Discrete Math Problems: V Set Differences, Unions, And Intersections

Set Differences, Unions, And Intersections


  1. $$A \setminus (B \setminus C) = (A \setminus B) \setminus C$$

    False . Let \( A = \{1,2\}, \; B = \{1\}, \; C = \{1\} \). Then \( B \setminus C = \varnothing \), so \( A \setminus (B \setminus C) = A = \{1,2\} \). On the other hand, \( A \setminus B = \{2\} \), and thus \( (A \setminus B)\setminus C = \{2\} \). Therefore, \( A \setminus (B \setminus C) \ne (A \setminus B)\setminus C \).


    However, we can show that $A \setminus (B \setminus C) \subseteq (A \setminus B) \setminus C$. Let \(x \in (A \setminus B)\setminus C\). Then \(x \in A\), \(x \notin B\), and \(x \notin C\). We want to show \(x \in A \setminus (B \setminus C)\), meaning that \(x \in A\) and \(x \notin (B \setminus C)\). We already have \(x \in A\). Now, \(x \in (B \setminus C)\) would require \(x \in B\), but we know that \(x \notin B\). So \(x \notin (B \setminus C)\). Therefore, \(x \in A \setminus (B \setminus C)\), which proves that \( (A \setminus B)\setminus C \subseteq A \setminus (B \setminus C)\).


  2. $$A \setminus (B \setminus C) = A \setminus (C \setminus B)$$

    True . Let \(x \in A \setminus (B \setminus C)\). Then \(x \in A\) and \(x \notin (B \setminus C)\). We want to show that \(x \in A \setminus (C \setminus B)\), meaning that \(x \in A\) and \(x \notin (C \setminus B)\). We already have \(x \in A\). Now, \(x \in (C \setminus B)\) would require that \(x \in C\) and \(x \notin B\). But since \(x \notin (B \setminus C)\), it is not the case that \(x \in B\) and \(x \notin C\), so \(x\) cannot satisfy both \(x \in C\) and \(x \notin B\). Hence \(x \notin (C \setminus B)\). Therefore, \(x \in A \setminus (C \setminus B)\), which proves that \(A \setminus (B \setminus C) \subseteq A \setminus (C \setminus B)\).


    By symmetry (interchanging \(B\) and \(C\)), the same argument shows that \(A \setminus (C \setminus B) \subseteq A \setminus (B \setminus C)\). Thus, \(A \setminus (B \setminus C) = A \setminus (C \setminus B)\).


  3. $$A \setminus (B \setminus C) = (A \setminus C) \setminus (A \setminus B)$$

    False. Let \( A = \{1,2\}, \; B = \{1\}, \; C = \{2\} \). Then \( B \setminus C = \{1\} \), so \( A \setminus (B \setminus C) = \{2\} \). On the other hand, \( A \setminus C = \{1\} \) and \( A \setminus B = \{2\} \), so \( (A \setminus C)\setminus (A \setminus B) = \{1\} \). Therefore, \( A \setminus (B \setminus C) \ne (A \setminus C)\setminus (A \setminus B) \).


    But we can show that the sets \(A \setminus (B \setminus C)\) and \((A \setminus C)\setminus (A \setminus B)\) are disjoint. Let \(x \in (A \setminus B)\cup (A \cap C)\). Then either \(x \in A \setminus B\), so \(x \in A\) and \(x \notin B\), or \(x \in A \cap C\), so \(x \in A\) and \(x \in C\). We want to show that \(x \notin A \cap (B \setminus C)\), meaning that it is not the case that \(x \in A\), \(x \in B\), and \(x \notin C\). We already have \(x \in A\). Now, if \(x \in A \setminus B\), then \(x \notin B\), which contradicts \(x \in B\). If \(x \in A \cap C\), then \(x \in C\), which contradicts \(x \notin C\). Therefore, \(x \notin A \cap (B \setminus C)\), which shows that the intersection is empty, so the sets are disjoint.


  4. $$(A \setminus B) \cup B = A$$

    False. Let \( A = \{1\}, \; B = \{2\} \). Then \( A \setminus B = \{1\} \), so \( (A \setminus B)\cup B = \{1\} \cup \{2\} = \{1,2\} \). On the other hand, \( A = \{1\} \). Therefore, \( (A \setminus B)\cup B \ne A \).


    However, we can show that \( (A \setminus B)\cup B = A \cup B \). ($\subseteq$) Let \(x \in (A \setminus B)\cup B\). Then either \(x \in A \setminus B\) or \(x \in B\). If \(x \in A \setminus B\), then \(x \in A\), so \(x \in A \cup B\). If \(x \in B\), then clearly \(x \in A \cup B\). Thus, \( (A \setminus B)\cup B \subseteq A \cup B \). ($\supseteq$) Conversely, let \(x \in A \cup B\). Then either \(x \in A\) or \(x \in B\). If \(x \in B\), then \(x \in (A \setminus B)\cup B\). If \(x \in A\) and \(x \notin B\), then \(x \in A \setminus B\), so again \(x \in (A \setminus B)\cup B\). Hence, \( A \cup B \subseteq (A \setminus B)\cup B \). Therefore, \( (A \setminus B)\cup B = A \cup B \).


  5. $$(A \setminus B) \setminus C = A \setminus (B \cup C)$$

    True. ($\subseteq$) Let \(x \in (A \setminus B)\setminus C\). Then \(x \in A\), \(x \notin B\), and \(x \notin C\). We want to show that \(x \in A \setminus (B \cup C)\), meaning that \(x \in A\) and \(x \notin (B \cup C)\). We already have \(x \in A\). Now, \(x \in (B \cup C)\) would require that \(x \in B\) or \(x \in C\), but we know that \(x \notin B\) and \(x \notin C\). So \(x \notin (B \cup C)\). Therefore, \(x \in A \setminus (B \cup C)\), which proves that \((A \setminus B)\setminus C \subseteq A \setminus (B \cup C)\). ($\supseteq$) Conversely, let \(x \in A \setminus (B \cup C)\). Then \(x \in A\) and \(x \notin (B \cup C)\). We want to show that \(x \in (A \setminus B)\setminus C\), meaning that \(x \in A\), \(x \notin B\), and \(x \notin C\). We already have \(x \in A\). Since \(x \notin (B \cup C)\), it follows that \(x \notin B\) and \(x \notin C\). Hence \(x \in (A \setminus B)\setminus C\), which proves that \(A \setminus (B \cup C) \subseteq (A \setminus B)\setminus C\). Therefore, $(A \setminus B) \setminus C = A \setminus (B \cup C)$.


  6. $$A \cup (B \setminus C) = (A \cup B) \setminus C$$

    False. Let \( A = \{1\}, \; B = \{2\}, \; C = \{1\} \). Then \( B \setminus C = \{2\} \), so \( A \cup (B \setminus C) = \{1,2\} \). On the other hand, \( A \cup B = \{1,2\} \), so \( (A \cup B)\setminus C = \{2\} \). Therefore, \( A \cup (B \setminus C) \ne (A \cup B)\setminus C \).


    However, we can show that \( (A \cup B)\setminus C \subseteq A \cup (B \setminus C) \). Let \(x \in (A \cup B)\setminus C\). Then \(x \in A \cup B\) and \(x \notin C\). We want to show \(x \in A \cup (B \setminus C)\), meaning that \(x \in A\) or \(x \in (B \setminus C)\). We already have \(x \in A \cup B\), so either \(x \in A\) or \(x \in B\). If \(x \in A\), then \(x \in A \cup (B \setminus C)\). If \(x \in B\) and \(x \notin C\), then \(x \in B \setminus C\), so again \(x \in A \cup (B \setminus C)\). Therefore, \(x \in A \cup (B \setminus C)\), which proves that \( (A \cup B)\setminus C \subseteq A \cup (B \setminus C)\).


  7. $$A \cap B = A \setminus (A \setminus B)$$

    True. ($\subseteq$) Let \(x \in A \cap B\). Then \(x \in A\) and \(x \in B\). We want to show that \(x \in A \setminus (A \setminus B)\), meaning that \(x \in A\) and \(x \notin (A \setminus B)\). We already have \(x \in A\). Now, \(x \in (A \setminus B)\) would require that \(x \in A\) and \(x \notin B\), but we know that \(x \in B\). So \(x \notin (A \setminus B)\). Therefore, \(x \in A \setminus (A \setminus B)\), which proves that \(A \cap B \subseteq A \setminus (A \setminus B)\). ($\supseteq$) Conversely, let \(x \in A \setminus (A \setminus B)\). Then \(x \in A\) and \(x \notin (A \setminus B)\). We want to show that \(x \in A \cap B\), meaning that \(x \in A\) and \(x \in B\). We already have \(x \in A\). Now, if \(x \notin B\), then \(x \in A \setminus B\), which contradicts \(x \notin (A \setminus B)\). Hence \(x \in B\). Therefore, \(x \in A \cap B\), which proves that \(A \setminus (A \setminus B) \subseteq A \cap B\). Therefore, $A \cap B = A \setminus (A \setminus B)$.


  8. $$A \setminus B = A \setminus (A \cap B)$$

    True. ($\subseteq$) Let \(x \in A \setminus B\). Then \(x \in A\) and \(x \notin B\). We want to show that \(x \in A \setminus (A \cap B)\), meaning that \(x \in A\) and \(x \notin (A \cap B)\). We already have \(x \in A\). Now, \(x \in (A \cap B)\) would require that \(x \in A\) and \(x \in B\), but we know that \(x \notin B\). So \(x \notin (A \cap B)\). Therefore, \(x \in A \setminus (A \cap B)\), which proves that \(A \setminus B \subseteq A \setminus (A \cap B)\). ($\supseteq$) Conversely, let \(x \in A \setminus (A \cap B)\). Then \(x \in A\) and \(x \notin (A \cap B)\). We want to show that \(x \in A \setminus B\), meaning that \(x \in A\) and \(x \notin B\). We already have \(x \in A\). Now, if \(x \in B\), then \(x \in A \cap B\), which contradicts \(x \notin (A \cap B)\). Hence \(x \notin B\). Therefore, \(x \in A \setminus B\), which proves that \(A \setminus (A \cap B) \subseteq A \setminus B\). Therefore, $A \setminus B = A \setminus (A \cap B)$.


  9. $$A \cap (B \setminus C) = (A \cap B) \setminus C$$

    True. ($\subseteq$) Let \(x \in A \cap (B \setminus C)\). Then \(x \in A\), \(x \in B\), and \(x \notin C\). We want to show that \(x \in (A \cap B)\setminus C\), meaning that \(x \in A \cap B\) and \(x \notin C\). We already have \(x \in A\) and \(x \in B\), so \(x \in A \cap B\). We also have \(x \notin C\). Therefore, \(x \in (A \cap B)\setminus C\), which proves that \(A \cap (B \setminus C) \subseteq (A \cap B)\setminus C\). ($\supseteq$) Conversely, let \(x \in (A \cap B)\setminus C\). Then \(x \in A \cap B\) and \(x \notin C\). So \(x \in A\), \(x \in B\), and \(x \notin C\). We want to show that \(x \in A \cap (B \setminus C)\), meaning that \(x \in A\) and \(x \in (B \setminus C)\). Since \(x \in B\) and \(x \notin C\), we have \(x \in B \setminus C\). Together with \(x \in A\), this gives \(x \in A \cap (B \setminus C)\). Therefore, \( (A \cap B)\setminus C \subseteq A \cap (B \setminus C)\). Hence, \(A \cap (B \setminus C) = (A \cap B)\setminus C\).


  10. $$A \setminus (B \setminus C) = (A \setminus B) \cup (A \cap C)$$

    True. ($\subseteq$) Let \(x \in A \setminus (B \setminus C)\). Then \(x \in A\) and \(x \notin (B \setminus C)\). We want to show that \(x \in (A \setminus B)\cup (A \cap C)\). We already have \(x \in A\). Now, \(x \notin (B \setminus C)\) means it is not the case that \(x \in B\) and \(x \notin C\), so either \(x \notin B\) or \(x \in C\). If \(x \notin B\), then \(x \in A \setminus B\). If \(x \in C\), then \(x \in A \cap C\). Hence in both cases \(x \in (A \setminus B)\cup (A \cap C)\), which proves \(A \setminus (B \setminus C) \subseteq (A \setminus B)\cup (A \cap C)\). ($\supseteq$) Conversely, let \(x \in (A \setminus B)\cup (A \cap C)\). Then either \(x \in A \setminus B\) or \(x \in A \cap C\). If \(x \in A \setminus B\), then \(x \in A\) and \(x \notin B\), so in particular \(x \notin (B \setminus C)\). If \(x \in A \cap C\), then \(x \in A\) and \(x \in C\), so \(x\) cannot be in \(B \setminus C\) (since that would require \(x \notin C\)). Thus in both cases \(x \in A \setminus (B \setminus C)\), which proves \((A \setminus B)\cup (A \cap C) \subseteq A \setminus (B \setminus C)\). Therefore, \(A \setminus (B \setminus C) = (A \setminus B)\cup (A \cap C)\).


  11. $$(A \cup B)\setminus C = (A \setminus C)\cap (B \setminus C)$$

    False. Let \( A = \{1\}, \; B = \{2\}, \; C = \{1\} \). Then \( A \cup B = \{1,2\} \), so \( (A \cup B)\setminus C = \{2\} \). On the other hand, \( A \setminus C = \varnothing \) and \( B \setminus C = \{2\} \), so \( (A \setminus C)\cap (B \setminus C) = \varnothing \). Therefore, \( (A \cup B)\setminus C \ne (A \setminus C)\cap (B \setminus C) \).


    But we can show that \((A \cup B)\setminus C = (A \setminus C)\cup (B \setminus C)\). ($\subseteq$) Let \(x \in (A \cup B)\setminus C\). Then \(x \in A \cup B\) and \(x \notin C\). We want to show that \(x \in (A \setminus C)\cup (B \setminus C)\). Since \(x \in A \cup B\), either \(x \in A\) or \(x \in B\). If \(x \in A\), then \(x \in A \setminus C\) because \(x \notin C\). If \(x \in B\), then \(x \in B \setminus C\) because \(x \notin C\). Hence \(x \in (A \setminus C)\cup (B \setminus C)\), which proves that \((A \cup B)\setminus C \subseteq (A \setminus C)\cup (B \setminus C)\). ($\supseteq$) Conversely, let \(x \in (A \setminus C)\cup (B \setminus C)\). Then either \(x \in A \setminus C\) or \(x \in B \setminus C\). If \(x \in A \setminus C\), then \(x \in A\) and \(x \notin C\), so \(x \in A \cup B\) and \(x \notin C\), hence \(x \in (A \cup B)\setminus C\). If \(x \in B \setminus C\), then \(x \in B\) and \(x \notin C\), so again \(x \in A \cup B\) and \(x \notin C\), hence \(x \in (A \cup B)\setminus C\). Therefore, \((A \setminus C)\cup (B \setminus C) \subseteq (A \cup B)\setminus C\).


  12. $$(A \cap B)\setminus C = (A \setminus C)\cup (B \setminus C)$$

    False. Let \( A = \{1,2\}, \; B = \{2,3\}, \; C = \{2\} \). Then \( A \cap B = \{2\} \), so \( (A \cap B)\setminus C = \varnothing \). On the other hand, \( A \setminus C = \{1\} \) and \( B \setminus C = \{3\} \), so \( (A \setminus C)\cup (B \setminus C) = \{1,3\} \). Therefore, \( (A \cap B)\setminus C \ne (A \setminus C)\cup (B \setminus C) \).


    But we can show that \((A \cap B)\setminus C = (A \setminus C)\cap (B \setminus C)\). ($\subseteq$) Let \(x \in (A \cap B)\setminus C\). Then \(x \in A \cap B\) and \(x \notin C\). We want to show that \(x \in (A \setminus C)\cap (B \setminus C)\). Since \(x \in A \cap B\), we have \(x \in A\) and \(x \in B\). Together with \(x \notin C\), this gives \(x \in A \setminus C\) and \(x \in B \setminus C\). Hence \(x \in (A \setminus C)\cap (B \setminus C)\), which proves that \((A \cap B)\setminus C \subseteq (A \setminus C)\cap (B \setminus C)\). ($\supseteq$) Conversely, let \(x \in (A \setminus C)\cap (B \setminus C)\). Then \(x \in A \setminus C\) and \(x \in B \setminus C\). So \(x \in A\), \(x \in B\), and \(x \notin C\). Hence \(x \in A \cap B\) and \(x \notin C\), so \(x \in (A \cap B)\setminus C\). Therefore, \((A \setminus C)\cap (B \setminus C) \subseteq (A \cap B)\setminus C\).


  13. $$A \setminus (B \cup C) = (A \setminus B)\cap (A \setminus C)$$

    True. ($\subseteq$) Let \(x \in A \setminus (B \cup C)\). Then \(x \in A\) and \(x \notin (B \cup C)\). We want to show that \(x \in (A \setminus B)\cap (A \setminus C)\), meaning that \(x \in A \setminus B\) and \(x \in A \setminus C\). We already have \(x \in A\). Now, \(x \notin (B \cup C)\) means \(x \notin B\) and \(x \notin C\). Hence \(x \in A \setminus B\) and \(x \in A \setminus C\). Therefore, \(x \in (A \setminus B)\cap (A \setminus C)\), which proves that \(A \setminus (B \cup C) \subseteq (A \setminus B)\cap (A \setminus C)\). ($\supseteq$) Conversely, let \(x \in (A \setminus B)\cap (A \setminus C)\). Then \(x \in A \setminus B\) and \(x \in A \setminus C\). So \(x \in A\), \(x \notin B\), and \(x \notin C\). Hence \(x \notin (B \cup C)\), so \(x \in A \setminus (B \cup C)\). Therefore, \((A \setminus B)\cap (A \setminus C) \subseteq A \setminus (B \cup C)\). Hence, $A \setminus (B \cup C) = (A \setminus B)\cap (A \setminus C)$ (DeMorgan's Laws).


  14. $$A \setminus (B \cap C) = (A \setminus B)\cup (A \setminus C)$$

    True. ($\subseteq$) Let \(x \in A \setminus (B \cap C)\). Then \(x \in A\) and \(x \notin (B \cap C)\). We want to show that \(x \in (A \setminus B)\cup (A \setminus C)\), meaning that \(x \in A \setminus B\) or \(x \in A \setminus C\). We already have \(x \in A\). Now, \(x \notin (B \cap C)\) means that it is not the case that \(x \in B\) and \(x \in C\), so \(x \notin B\) or \(x \notin C\). If \(x \notin B\), then \(x \in A \setminus B\). If \(x \notin C\), then \(x \in A \setminus C\). Hence \(x \in (A \setminus B)\cup (A \setminus C)\), which proves that \(A \setminus (B \cap C) \subseteq (A \setminus B)\cup (A \setminus C)\). ($\supseteq$) Conversely, let \(x \in (A \setminus B)\cup (A \setminus C)\). Then either \(x \in A \setminus B\) or \(x \in A \setminus C\). If \(x \in A \setminus B\), then \(x \in A\) and \(x \notin B\), so \(x \notin (B \cap C)\), hence \(x \in A \setminus (B \cap C)\). If \(x \in A \setminus C\), then \(x \in A\) and \(x \notin C\), so again \(x \notin (B \cap C)\), hence \(x \in A \setminus (B \cap C)\). Therefore, \((A \setminus B)\cup (A \setminus C) \subseteq A \setminus (B \cap C)\). Hence, $A \setminus (B \cap C) = (A \setminus B)\cup (A \setminus C)$ (DeMorgan's Laws).


  15. $$(A \cap B) \cup C = (A \cup B) \cap C$$

    False. Let \( A = \{1,2\}, \; B = \{2,3\}, \; C = \{3\} \). Then \( A \cap B = \{2\} \), so \( (A \cap B)\cup C = \{2,3\} \). On the other hand, \( A \cup B = \{1,2,3\} \), so \( (A \cup B)\cap C = \{3\} \). Therefore, \( (A \cap B)\cup C \ne (A \cup B)\cap C \).


    But we can show that \((A \cap B)\cup C = (A \cup C)\cap (B \cup C)\). ($\subseteq$) Let \(x \in (A \cap B)\cup C\). Then either \(x \in A \cap B\) or \(x \in C\). If \(x \in C\), then clearly \(x \in A \cup C\) and \(x \in B \cup C\). If \(x \in A \cap B\), then \(x \in A\) and \(x \in B\), so again \(x \in A \cup C\) and \(x \in B \cup C\). Hence \(x \in (A \cup C)\cap (B \cup C)\), which proves that \((A \cap B)\cup C \subseteq (A \cup C)\cap (B \cup C)\). ($\supseteq$) Conversely, let \(x \in (A \cup C)\cap (B \cup C)\). Then \(x \in A \cup C\) and \(x \in B \cup C\). If \(x \in C\), then \(x \in (A \cap B)\cup C\). If \(x \notin C\), then from \(x \in A \cup C\) we get \(x \in A\), and from \(x \in B \cup C\) we get \(x \in B\). Hence \(x \in A \cap B\), so \(x \in (A \cap B)\cup C\). Therefore, \((A \cup C)\cap (B \cup C) \subseteq (A \cap B)\cup C\). Hence, \((A \cap B)\cup C = (A \cup C)\cap (B \cup C)\).


  16. $$(A \cup B) \cap C = (A \cap B) \cup C$$

    False. Let \( A = \{1,2\}, \; B = \{2,3\}, \; C = \{3\} \). Then \( A \cup B = \{1,2,3\} \), so \( (A \cup B)\cap C = \{3\} \). On the other hand, \( A \cap B = \{2\} \), so \( (A \cap B)\cup C = \{2,3\} \). Therefore, \( (A \cap B)\cup C \ne (A \cup B)\cap C \).


    But we can show that \((A \cup B)\cap C = (A \cap C)\cup (B \cap C)\). ($\subseteq$) Let \(x \in (A \cup B)\cap C\). Then \(x \in A \cup B\) and \(x \in C\). We want to show that \(x \in (A \cap C)\cup (B \cap C)\). Since \(x \in A \cup B\), either \(x \in A\) or \(x \in B\). If \(x \in A\), then \(x \in A \cap C\) because \(x \in C\). If \(x \in B\), then \(x \in B \cap C\) because \(x \in C\). Hence \(x \in (A \cap C)\cup (B \cap C)\), which proves that \((A \cup B)\cap C \subseteq (A \cap C)\cup (B \cap C)\). ($\supseteq$) Conversely, let \(x \in (A \cap C)\cup (B \cap C)\). Then either \(x \in A \cap C\) or \(x \in B \cap C\). If \(x \in A \cap C\), then \(x \in A\) and \(x \in C\), so \(x \in A \cup B\) and \(x \in C\), hence \(x \in (A \cup B)\cap C\). If \(x \in B \cap C\), then \(x \in B\) and \(x \in C\), so again \(x \in A \cup B\) and \(x \in C\), hence \(x \in (A \cup B)\cap C\). Therefore, \((A \cap C)\cup (B \cap C) \subseteq (A \cup B)\cap C\). Hence, \((A \cup B)\cap C = (A \cap C)\cup (B \cap C)\).

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