Title: Sets
Course: Demonstrating STACK (v4.3.7)

If \(X\), \(Y\) are sets containing \([value]\), \([value]\) elements respectively, how many elements are there in the cartesian product, \(X\times Y\), of the two sets?

\(| X \times Y| = \)
Feedback: The cartesian product of sets \(X\) and \(Y\), \(X \times Y\) is the set containing all ordered pairs with as first member an element of \(X\) and as second member an element of \(Y\). For each element of \(X\), one can construct \([value]\) different pairs, with \([value]\) being the number of elements in \(Y\). Since there are \([value]\) elements in \(X\), this will mean a total of \([value]*[value]\) pairs. Hence the cardinality of \(X \times Y\) is \([value]\).

Determine the cardinality of the set \(A\), i.e. \(|A|\), when \[ A =[value] \] \(|A|\) =
Feedback: The cardinality of a set is the number of different elements in a set, i.e. \[ |[value]| = |[value]| = [value].\]

Determine \( A \cap B\), when \[ A = [value]\, \mbox{and}\ B = [value] .\] \( A \cap B\) =
Feedback: The intersection of two sets is the set of all elements that are present in both sets, i.e., \[[value] \cap [value] = [value]\]

Let \[A=[value], B=[value]\mbox{ and } C=[value]. \] Work out, using only the definition of union, intersection and complement, \[(A \cap B) \cup (A \cap C)\]
Note: \([value]\) is entered as `delta', \([value]\) is entered as `beta' and \([value]\) is entered as `gamma'. Sets are entered using {}, e.g. {1,2,3}.

\((A \cap B) \cup (A \cap C)\)=
Feedback: The intersection of \(A\) and \(B\) is the set of all elements contained in both \(A\) and \(B\), hence \[ A \cap B = [value].\] Similarly, the intersection of \(A\) and \(C\) is the set of all elements contained in both \(A\) and \(C\), hence \[ A \cap C = [value].\] The union of two sets is the set containing all elements that are in either or both of the two sets, hence, \[ (A \cap B) \cup (A \cap C) = [value]\cup[value]=[value].\]

Express the following set as an interval: \[ \left[ 1, [value] \right] \cap \left[ [value], [value] \right]. \] Note: Give your answer as an inequality such as \( x<= -7\) and \(x>5\).
Feedback: \(\left[ 1, [value] \right] \cap \left[ [value], [value] \right]\mbox{ is the set containing all values in both }\left[ 1, [value] \right] \mbox{ and in }\left[ [value], [value] \right],\mbox{ hence }\left[[value], [value] \right]\). This is entered as \(x>=[value]\mbox{ and }x<=[value]\).

Express the following set as an interval: \[ \left( [value], [value] \right]\ \backslash\ \left[ [value], 55 \right). \]
Note: Give your answer as an inequality such as \( x>= -7\) and/or \(x<5\).
Feedback: \(\left( [value], [value] \right]\ \backslash\ \left[ [value], 55 \right) \mbox{ are all the values that are in }\left( [value], [value] \right] \mbox{ but not in }\left[ [value], 55 \right),\mbox{ i.e. it is the interval }\left( [value], [value] \right),\mbox{ as }[value] \mbox{ is included in the second interval}.\) This is entered as \(x>[value]\mbox{ and }x<[value]\).

Give a proper strict subset of A, i.e. a subset B such that \[ B \subset A \] where \[ A = [value] . \] B =
Feedback: To obtain a proper strict subset, you remove one or more elements so, e.g., \[[value]\] is a proper strict subset.