0
$\begingroup$

$$S=\bigcap_{n=1}^\infty \left( \left[0, \frac{1}{2n+1}\right]\cup\left[\frac{1}{2n},1\right]\right)$$ 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?

$\endgroup$
4
  • $\begingroup$ $S$ 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. $\endgroup$
    – JMoravitz
    Sep 28 '20 at 16:45
  • $\begingroup$ 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 $A_2=B_1=\emptyset$ $\endgroup$
    – JMoravitz
    Sep 28 '20 at 16:48
  • $\begingroup$ That was a silly mistake of mine. I did intend to put intersections. $\endgroup$ Sep 28 '20 at 16:56
  • $\begingroup$ HINT: Rewrite $A_n\cup B_n$ as $[0,1]\setminus\left(\frac1{2n+1},\frac1{2n}\right)$ and apply De Morgan’s law. $\endgroup$ Sep 28 '20 at 17:00
0
$\begingroup$

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 $$ S = [0,1]\setminus \left(\bigcup_{n=1}^\infty \left(\frac{1}{2n+1},\frac{1}{2n}\right)\right). $$

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.