Now suppose I begin with rational $y$ whose index is $m_k$ such that $r_{m_{k+1}}$ is boundary point (always on left side) of $B(r_{m_k}; 1/2^{m_k})$ and I end up with same irrational number $r$ as above.

Does such $y$ exist, and disjoint sequence $(n_k), (m_k)$ exist?

• @HennoBrandsma This is not duplicate. I couldn't find it's answer on the marked question. There are other constructions not this one. This one is specific to method. – Hash Nuke Jul 17 '18 at 21:25
• It's not clear to me how the answer to this question changes when one considers that there are many possible enumerations $\{r_n\}$ of the rationals. Is it clear that some exist where the set is perfect or not perfect? – Ingix Jul 18 '18 at 11:07
• I think I can give such an enumeration. The construction is technical, so I need more time (which I don't have right now) to write it down with clarity. I hope I can continue to do this tomorrow. – Ingix Jul 19 '18 at 14:58
• @DanielWainfleet: A simpler abstract proof uses the Cantor-Bendixson theorem --- Remove an appropriate scattered set (which will be countable when the metric space is separable, but we don't need countability here) from the closed set the OP has, leaving a perfect set. – Dave L. Renfro Jul 20 '18 at 12:30

A proof that an enumaration of the rationals exists such that the resulting $F$ has an isolated irrational point $x$, which can be chosen freely.

This proof has 3 parts:

1. Finding a sequence of different rationals (to be placed as $\{r_{4n}\}$) such that $\bigcup_{n=1}^\infty B(\epsilon_{4n};r_{4n})$ is a cover for the half-open interval $[a_1, x)$, with some $a_1 < x$ and that cover not containing $x$.

2. Similiarly, finding a sequence of different rationals (to be placed as $\{r_{4n-2}\}$) such that $\bigcup_{n=1}^\infty B(\epsilon_{4n-2};r_{4n-2})$ is a cover for the half-open interval $(x, b_1]$, with some $b_1 > x$ and that cover not containing $x$.

3. Finally, putting all the remaining rationals into the odd spots of the enumaration ($\{r_{2n-1}\}$), such that $B(\epsilon_{2n-1};r_{2n-1})$ does not contain $x$.

If all of this is done, then $x$ is the only point from the interval $[a_1, b_1]$ that lies in $F$ (and it is not a boundary point), so $x$ is isolated.

Step 1:

Given any real number $a_1$ and a positive $d$, one can construct real numbers $$d_1=0.9d, d_{n+1}=\frac1{16}d_n, n=1,2,\ldots$$

From this, one can construct $a_2, a_3,\ldots$ such that $a_{n+1} - a_n =d_n$ for $n=1,2,\ldots$ We have $a_n = a_1+ \sum_{i=1}^{n-1}d_i = a_1+ \sum_{i=1}^{n-1}\frac{d_1}{2^{4(i-1)}}$ and thus $\lim_{n \to \infty} a_n = a_1+\frac{16}{15}d_1$.

Let the closed intervals $[a_n,a_{n+1}]$ be denoted as $I_n, n=1,2,\ldots$ The length of each $I_n$ is equal to $d_n$, which is equal to $0.9\frac{d}{2^{4n-4}}$.

Set $d=\frac{2}{2^4}$. This makes $d=2\epsilon_4$, so $d$ is the length of the open interval $B(\epsilon_4;r_4)$. It's easy to see that generally $\frac{d}{2^{4n-4}}$ is the length of the open interval $B(\epsilon_{4n};r_{4n})$. So we have that the length of $I_n$ is 90% of the length of the open interval interval $B(\epsilon_{4n};r_{4n})$.

Finally, choose $a_1 = x-\frac{16}{15}d_1$. This means the $\{a_n\}$ form an increasing sequence whose limit is $x$.

Now we pick rationals $r_{4n}$ in increasing sequence, starting with $r_4$. This is done in such a way that

1. $B(\epsilon_{4n};r_{4n})$ covers $I_n$ (easily done with enough 'wiggle room' to have an infinite choice of potential $r_{4n}$),
2. $B(\epsilon_{4n};r_{4n})$ does not contain $x$ ($x>a_{n+1}$ makes sure there is 'space' to find for the upper boundary of $B(\epsilon_{4n};r_{4n}))$, and
3. $r_{4n}$ is different from all previously chosen $r_{4i}$ (possible because condition 1 and 2 still leave an infinite amount of possible choices).

Obviously, we have $r_{4n} < x$ for all $n=1,2,\ldots$ Additionally, we have $\lim_{n\to \infty}r_{4n} = x$. That means any interval whose closure is not containing $x$ only has a finte number of $r_{4n}$ in them.

This concludes step 1 of our prove, as $\bigcup_{n=1}^{\infty}I_n=[a_1,x)$ and $I_n \subset B(\epsilon_{4n};r_{4n}), n=1,2,\ldots$

Step 2 can be done analogously. The $\{b_i\}$ will decrease this time instead of the incereasing $\{a_i\}$, and $d=\frac{2}{2^2}$ is chosen such that $d$ becomes the length of $B(\epsilon_{2};r_{2})$. The chosen $r_{4n-2}$ are all greater than $x$, so they are all different from the chosen $r_{4n}$.

For step 3, we partition the rationals $\mathbb Q$ into classes $\mathbb Q_1, \mathbb Q_2, \mathbb Q_3 \ldots$ such that

$$\mathbb Q_1 = \{r \in \mathbb Q: \epsilon_1 < |x-r|\}$$ $$\mathbb Q_n = \{r \in \mathbb Q: \epsilon_{2n-1} < |x-r| < \epsilon_{2n-3}\}, n=2,3,\ldots$$

This is a partition as any $|x-r|$ is irrational, so cannot be $0$ or any $\epsilon_k$. The lower bound for $\mathbb Q_n$ is the upper bound for $\mathbb Q_{n+1}$, so there are no gaps. Let $R$ be set set of rationals not already chosen for $r_{4n-2}$ or $r_{4n}$. The closure (in the reals) of each $\mathbb Q_n$ is a finite union of intervals not containing $x$, so each $\mathbb Q_n$ only contains a finite number of rationals from $\mathbb Q\backslash R$. That means each

$$R_n = \mathbb Q_n \cap R, n=1,2,\ldots$$

contains an infinite number of rationals.

Each $R_n$ can be enumerated:

$$R_n = \{r_{n1}, r_{n2},\ldots\}, n=1,2,\ldots$$

These enumerations can be combined to form a full enumeration of $R$:

$$R = \{r_{11}, r_{12},r_{21}, r_{13}, r_{22}, r_{31}, r_{14}, \ldots \}$$

This is the classical diagonal enumeration used in the prove that an infinte countable union of infinite countable sets its again countable. We fill the gaps in our enumeration of all rationals (the odd indices) by the elements of $R$ in that order: $r_1=r_{11}, r_3=r_{12}, r_5=r_{21},\ldots$.

So the first time an element of $R_n$ comes up in the enumaration of $R$ is at position $\frac{n(n+1)}{2}$. That element becomes $r_{n(n+1)-1}$. By the definition of $R_n \subseteq \mathbb Q_n$ we have $|x-r_{n(n+1)-1}| > \epsilon_{2n-1}$. It is easily seen that $n(n+1)-1 \ge 2n-1$ for all $n \ge 1$ (equality holds just for $n=1$). Since the $\epsilon_k$ are decreasing it follows that $x \notin B(\epsilon_{n(n+1)-1};r_{n(n+1)-1})$. Any element of $R_n$ that apears later will have an even higher index $k>n(n+1)-1$ in $\{r_k\}$, which corresponds to an even lower $\epsilon_k$ and thus $x \notin B(\epsilon_k;r_k)$ for all odd $k$ follows.

This concludes the proof.

• I am reading the proof .. for now I will mark it as accepted. Thank you for your effort – Hash Nuke Jul 20 '18 at 16:38
• If anything is unclear, feel free to ask. I know that often it is hard to understand the idea of a proof if it contains much technical detail. – Ingix Jul 24 '18 at 7:17