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.


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 $A \cap B = \{ 0 \}$?

  • 1
    $\begingroup$ I agree with the last point. The best thing to do to learn to solve questions like these is to draw everything always, until you can do the same in your head, without the paper. Visualization is key in any part of mathematics that has to do with... Actually, basically any part of mathematics at all. $\endgroup$ – Arthur Jul 3 '13 at 13:27
  • $\begingroup$ Thank you for your answer! I will look into this now... $\endgroup$ – Lukas Arvidsson Jul 3 '13 at 13:39

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