Combining Set Operations
One can combine a variety of set operations in much the same way you can combine differences, sums, products, and quotients of numbers.
To indicate which operations come first in terms when computing the result of several set operations, we use parentheses!
For example, if you wanted to compute \(A\cap (B\cup C)\), you would first compute the result of \(B\cup C\), and then find the intersection of the result with \(A\).
When dealing with set complements we tend not to put parentheses around expressions like \(\overline{A}\), so make a note to evaluate complements like \(\overline{A}\) first in more complicated problems involving complements.
\(A\cup (B\cap C)=\{1,3,5,7\}\)
First find \(B\cap C\), which equals \(\{5\}\). Then you have \(A\cup \{5\}\) which gives you \(\{1,3,5,7\}=A\). The parentheses indicated that the intersection should be computed first.
\((A\cup B)\cap C=\{1,2,3,5,7\}\cap C=\{5,7\}\).
First compute the union inside parentheses, then see what that has in common with \(C\).
\(A\cap B\cap C=\{5,7\}\).
This can be done in one of two ways. The first way is by recognizing that this problem is asking “what elements do \(A\), \(B\), and \(C\) all have in common?” The second way would be to first compute \(A\cap B\) or \(B\cap C\), then seeing what those have in common with \(C\) or \(A\), respectively.
With all these sorts of problems, if there are no parentheses given, you can use your own to help you organize your work.
\((A\cap B)\cup (C\cap D)=\{2,3,4,5,6,7\}\).
Notice that \((A\cap B)=\{2,3,4,5\}\) and \((C\cap D)=\{5,6,7\}\). So \((A\cap B)\cup (C\cap D)=\{2,3,4,5\}\cup \{5,6,7\}\). Union means “dump the two sets together into one big set” so the final result is \(\{2,3,4,5,6,7\}\).
\(\overline{A}\cap (B\cup C)=\{1,4,8,9\}\)
First compute \((B\cup C)\) and \(\overline{A}\) separately. This gives \(B\cup C=\{1,2,4,5,8,9\}\) and \(\overline{A}=\{1,4,6,8,9,10\}\). Then, to compute the final answer, compute the intersection of these two sets by determining what they have in common.
\((A\cap B)\cap \overline{C}=\{2,5\}\)
First compute the result of \(A\cap B\) and \(\overline{C}\), which gives \(A\cap B=\{2,5\}\) and \(\overline{C}=\{0,1,2,3,4,5,6,7,10\}\) respectively. Finally, compute the intersection of these two sets by determining what they have in common. This gives the final answer you see.
\(A\cap(C\setminus B)=\{2,4,8\}\)
First compute \(C \setminus B=\{2,4,8\}\). Then, compute the final answer by intersecting \(\{2,4,8\}\), by determining what \(A\) and this set have in common.
\((C\setminus B)\cup (C\setminus A)=\{1,2,4,5,7,8\}\)
Note that \(C\setminus B=\{2,4,8\}\) and \(C\setminus A=\{1,5,7\}\). To find the final answer, union these together by essentially dumping the elements of these two sets together into one big set.
\(A\setminus B=\emptyset\).
By computing set difference \(A\setminus B\), I am looking to remove any elements of \(B\) that I can from \(A\). Since \(A\subseteq B\), if we remove any elements of \(B\) from \(A\), this means we remove all elements of \(A\) as well. So, nothing remains. We indicate that with the empty set symbol \(\emptyset\).