Question

Each of the following statements is either true or false. If a statement is true, prove it. If a statement is false, disprove it. These exercises are cumulative, covering all topics addressed in Chapters \(1-9 .\) If \(A\) and \(B\) are sets, then \(\mathscr{P}(A) \cap \mathscr{P}(B)=\mathscr{P}(A \cap B)\).

Short answer

The statement is false. There are cases where \(\mathscr{P}(A) \cap \mathscr{P}(B)\) does not equal to \(\mathscr{P}(A \cap B)\).

Step-by-step solution

Understand the Terms

Power set \(\mathscr{P}(A)\) of a set \(A\) is a set of all of its subsets. Thus, it contains every subset that can be created from set \(A\). The intersection of two sets \(A\) and \(B\) (denoted \(A \cap B\)) is the set that contains all elements that \(A\) and \(B\) have in common.

Evaluate Left Side of the Equation

\(\mathscr{P}(A) \cap \mathscr{P}(B)\) represents the intersection of the power sets of \(A\) and \(B\), which can be interpreted as the set of all common subsets between \(A\) and \(B\).

Evaluate Right Side of the Equation

\(\mathscr{P}(A \cap B)\) represents the power set of the intersection of \(A\) and \(B\), which can be interpreted as the set of every subset that can be formed from the common elements between \(A\) and \(B\).

Compare Both Sides of the Equation

Comparing the interpretations from Step-2 and Step-3, it is clear that \(\mathscr{P}(A) \cap \mathscr{P}(B)\) is not always equal to \(\mathscr{P}(A \cap B)\). For a simple disproof, consider \(A = \{1\}\) and \(B = \{2\}\). Then \(\mathscr{P}(A) \cap \mathscr{P}(B) = \{\emptyset\}\) but \(\mathscr{P}(A \cap B) = \{\emptyset, \{1, 2\}\}\). Since \(\{\emptyset\} ≠ \{\emptyset, \{1, 2\}\}\), the statement is false.