![enter image description here][1]I pretty much know how operations on sets work and I do have idea how to mark them on an axis but i find this a little different so if somebody could explain that to me please.

We have set $A = <0,1) \cup \{2\}, B = [0,5,2)$

find $A \cap B$, $A \cup B$, $A \setminus B$, $B \setminus A$, $A'$, $B'$ where $U = \mathbb{R}$

how do i treat these: <, [ ?

Please take a look here: http://postimg.org/image/tajpuc4gv/

That's exactly how it is. That's a picture from my last exam I managed to do. it says exactly :

b) For sets A=.... and B =.... find: .....

  • $\begingroup$ These are intervals: $[a,b)$ means $(a,b) \cup \{a\}$, in other words $a$ is included inside the interval. I'm not sure about $B$, did you want to write $B= [0.5, 2)$? $\endgroup$ – Blex Jun 7 '15 at 10:44
  • $\begingroup$ The curly are used for "listing" elements in a set like : $X = \{ 1,2, 3 \}$ while square and normal ones are used to describe intervals in $\mathbb R$ : closed : $[0,1]$ ($1$ is "in") or open : $[0,1)$ ($1$ is "out"). $B = [0,5,2)$ makes no sense. $\endgroup$ – Mauro ALLEGRANZA Jun 7 '15 at 10:46
  • $\begingroup$ Where have you seen these?. If it had said $[0,1)$ then it might mean all numbers greater than or equal to $0$ and strictly less than $1$. So $[0,1)\cup\{2\}$ might mean those numbers and the number $2$. $\endgroup$ – Henry Jun 7 '15 at 10:47
  • $\begingroup$ Please take a look at the picture I've edited my post with. It sure enough says [0,5,2) $\endgroup$ – imdumb Jun 7 '15 at 10:58
  • $\begingroup$ Do you guys reckon i should treat < exactly the same as [ ? $\endgroup$ – imdumb Jun 7 '15 at 12:29

That looks as if it might be Polish, and I believe that Polish, like German, uses the decimal comma rather than the decimal point. Thus, what looks like $[0,5,2)$ may be what I would write $[0,5.2)$. I don’t know what the left angle bracket in $\langle 0,1)$ means, though, unless it’s just a sloppy parenthesis, and what’s intended is simply the open interval $(0,1)$. In that case $A\cap B$ would be $(0,1)\cup\{2\}$ and $A\setminus B$ would be $\varnothing$, for instance.

  • $\begingroup$ Well, the guy who was getting those exams ready for us is from Sydney. No idea what is that all about. $\endgroup$ – imdumb Jun 8 '15 at 15:41

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