You can think of a universal set \(U\) as an “ambient set” in which all other sets in a problem are subsets of \(U\). In other words, \(U\) is usually a “biggest set” that contains all elements that we care to talk about.
In other words, \(\overline{A}\) is the set of all elements NOT in \(A\) (relative to \(U\)).
Set intersection and set complements allow us to define a new operation called set difference.
You can think of the difference of two sets \(A\) and \(B\) to be all the stuff in \(A\) but with all the \(B\)-stuff removed (hence the idea of ‘difference’ or ‘subtraction’).
\(\overline{A}=\{1,4,6,8,9\}\).
Remember, we are looking for the set of all elements that are NOT in A, but ARE in the universal set given.
\(\overline{B}=\{2,3,4,5,7,8\}\).
Remember that we are looking for the set of all elements that are NOT in \(B\), but ARE in the universal set \(U\).
\(\overline{C}=\{n\ :\ n\) is an odd natural number\(\}\)
Again, we are looking for the set of all things NOT in \(C\) that ARE within our context universal set \(\mathbb{N}\), i.e. the set of all natural numbers.
Pro-Tip: One could have also answered this question by simply saying “the set of all odd natural numbers” as opposed to writing it in set-builder notation above and still be mathematically correct (although the person grading your work probably wants the set builder notation, but you COULD argue with them on this point if they don’t tell you what notation to use!).
\(\overline{A}=\{1,2,3,4,5\}\) or, equivalently, \(\overline{A}=\{x\ :\ x\) is a natural number and \(x\leq 5\}\).
The original set \(A\) specified that the elements therein must be natural numbers greater than 5. The opposite of \(A\), or rather the complement of \(A\) is therefore the set of all natural numbers less than or equal to 5. The notations for this set are as you see above.
Often, it is more useful to first determine what your set contains before trying to determine which notation to use to represent your set. Unless otherwise specified, you have the choice as to which notation you want to use.
\(A\setminus B=\{0,6,7\}\) and \(B\setminus A=\{4\}\)
Here, when dealing with set differences, we are looking for the set of all elements that ARE in the first set (before the backslash) that are NOT ALSO in the second set (after the backslash). Put another way, for \(A\setminus B\), we are looking to essentially “remove” any elements of \(B\) that we see. Similarly for \(B\setminus A\), where we are looking to “remove” any elements of \(A\) that we see from \(A\)’s list.
\(B\setminus C=\{\}=\emptyset\) and \(C\setminus B=\{0,4\}\)
For \(B \setminus C\), note that there are NO elements in \(B\) that are not also in C. So, if we remove all the \(C\)-stuff from \(B\), there isn’t anything left.
For \(C \setminus B\), if we take the \(B\)-stuff out of \(C\), we end up with only a 0 and 4 remaining in \(C\). Hence the answer you see.
\(A\setminus B= \{11, 12, 13, 14, 15, 16, 17,18,19,20\}\)\(A\setminus B= \{11, 12, 13, 14, 15, 16, 17,18,19,20\}\)
Here its a good idea to “decode” what is in each set \(A\) and \(B\) before trying to find the difference of the two.
\(A\) is the set of all natural numbers greater than 10, i.e. \(\{11,12,13,14,15,…\}\). For \(B\) we are looking at the set \(B=\{21,22,23,24,25,26…\}\). So, to find \(A\setminus B\), or “the set of all things in \(A\) with the \(B\)-stuff removed” is the set containing the natural numbers between 11 and 20 including the 11 and 20.