Determine which of the following statements are true for all sets A, B, C, and D. If a double implication fails, determine whether one or the other of the possible implications holds. If an equality fails, determine whether the statement becomes true if the “equals” symbol is replaced by one or the other of the inclusion symbols \(\subset\) or \(\supset\).
\(A \subset B\) and \(A \subset C \Leftrightarrow A \subset (B \cup C)\) .
\(A \subset B\) or \(A \subset C \Leftrightarrow A \subset (B \cup C)\) .
\(A \subset B\) and \(A \subset C \Leftrightarrow A \subset (B \cap C)\) .
\(A \subset B\) or \(A \subset C \Leftrightarrow A \subset (B \cap C)\) .
\(A - (A - B) = B\) .
\(A - (B - A) = A - B\) .
\(A \cap (B - C) = (A \cap B) - (A \cap C)\) .
\(A \cup (B - C) = (A \cup B) - (A \cup C)\) .
\((A\cap B)\cup (A - B) = A.\)
\(A \subset C\) and \(B \subset D \Rightarrow (A \times B) \subset (C \times D)\) .
The converse of (j).
The converse of (j), assuming that \(A\) and \(B\) are nonempty.
\((A\times B)\cup (C\times D) = (A\cup C)\times (B\cup D).\)
\((A\times B)\cap (C\times D) = (A\cap C)\times (B\cap D).\)
\(A \times (B - C) = (A \times B) - (A \times C)\) .
\((A - B)\times (C - D) = (A\times C - B\times C) - A\times D.\)
\((A\times B) - (C\times D) = (A - C)\times (B - D).\)
Solution
False as an equivalence. The forward implication holds: \[ A\subset B\text{ and }A\subset C\Longrightarrow A\subset B\cup C. \] The converse fails, for example with \(A=\{1\}\), \(B=\{1\}\), \(C=\varnothing\).
False as an equivalence. The forward implication holds. The converse fails for \[ A=\{1,2\},\qquad B=\{1\},\qquad C=\{2\}. \]
True: \[ A\subset B\text{ and }A\subset C\iff A\subset B\cap C. \]
False. If \(A\subset B\cap C\), then in fact \(A\subset B\) and \(A\subset C\), so the right-to-left implication holds. The other implication fails, e.g. \(A=B=\{1\}\), \(C=\varnothing\).
False. In fact \[ A-(A-B)=A\cap B\subset B, \] and the reverse inclusion need not hold.
False. Since \(B-A\) is disjoint from \(A\), \[ A-(B-A)=A\supset A-B, \] and equality need not hold.
True: \[ A\cap(B-C)=A\cap B\cap C^c=(A\cap B)-(A\cap C). \]
False. One always has \[ A\cup(B-C)\supset (A\cup B)-(A\cup C), \] since the right-hand side is \(B\cap A^c\cap C^c\). Equality fails whenever \(A\ne\varnothing\).
True, since \(A\cap B\) and \(A-B\) partition \(A\): \[ (A\cap B)\cup(A-B)=A. \]
True. If \((a,b)\in A\times B\), then \(a\in A\subset C\) and \(b\in B\subset D\), hence \((a,b)\in C\times D\).
False. For example, if \(A=\varnothing\), then \(A\times B=\varnothing\subset C\times D\) regardless of whether \(B\subset D\).
True. Suppose \(A,B\ne\varnothing\) and \(A\times B\subset C\times D\). Fix \(b_0\in B\). For every \(a\in A\), \[ (a,b_0)\in C\times D, \] so \(a\in C\), hence \(A\subset C\). Similarly, fixing \(a_0\in A\) gives \(B\subset D\).
False. Always \[ (A\times B)\cup(C\times D)\subset (A\cup C)\times(B\cup D), \] and the right-hand side also contains \(A\times D\) and \(C\times B\), which need not lie in the left-hand side. For example take \(A=D=\{1\}\) and \(B=C=\varnothing\).
True: \[ (A\times B)\cap(C\times D)=(A\cap C)\times(B\cap D). \]
True: \[ A\times(B-C)=(A\times B)-(A\times C). \]
True. First \[ (A\times C)-(B\times C)=(A-B)\times C. \] Subtracting \(A\times D\) then gives \[ ((A-B)\times C)-(A\times D)=(A-B)\times(C-D). \]
False. One always has \[ (A-C)\times(B-D)\subset (A\times B)-(C\times D), \] but equality need not hold. For example, \((A\times B)-(C\times D)\) contains \((A-C)\times(B\cap D)\), which is disjoint from \((A-C)\times(B-D)\).