# What does this set actually look like?

$$S=\bigcap_{n=1}^\infty \left( \left[0, \frac{1}{2n+1}\right]\cup\left[\frac{1}{2n},1\right]\right)$$ Generally if we put down the values of $$n$$ and compute, then my value comes out $$\{0\} \cup \left[\frac{1}{2},1\right]$$ I think. Is there any way to compute this easily? So I did try to compute this. Let $$A_n=[0, \frac{1}{2n+1}]$$ and $$B_n=[\frac{1}{2n},1]$$. So that $$S=\bigcap_{n=1}^\infty ( A_n\cup B_n)$$ $$=({A_1}\cup{B_1}) \cap(A_2 \cup B_2) \cap(A_3 \cup B_3) . . .$$ $$=(A_1\cap A_2 \cap A_3 . . .)\cup(B_1 \cap B_2 \cap B_3 . . .)$$ And so the result follows. Is is wrong to write like this?

• $S$ will be a subset of $[0,1]$ containing all elements who are not strictly between $\frac{1}{2n+1}$ and $\frac{1}{2n}$ for any natural $n$. In particular $S$ will contain all values in $[\frac{1}{4},\frac{1}{3}]$ among others which were missing from your attempt. Sep 28 '20 at 16:45
• As for "Is it wrong to write it like this" even adjusting the union into an intersection typo you made, $(A_1\cup B_1)\cap (A_2\cup B_2)\neq (A_1\cup A_2)\cap (B_1\cup B_2)$. Whatever expansion you attempted with $(A_1\cup B_1)\cap (A_2\cup B_2)\cap \dots$ appears invalid. With the edit swapping intersections with unions and vice versa, it is still invalid since $(A_1\cup B_1)\cap (A_2\cup B_2)\neq (A_1\cap A_2)\cup (B_1\cap B_2)$. Consider for example $A_1 = B_2 = \{1\}$ and $A_2=B_1=\emptyset$ Sep 28 '20 at 16:48
• That was a silly mistake of mine. I did intend to put intersections. Sep 28 '20 at 16:56
• HINT: Rewrite $A_n\cup B_n$ as $[0,1]\setminus\left(\frac1{2n+1},\frac1{2n}\right)$ and apply De Morgan’s law. Sep 28 '20 at 17:00

Let us consider the complement in $$I=[0,1]$$ instead: \begin{align*} I\setminus S &=I\setminus \bigcap_{n=1}^\infty \left( \left[0, \frac{1}{2n+1}\right]\cup\left[\frac{1}{2n},1\right]\right) \\ &=\bigcup_{n=1}^\infty I\setminus\left( \left[0, \frac{1}{2n+1}\right]\cup\left[\frac{1}{2n},1\right]\right) \\ &=\bigcup_{n=1}^\infty \left(\frac{1}{2n+1},\frac{1}{2n}\right). \end{align*} Note that this union is disjoint, so the complement of $$S$$ in $$I$$ consists of countable infinitely many disjoint open intervals.
We may of course also now write $$S = [0,1]\setminus \left(\bigcup_{n=1}^\infty \left(\frac{1}{2n+1},\frac{1}{2n}\right)\right).$$