Question

Suppose \(A=\\{0,1\\}\) and \(B=\\{1,2\\} .\) Find: (a) \((A \times B) \cap(B \times B)\) (b) \((A \times B) \cup(B \times B)\) (c) \((A \times B)-(B \times B)\) (d) \((A \cap B) \times A\) (e) \((A \times B) \cap B\) (f) \(\mathscr{P}(A) \cap \mathscr{P}(B)\) \((\mathbf{g}) \mathscr{P}(A)-\mathscr{P}(B)\) (h) \(\mathscr{P}(A \cap B)\) (i) \(\mathscr{P}(A \times B)\)

Short answer

(a) \{(1,1), (1,2)\} (b) \{(0,1), (0,2), (1,1), (1,2), (2,1), (2,2)\} (c) \{(0,1), (0,2)\} (d) \{(1,0), (1,1)\} (e) \(\emptyset\) (f) \{\emptyset, \{1\}\} (g) \{\{0\}, \{0, 1\}\} (h) \{\emptyset, \{1\}\} (i) \(\mathscr{P}(A \times B)\)

Step-by-step solution

(a) Find (A x B) ∩ (B x B)

First, find the Cartesian product \(A \times B = \{(0,1), (0,2), (1,1), (1,2)\}\) and \(B \times B = \{(1,1), (1,2), (2,1), (2,2)\}\). The intersection of these two sets is \((A \times B) \cap (B \times B) = \{(1,1), (1,2)\}\).

(b) Find (A x B) ∪ (B x B)

The union is calculated as \((A \times B) \cup (B \times B) = \{(0,1), (0,2), (1,1), (1,2), (2,1), (2,2)\}\).

(c) Find (A x B) - (B x B)

The difference is \((A \times B) - (B \times B) = \{(0,1), (0,2)\}\).

(d) Find (A ∩ B) x A

The intersection \(A \cap B = \{1\}\). Thus, \((A \cap B) \times A = \{(1,0), (1,1)\}\).

(e) Find (A x B) ∩ B

Here the result is the empty set as no pair from \(A \times B\) is found in list B. Hence, \((A \times B) \cap B = \emptyset\).

(f) Find 𝒫(A) ∩ 𝒫(B)

First, find the power sets: \(\mathscr{P}(A) = \{\emptyset, \{0\}, \{1\}, \{0, 1\}\}\), \(\mathscr{P}(B) = \{\emptyset, \{1\}, \{2\}, \{1, 2\}\}\). The intersection is \(\mathscr{P}(A) \cap \mathscr{P}(B) = \{\emptyset, \{1\}\}\).

(g) Find 𝒫(A) - 𝒫(B)

\(\mathscr{P}(A) - \mathscr{P}(B) = \{\{0\}, \{0, 1\}\}\).

(h) Find 𝒫(A ∩ B)

The intersection \(A \cap B = \{1\}\). It's power set \(\mathscr{P}(A \cap B) = \{\emptyset, \{1\}\}\).

(i) Find 𝒫(A x B)

The Cartesian product \(A \times B = \{(0,1), (0,2), (1,1), (1,2)\}\). It's power set: \(\mathscr{P}(A \times B) = \{\emptyset, \{(0,1)\}, \{(0,2)\}, \{(1,1)\}, \{(1,2)\}, \{(0,1), (0,2)\}, \ \{(0,1), (1,1)\}, \{(0,1), (1,2)\}, \{(0,2), (1,1)\}, \{(0,2), (1,2)\}, \ \{(1,1), (1,2)\}, \{(0,1), (0,2), (1,1)\}, \{(0,1), (0,2), (1,2)\}, \{(0,1), (1,1), (1,2)\}, \ \{(0,2), (1,1), (1,2)\}, \{(0,1), (0,2), (1,1), (1,2)\}\}\).

Key concepts

Cartesian Product

The Cartesian product is the foundation for understanding more complex set operations. Imagine you have two sets, say Set A and Set B. The Cartesian product, denoted by \(A \times B\), is the set of all possible ordered pairs where the first element is from Set A and the second is from Set B. For example, if \(A = \{0, 1\}\) and \(B = \{1, 2\}\), then the Cartesian product \(A \times B\) is \(\{(0,1), (0,2), (1,1), (1,2)\}\). Visualizing each pair as a coordinate point can help you grasp this concept better, as it's like plotting points on a two-dimensional grid.

It's crucial to remember that the order matters in ordered pairs; \((1, 2)\) is not the same as \((2, 1)\). This distinction is particularly important when considering the associative and commutative properties which do not hold for Cartesian products — \(A \times B\) is not the same as \(B \times A\) if the sets are different.

Set Intersection

When you're looking at the common elements between two sets, you're dealing with set intersection, represented by the symbol \(\cap\). Let's say you've computed the Cartesian product of two sets and you want to find common elements with another set. You're essentially looking for overlapping elements that exist in both sets. For example, in our exercise, finding \((A \times B) \cap (B \times B)\) involves finding ordered pairs that are present in both Cartesian products. The result, \(\{(1,1), (1,2)\}\), shows the pairs that are common to both \(A \times B\) and \(B \times B\). An intersection can sometimes result in an empty set, indicating no common elements, which is denoted by \(\emptyset\).

Set Union

Unlike intersection, which is about commonality, the union of two sets captures the entirety of their elements. It's denoted by \(\cup\) and essentially merges both sets, excluding any duplicates to maintain the definition of a set where each element is unique. If you consider our example, \( (A \times B) \cup (B \times B)\), the union included all ordered pairs from both Cartesian products without repeating any. The resulting set is \(\{(0,1), (0,2), (1,1), (1,2), (2,1), (2,2)\}\), showcasing a full combination of both sets' elements. Understanding the concept of union is helpful in visualizing the collective items from different sets.

Power Set

The power set is a concept that can initially seem overwhelming but is quite straightforward. It is the set of all subsets of a given set, including the empty set and the set itself. Represented using the notation \(\mathscr{P}(A)\), it includes everything from no elements to all possible elements. Considering our set \(A\), the power set \(\mathscr{P}(A)\) is \(\{\emptyset, \{0\}, \{1\}, \{0, 1\}\}\). It's like exploring all possible combinations of the set's elements. The size of the power set is always \(2^n\), where \(n\) is the number of elements in the original set, reflecting the binary decision of including or excluding each element.

Set Difference

Set difference is a little like subtraction for sets. Denoted by a minus sign \(-\), it involves taking elements of one set and removing any that are also in another set. In our context, finding \((A \times B) - (B \times B)\) means you're stripping away the ordered pairs from \(A \times B\) that also appear in \(B \times B\), leaving you with \(\{(0,1), (0,2)\}\). It's a handy operation for isolating unique elements in a set, and it underscores the fact that order and uniqueness within the original sets are crucial when it comes to set operations. Set difference operates strictly in one direction — unlike intersection and union, it's not commutative. The difference of \(A - B\) is not the same as \(B - A\), unless both sets are equal.