Intersection of union of crazy intervals in $\mathbb{R}$

Question

I am looking at two sets $X:=[0,1]$ and $V:= X \cap \mathbb{Q}= \{v_1,v_2,...\}$. For each $n,k \in \mathbb{N}_{\ge1}$ I define an interval $I_{n,k}:= X \cap (v_n-2^{-(n+k)},v_n+2^{-(n+k)}) $. Now I want to understand this crazy intersection: $$ D := \bigcap_{k\ge1} \bigcup_{n\ge1} I_{n,k} \ \ \ .$$

I think I could already prove that $V \subseteq D$ (please correct me if I'm wrong). Is this a strict inclusion or is there is equality between $V$ and $D$? My main problem is, that I cannot make proper intuitive sense of the set $D$. Thanks for any idea!

Answer 1

Here's an explicit construction of a number that is in $D$ but not in $V$:

Start by setting $a_0=0, b_0=1$, and repeat the following steps for each $k\ge 1$:

• Let $n$ be the smallest $n$ such that $a_{k-1}<v_n<b_{k-1}$ and $v_n\ne v_k$.
• Let $\varepsilon = \min(2^{-(k+n_k)}, |v_n-v_k|, v_n-a_{k-1}, b_{k-1}-v_n)$.
• Let $a_k = v_n-\varepsilon/2$ and $b_k = v_n+\varepsilon/2 $.

Then $[a_k,b_k]$ is a strictly decreasing sequence of closed intervals such that

• $[a_k, b_k]\subseteq \cup_n I_{n,k} $
• For every $k\ge 1$ we have $v_k \notin [a_k,b_k]$

The number $x=\lim_{k\to \infty} a_k$ is in every $[a_k,b_k]$, and so it must be in $D$ (due to the first of the above properties), but cannot be in $V$ (due to the second property).

It is easy to adapt this construction to produce $2^{\aleph_0}$ different $x\in D\setminus V$, such as by replacing $[a_k,b_k]$ with its left or right third after each round.

Answer 2

(1)Look up a proof of the Baire Category Theorem for completely metrizable spaces. It's quite simple.(2) A corollary is that if $F=\{f_n :n\in N\}$ is a non-empty countable family of dense open subsets of $[0,1]$, then $\cap F$ is uncountable, because if $S=\{r_n : n\in N\}$ is any countable set then $G=\{f_n\backslash \{r_n\} \}$ is also a non-empty countable family of dense open sets so $\cap G$ is dense in $[0,1]$. So $\phi\ne \cap G=(\cap F)\backslash S$ so $\cap F\ne S.$ In your Q, the set $D$ is uncountable so $D\ne [0,1]\cap Q$.