Under what conditions are the variables $a, b$ satisfying $A \cap B = \emptyset$ and $A \cap B = \{0\}$ I am struggling with the following question:
You have the two sets $A={x∈R; |x−a|≤1}$ och $B={x∈R; |x−b|≥2}.$ Give the conditions for when the variables: $a$ and $b$ are satisfying $A \cap B = \emptyset$ and $A \cap B =\{0\}.$
I am not sure how to think here. But here is my start:
$0 \leq| x−a | ≤ 1$ and  $ 2 \leq | x−b |$
Which means that for $A \cap B = \emptyset$, the sets can not "overlap" and they wont do that when $|a - b| \geq 1$.
This is apparently wrong and the correct answer is: $|a - b| < 1,$ $b-1 < a < b + 1$ (according to my textbook).
For the second question when  $A \cap B =\{0\}.$ I know that the only element after the set union operation should be $0$. 
I am not sure how to get to the correct answer which (according to my textbook) is: $|a − b| = 3$, $a = b \pm 3$
I would be very happy if someone can help me out and explain how to think to get these answers.
Thank you.
 A: I'll give you some hints on how to tackle this problem. If you find something unclear, just leave a comment, and I'll be more than happy to help you.


*

*One thing you should note is that $|a - b|$ tell you how far apart $a$, and $b$ are.


*

*For example: $|5 - 9| = |-4| = 4$, that means 5, and 9 are 4 (units) apart. Another example is: $|-2 - (-7)| = |-2 + 7| = |5| = 5$, which means that -2; and -7 are 5 units apart. (in the latter example, $a = -2$, and $b = -7$).

*If you still cannot imagine it, get some crap paper, a pencil, draw a real line, and mark the points according to the numbers on it.


*So all values of $x$ that satisfy $\color{red}{|x - 5|} \color{blue}{< 1}$ are those whose distance from 5 is less than 1 ($\color{red}{|x - 5|}$ means that 'the distance from $x$ to 5'; and $\color{blue}{< 1}$ basically means that 'less than 1'), another way to put it, it's all number on the interval $(4; 6)$. If you cannot see why, use the real line.

*Analogously, all numbers $x$ satisfy $|x - 2| \ge 3$ are those whose distance from 2 is greater than or equal to 3, it's of course the interval $(-\infty; -1] \cup [5; +\infty)$. Note that 1, and 5 are included, since there distance from 2 is equal to 3.
Also note that, in the above example, the numbers 4; and 6 do NOT satisfy our requirement (i.e, the requirement that the distance from to 5 is less than 1. In fact, their distance from 5 is indeed 1).

*Ok, back to your problem, mark some arbitrary number $a$ on the real line, find the set $A$, i.e the set of all numbers, whose distance from $a$ is less than or equal to 1. Do the same for set $B$. So what conditions that $a$, and $b$ must have so that $A \cap B = \emptyset$; and $A \cap B = \{ 0 \}$?
