I'm having trouble figuring out how to compute the difference of a set and union/intersection of sets. For example:

$$ [0,1] \setminus \bigcap_{i = 3}^\infty \left[\frac{1}{n+1}, 1-\frac{1}{n+1}\right] $$

For this, I get $[0,1] \setminus \left[\frac{1}{4}, \frac{3}{4}\right]$. From here, logic dictates that the answer would just simply be $[0,1]$ because the elements of B don't exist at all in A, but that can't possibly be correct, right? It would have to be $\left[0,\frac{1}{4}) \cup (\frac{3}{4},1\right]$?


$$ \bigcap_{i \in ℕ} \left([0,1] \setminus \left[\frac{1}{n+1}, 1-\frac{1}{n+1}\right]\right) $$

confuses me.

Thank you.

  • $\begingroup$ For the first formula, you're right. For the second, consider independently the intersection of the left intervals and and the intersection of the right intervals. $\endgroup$
    – Bernard
    Mar 1 '19 at 22:58

I think you might be confused by notation $[a, b]$. Remember that $[a, b] = \{x\in \mathbb{R} : a\leq x \leq b\}$.

Your first answer is good up to $[0,1] \setminus \left[\frac{1}{4}, \frac{3}{4}\right]$. In this case we wouldn't have $[0,1] \setminus \left[\frac{1}{4}, \frac{3}{4}\right] = [0, 1]$ since $\left[\frac{1}{4}, \frac{3}{4}\right]$ is in $[0, 1]$ ($\frac{1}{4}>0$ and $\frac{1}{4}<1$, $\frac{3}{4}>0$ and $\frac{3}{4}<1$ means $[\frac{1}{4}, \frac{3}{4}]$ is contained in $[0,1]$).

It helps to imagine $[0,1]$ and $\left[\frac{1}{4}, \frac{3}{4}\right]$ on the number line: the parts contained in $[0,1] \setminus \left[\frac{1}{4}, \frac{3}{4}\right]$ are the parts of $[0, 1]$ that don't overlap with $\left[\frac{1}{4}, \frac{3}{4}\right]$

The actual answer is $[0,1] \setminus \left[\frac{1}{4}, \frac{3}{4}\right] = [0, \frac{1}{4}) \cup (\frac{3}{4}, 1]$

For $\bigcap_{n\in \mathbb{N}}[0,1]\setminus [\frac{1}{n+1}, 1 - \frac{1}{n+1}]$ we have the following:

$$ \begin{align} \bigcap_{n\in \mathbb{N}}[0,1]\setminus [\frac{1}{n+1}, 1 - \frac{1}{n+1}] &= \bigcap_{n\in \mathbb{N}}[0, \frac{1}{n+1}) \cup (1 - \frac{1}{n+1}, 1] \\ &= \{0, 1\} \end{align} $$

  • $\begingroup$ Okay, good. I was thinking the "right" way then in doing the number line method. I was second-guessing myself since online calculators assumed it was just doing set difference, which is where the $[0,1]$ was coming from. On the second, I'm still confused as to what happened in the intermediary step. Did you "swap" the segments? On a number line, $[\frac{1}{n+1}, 1-\frac{1}{n+1] quickly "puckers". Is that why? $\endgroup$ Mar 1 '19 at 23:36
  • $\begingroup$ First I took the set difference between $[0, 1]$ and $[\frac{1}{n+1}, 1- \frac{1}{n+1}]$ to get $[0,1]\setminus [\frac{1}{n+1}, 1 - \frac{1}{n+1}] = [0, \frac{1}{n+1}) \cup (1 - \frac{1}{n+1}, 1]$. Then you take the intersection of all $[0, \frac{1}{n+1}) \cup (1 - \frac{1}{n+1}, 1]$ to get $\{0, 1\}$. To see that $\{0, 1\}$ is the right answer, note that $[0, \frac{1}{n+1}) \cup (1 - \frac{1}{n+1}, 1]$ gets closer to $\{0, 1\}$ as $n$ gets larger, so if you drew some line segments for n = 1 through n = 10, you can see that the overlapping portions will eventually only contain 0 and 1. $\endgroup$
    – user648059
    Mar 1 '19 at 23:41

Let $A_n$ denote the interval $A_n:=\left[\frac{1}{n+1}, 1-\frac{1}{n+1}\right]$, then $A_n \subset A_{n+1}$ for all $n \in \mathbb N$. Therefore, $$ \bigcap_{n = 3}^\infty A_n = A_3 \text{ and therefore } [0,1] \setminus \bigcap_{n=3}^\infty A_n = [0,1] \setminus A_3 $$

For the other set, from $A_n \subset A_{n+1}$ it follows that $[0,1] \setminus A_{n+1} \subset [0,1] \setminus A_n$, so the intersections get smaller when $n$ increases: $$ \bigcap_{n = 1}^m \left([0,1] \setminus A_n \right) = [0,1] \setminus A_m $$ As $A_n$ can get arbitrary close to $]0, 1[$ (for any $\epsilon > 0$ we can find an $m \in \mathbb N$ such that $\frac{1}{n+1} < \epsilon$ for all $n>m$) but never will reach $[0,1]$, we have $$ \bigcap_{n \in \mathbb N} \left([0,1] \setminus A_n \right) = [0,1] \,\setminus \,]0,1[ = \{0, 1\} $$


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