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Considere el conjunto universal como $ U = \\var {univ} $ y sean $ A $ y $ B $ subconjuntos de $ U $
\ntales que $ A = \\var {set1} $ y $ B = \\var {set2} $.
\nEl conjunto universal para el producto cartesiano $ A \\times B $ es $ U \\times U $.
\nListar los elementos de los siguientes conjuntos.
\nIndicaciones.
\n$ A ^ c = UA $, es el conjunto de todos los elementos de $ U $ y no están en $ A $, así $ A ^ c = \\ var {univ - set1} $.
\n$ B ^ c = UB $, es el conjunto de todos los elementos de $ U $ y no están en $ B $, así $ B ^ c = \\ var {univ - set1} $.
\n$ A ^ c \\ cap B ^ c $, es el conjunto de todos los elementos que se encuentran en $ A ^ c $ y $ B ^ c $.
\nEsto es equivalente al conjunto de todos los elementos ni en $ A $ ni en $ B $, es decir, $ \\ var {(univ-set1) y (univ-set2)} $.
\n$ A \\ cap B $ Es el conjunto de todos los elementos presentes en $ A $ y $ B $, es decir, $ \\ var {set1 y set2} $.
\nAsí que $ (A ^ c \\ cap B ^ c) \\ times (A \\ cap B) $ es el conjunto de todos los pares $ (x, y) $, donde $ x $ está en $ A ^ c \\ cap B ^ c $, y $ y $ está en $ A \\ cap B $.
\n$ (U \\ times A) ^ c $ es el conjunto de todos los pares $ (x, y) $ que se encuentra en $ U \\ times U $ que no están en $ U \\ times A $. Ya que $ U $ es el conjunto universal, esto es equivalente a $ U \\ times (A ^ c) $, el producto de $ U $ con el conjunto de elementos no están en $ A $.
\nDel mismo modo, $ (U \\ times B) ^ c $ es equivalente a $ U \\ times (B ^ c) $.
\nOtra vez, $ U $ es el conjunto universal, $ (U \\ times A) ^ c \\ cap (U \\ times B) ^ c = U \\ times (A ^ c \\ cap B ^ c) $.
\nPor un argumento similar, $ (A \\ times U) ^ c \\ cap (B \\ times U) ^ c = (A ^ c \\ cap B ^ c) \\ times U $.
\nAsí, $ (U \\ times A) ^ c \\ cap (U \\ times B) ^ c \\ cap (A \\ times U) ^ c \\ cap (B \\ times U) ^ c $ es equivalente a $ (A ^ c \\ cap B ^ c) \\ times (A ^ c \\ cap B ^ c) $. Esto es, el conjunto de todos los pares que no están en $ A $ o $ B $.
\n$ AB $ es el conjunto de elementos que se encuentran en $ A $ pero no $ B $, es decir, $ \\ var {set1-set2} $.
\n$ (A \\ cup B) ^ c $ es el conjunto de elementos $ U - (A \\ cup B) $, los problemas están en $ U $ y no está $ A \\ cup B $, así $ (A \\ cup B ) ^ c = \\ var {univ- (set1 o set2)} $.
\n$ [(A \\ cup B) \\ times U] ^ c $ es equivalente a $ (A \\ cup B) ^ c \\ times U $.
\nAsí $ [(A \\ cup B) \\ times U] ^ c \\ cap [U \\ times (A \\ cap B)] = (A \\ cup B) ^ c \\ times (A \\ cap B) $.
\n$ A ^ cB $ es el conjunto de todos los elementos que se encuentran en $ A ^ c $ pero no $ B $. Eso es equivalente al conjunto de elementos que no están en $ A $ ni en $ B $, es decir, $ (A \\ cup B) ^ c = \\ var {univ- (set1 o set2)} $.
\nDel mismo modo, $ B ^ c - A = (B \\ cup A) ^ c = (A \\ cup B) ^ c = \\ var {univ- (set1 o set2)} $.
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