Finding the difference between a set and a union of sets? 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 $\left[0,\frac{1}{4}) \cup (\frac{3}{4},1\right]$?
Similarly,
$$
\bigcap_{i \in ℕ} \left([0,1]  \setminus  \left[\frac{1}{n+1}, 1-\frac{1}{n+1}\right]\right)
$$
confuses me.
Thank you.
 A: 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:
$$
\begin{align}
\bigcap_{n\in \mathbb{N}}[0,1]\setminus [\frac{1}{n+1}, 1 - \frac{1}{n+1}] &= \bigcap_{n\in \mathbb{N}}[0, \frac{1}{n+1}) \cup (1 - \frac{1}{n+1}, 1] \\
&= \{0, 1\}
\end{align}
$$
A: 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
$$
\bigcap_{n  \in \mathbb N} \left([0,1] \setminus A_n \right) = [0,1] \,\setminus \,]0,1[ = \{0, 1\}
$$
