1) Demuestre $A \subseteq B \iff \wp(A) \in \wp(B)$
Demostración
Por definición $A \in \wp(A) \wedge B \in \wp(B)$
Si $A \subseteq B$
$\Rightarrow A \subseteq B \in \wp(B)$
$\Rightarrow A \in \wp(B)$
Sea $A_{k} \subseteq A$
$\Rightarrow A_{k} \in \wp(B)$
$\Rightarrow \wp(A) \subseteq \wp(B)$
Ahora bien,
si $\wp(A) \subseteq \wp(B)$
$\Rightarrow A \in \wp(A) \subseteq \wp(B)$
$\Rightarrow A \in \wp(B)$
$\Rightarrow A\subseteq B$
2) Sean $(A \wedge B) \in \wp(E)$
Demuestre $\wp(A) \cup \wp(B) = \wp(A \cup B) \Rightarrow (A \subseteq B) \vee (B \subseteq A)$
Demostración
Por la contrapositiva
$\neg [(A \subseteq B) \vee (B \subseteq A)] \Rightarrow \neg [\wp(A) \cup \wp(B) = \wp(A \cup B)]$
$\neg [(A \subseteq B) \vee (B \subseteq A)] \Rightarrow [\wp(A) \cup \wp(B) \not = \wp(A \cup B)]$
$(A \not \subseteq B) \wedge (B \not \subseteq A) \Rightarrow [\wp(A) \cup \wp(B) \not = \wp(A \cup B)]$
$\Rightarrow \exists a,b (\{a, b\} \not \in \wp(A) \cup \wp(B) \wedge \{a, b\} \in \wp(A \cup B))$
$\Rightarrow \wp(A) \cup \wp(B) \not = \wp(A \cup B)$
$\therefore \neg [(A \subseteq B) \vee (B \subseteq A)] \Rightarrow \neg [\wp(A) \cup \wp(B) = \wp(A \cup B)] $
$\therefore \wp(A) \cup \wp(B) = \wp(A \cup B) \Rightarrow (A \subseteq B) \vee (B \subseteq A)$
3)Sea $\{E_{i}: i \in I \}$ una familia de subconjuntos de $E$, donde $E, I$ son conjuntos.
Demuestre que:
a-) $\wp(\bigcap_{i \in I} E_{i}) = \bigcap_{i \in I} \wp(E_{i})$
b-) $\bigcup_{i \in I} \wp(E_{i}) \subseteq \wp(\bigcup_{i \in I} E_{i})$ y, en general, la igualdad no se cumple.
Solución
a-)
Sea $A \in \bigcap_{i \in I} \wp(E_{i})$
$\Rightarrow A \in \wp(E_{i}) \forall i \in I$
$\Rightarrow A \subseteq E_{i} \forall i \in I$
$\Rightarrow A \subseteq E_{i} \subseteq \bigcap_{i \in I} E_{i}$
$\Rightarrow A \in \wp(\bigcap_{i \in I} E_{i})$
$\Rightarrow \bigcap_{i \in I} \wp(E_{i}) \subseteq \wp(\bigcap_{i \in I} E_{i})$
Análogamente
Sea $A \in \wp(\bigcup_{i \in I} E_{i})$
$\Rightarrow A \subseteq E_{i} \subseteq \bigcap_{i \in I} E_{i}$
$\Rightarrow A \subseteq E_{i} \forall i \in I$
$\Rightarrow A \in \wp(E_{i}) \forall i \in I$
$\Rightarrow A \in \bigcap_{i \in I} \wp(E_{i})$
$\Rightarrow \wp(\bigcap_{i \in I} E_{i}) \subseteq \bigcap_{i \in I} \wp(E_{i}) $
$\therefore \wp(\bigcap_{i \in I} E_{i}) = \bigcap_{i \in I} \wp(E_{i})$
b-)
b.1-)
Sea $A \in \bigcup_{i \in I} \wp(E_{i})$
$\Rightarrow \exists i (A \in \wp(E_{i}) \wedge A \subseteq E_{i})$
$\Rightarrow A \subseteq \bigcup_{i \in I} E_{i}$
$\Rightarrow A \in \wp(\bigcup_{i \in I} E_{i})$
$\therefore \bigcup_{i \in I} \wp(E_{i}) \subseteq \wp(\bigcup_{i \in I} E_{i})$
b.2-)
En virtud de lo anterior, $\bigcup_{i \in I} \wp(E_{i}) \subseteq \wp(\bigcup_{i \in I} E_{i})$
Entonces por hipótesis,
debe haber al menos un caso en el que $\bigcup_{i \in I} \wp(E_{i}) \not \supseteq \wp(\bigcup_{i \in I} E_{i})$
Considérese el caso en el que $\exists a, b, i, j (a \in E_{i} \wedge a \not \in E_{j} \wedge b \not \in E_{i} \wedge b \in E_{j})$
Para dicho caso, $\{a, b\} \not \in \bigcup_{i \in I} \wp(E_{i}) \wedge \{a, b\} \in \wp(\bigcup_{i \in I} E_{i})$
Nota: $\bigcup_{i \in I} \wp(E_{i}) \supseteq \wp(\bigcup_{i \in I} E_{i}) \iff \exists j \forall i (E_{i} \subseteq E_{j})$
4) Demuestre que $\forall D \in \wp(E), (D \subseteq A) \wedge (D \subseteq B) \Leftrightarrow D \subseteq A \cap B$
Solución
Parta de $D \subseteq A \cap B$
Recuerde que $(A \cap B \subseteq A) \wedge (A \cap B \subseteq B)$
$\Rightarrow (D \subseteq A \cap B \subseteq A) \wedge (D \subseteq A \cap B \subseteq B)$
$\Rightarrow (D \subseteq A) \wedge (D \subseteq B)$
Para $(D \subseteq A) \wedge (D \subset B) \Rightarrow D \subseteq A \cap B$
Por la contrapositiva
$\neg [D \subseteq A \cap B] \Rightarrow \neg [(D \subseteq A) \wedge (D \subseteq B)]$
$D \not \subseteq A \cap B \Rightarrow (D \not \subseteq A) \vee (D \not \subseteq B)$
$D \not \subseteq A \cap B$
$\Rightarrow (D \not \subseteq A \cap B \subseteq A) \wedge (D \not \subseteq A \cap B \subseteq B)$
$\Rightarrow (D \not \subseteq A) \wedge (D \not \subseteq B)$
5) Si $A \subseteq E \wedge B \subseteq E$, $A \subseteq B \Leftrightarrow \complement_{E}B \subseteq \complement_{E}A$
Solución
Sea $A \subset B$
$\Rightarrow \exists x (x \in B: x \not \in A)$
Sea $x_{1} (x_{1} \in B: x_{1} \not \in A)$
$\Rightarrow x_{1} \in \complement_{E}A \wedge x_{1} \not \in \complement_{E}B$
$\Rightarrow \complement_{E}B \subseteq \complement_{E}A$
Para el recíproco
$\complement_{E}B \subseteq \complement_{E}A \Rightarrow A \subseteq B$
$\Rightarrow \exists \omega (\omega \in \complement_{E}A: \omega \not \in \complement_{E}B)$
Sea $\omega_{1} (\omega_{1} \in \complement_{E}A: \omega_{1} \not \in \complement_{E}B)$
$\Rightarrow \omega_{1} \in \complement_{E}(\complement_{E}B) \wedge \omega_{1} \not \in \complement_{E}(\complement_{E}A)$
$\Rightarrow \omega_{1} \in B \wedge \omega_{1} \not \in A$
$\Rightarrow A \subseteq B$
6) Demuestre que $A \subset B \Leftrightarrow A - B = \emptyset$
Solución
Considere $A \subset B \Rightarrow A - B = \emptyset$
Por la contrapositiva
$A - B \not = \emptyset \Rightarrow A \subset B$
Como $A - B \not = \emptyset \Rightarrow x \in A - B$
$\Rightarrow x \in A \wedge x \not \in B$
$\Rightarrow A \subset B$
Considere $ A - B = \emptyset \Rightarrow A \subset B$
Asuma $A \not \subset B$ y $A - B = \emptyset$
$\Rightarrow \exists x (x \in A : x \not \in B)$
$\Rightarrow A - B \not = \emptyset$
$\Rightarrow \Leftarrow$
$\therefore A - B = \emptyset \Rightarrow A \subset B$
7) Con $ A \subseteq E \wedge B \subseteq E$, demuestre que $ A - B = A \cap \complement_{E}B$
Solución
Considere $A - B \not = \emptyset$
Sea $x \in A - B$, es decir $x \in A \wedge x \not \in B$
Como $x \not \in B$
$\Rightarrow x \in \complement_{E}B$
Como $x \in A$
$\Rightarrow x \in A \cap \complement_{E}B$
Considere $A - B = \emptyset$, es decir $A \subset B$
$\Rightarrow \exists \omega (\omega \in B \wedge \omega \not \in A)$
$\Rightarrow \omega \not \in \complement_{E}B \wedge \omega \not \in A$
$\Rightarrow A \cap \complement_{E}B = \emptyset$
8) Muestre que $A - (A - B) = (A \cap B)$
Solución
Observe dos casos
(1) $A - B = \emptyset$
(2) $A - B \not = \emptyset$
Considere (1)
$\Rightarrow \exists Z (Z \not \subseteq A \wedge Z \subseteq B)$
$\Rightarrow Z \not \subseteq A - B \wedge Z \not \subseteq A \cap B$
$\Rightarrow A \not \subseteq A - (A - B)$
Considere (2)
$\Rightarrow \exists Z (Z \subseteq A \wedge Z \not \subseteq B)$
$\Rightarrow Z \subseteq A - B \wedge Z \not \subseteq A \cap B$
$\Rightarrow Z \not \subseteq A - (A - B)$
9) Muestre que $A \cap (B - C) = (A \cap B) - (A \cap C)$
Solución
Sea $Z \subseteq A \cap (B - C)$
$\Rightarrow Z \subseteq A \wedge Z \subseteq B - C$
$\Rightarrow Z \subseteq B \wedge Z \not \subseteq C$
$\Rightarrow Z \subseteq A \cap B \wedge Z \not \subseteq A \cap C$
$ \Rightarrow Z \subseteq (A \cap B) - (A \cap C)$
$ \Rightarrow A \cap (B - C) \subseteq (A \cap B) - (A \cap C)$
Considérese ahora $Z \subseteq (A \cap B) - (A \cap C)$
$\Rightarrow Z \subseteq (A \cap B) \wedge Z \not \subseteq (A \cap C)$
$\Rightarrow Z \subseteq A \wedge Z \subseteq B \wedge Z \not \subseteq C$
$ \Rightarrow Z \subseteq B - C$
$\Rightarrow \subseteq A \cap (B - C)$
$ \Rightarrow (A \cap B) - ( A \cap C) \subseteq A \cap (B - C)$
$\therefore A \cap (B - C) = (A \cap B) - (A \cap C)$
10) Halle $\complement_{E}[((\complement_{E}A) \cup B) \cap (C \cup \complement_{E}D)]$
Solución
Por De Morgan
$= \complement_{E}((\complement_{E}A) \cup) \cup \complement_{E}(C \cup \complement_{E}D)$
Por De Morgan
$= (\complement_{E}(\complement_{E}A) \cap \complement_{E}B) \cup (\complement_{E}C \cap \complement_{E}(\complement_{E}D))$
$= (A \cap \complement_{E}B) \cup (\complement_{E}C \cap D)$
$= (A - B) \cup (D - C)$
11) Sean $ A, B \in \wp(E)$
Demuestre que $A \Delta B = (A \cup B) - (A \cap B)$
Solución
$(A \cup B) - (A \cap B)$
$\Rightarrow (A \cup B) \cap \complement_{E}(A \cap B)$
Por De Morgan
$\Rightarrow (A \cup B) \cap (\complement_{E}A \cup \complement_{E}B)$
Por ley distributiva
$\Rightarrow [(A \cup B) \cap \complement_{E}A] \cup [(A \cup B) \cap \complement_{E}B]$
Por ley distributiva
$\Rightarrow (A \cap \complement_{E}A) \cup (B \cap \complement_{E}A) \cup (A \cap \complement_{E}B) \cup (B \cap \complement_{E}B)$
$\emptyset \cup (B \cap \complement_{E}A) \cup (A \cap \complement_{E}B) \cup \emptyset$
$\Rightarrow (B \cap \complement_{E}A) \cup (A \cap \complement_{E}B)$
$\Rightarrow (B - A) \cup (A - B)$
$\Rightarrow A \Delta B$
12) Muestre que $(A \Delta B) \cap C = (A \cap C) \Delta (B \cap C)$
Solución
$\Rightarrow ((A - B) \cup (B - A)) \cap C$
Por ley distributiva
$\Rightarrow (C \cap (A - B)) \cup ( C \cap (B - A))$
$\Rightarrow [(C \cap A) - (C \cap B)] \cup [(C \cap B) - (C \cap A)]$
$\Rightarrow (A \cap C) \Delta (B \cap C)$
13) Muestre que $A \times (B \cup C) = (A \times B) \cup (A \times C) $
Solución
Tome $A \times (B \cup C)$
Sea $ a \in A, b \in B, c \in C$
Observe $b, c \in B \cup C$
$\Rightarrow \forall k \in A \times (B \cup C) [k = (a, b) \veebar (a, c)]$
Ahora tome $(A \times B) \cup (A \times C)$
$A \times B = \{(a, b)\} \wedge A \times C = \{a, c\}$
$\Rightarrow \forall k \in (A \times B) \cup (A \times C) [k = (a, b) \veebar (a, c)]$
$\therefore A \times (B \cup C) = (A \times B) \cup (A \times C)$
14) Sea $E$ un conjunto y sean $A, B, C$ en $\wp(E)$.
Demuestre que $(A \Delta B) - (A \Delta C) = ((A \cap C) - B) \cup (B - (A \cup C))$
Solución
Considere $(A \Delta B) - (A \Delta C)$
$\Rightarrow [(A - B) \cup (B - A)] - [(A - C) \cup (C - A)]$
Por De Morgan, involución y De Morgan
$\Rightarrow [(A \cap \complement_{E}B) \cup (B \cap \complement_{E}A)] \cap [(\complement_{E}A \cup C) \cap (\complement_{E}C \cup A)]$
Por distributiva
$\Rightarrow [(\complement_{E}A \cup C) \cap (\complement_{E}C \cup A)] \cap A \cap \complement_{E}B] \cup [(\complement_{E}A \cup C) \cap (\complement_{E}A \cup C) \cap (\complement_{E}C \cup A) \cap B \cap \complement_{E}A]$
Por distributiva
$\Rightarrow [((A \cap \complement_{E}A) \cup (C \cap A)) \cap (\complement_{E}C \cup A) \cap \complement_{E}B] \cup [((\complement_{E}A \cap A) \cup (\complement_{E}A \cap \complement_{E}C)) \cap (\complement_{E}A \cup C) \cap B]$
$\Rightarrow [(C \cap A) \cap (\complement_{E}C \cup A) \cap \complement_{E}B] \cup [(\complement_{E}A \cap \complement_{E}C) \cap (\complement_{E}A \cup C) \cap B]$
Por distributiva
$\Rightarrow [((C \cap \complement_{E}C) \cup (C \cap A)) \cap A \cap \complement_{E}B] \cup [(\complement_{E}C \cap C) \cup (\complement_{E}C \cap \complement_{E}A) \cup \complement_{E}A \cap B]$
$\Rightarrow [(C \cap A) \cap A \cap \complement_{E}B] \cup [(\complement_{E}C \cap \complement_{E}A) \cap \complement_{E}A \cap B]$
$\Rightarrow [(C \cap A) - B] \cup [( \complement_{E}C \cap \complement_{E}A) \cap B]$
Por De Morgan
$\Rightarrow [(A \cap C) - B] \cup [B \cap \complement_{E}(A \cup C)]$
$\Rightarrow [(A \cap C) - B] \cup [B - (A \cup C)]$
15) Demuestre que para cualesquiera tres conjuntos, no vacíos, $A, B$ y $C$, se cumple $A \subseteq B \Rightarrow [C - A \not = \emptyset \veebar A \cap B = A \cup C]$
Solución
Suponga $Z \subseteq A$
$\Rightarrow Z \subseteq A \subseteq B$
$\Rightarrow Z \subseteq B$
$\Rightarrow Z \subseteq A \cap B$
Observe dos casos, $C - A \not = \emptyset \veebar C - A = \emptyset$
Si $C - A = \emptyset$
$\Rightarrow C \subset A$
$\Rightarrow Z \subseteq C \vee C \subseteq Z$
$\Rightarrow A \cap B = A \cap C$
Si $C - A \not = \emptyset$
$\Rightarrow C \supset A$
$\Rightarrow A \cap B \not = A \cup C$
$\therefore A \subseteq B \Rightarrow [C - A \not = \emptyset \veebar A \cap B = A \cup C]$
16) Si $A, B$ y $D$ están en $\wp(E)$, demuestre que $(A \Delta B) - D = (A \cup D) \Delta (B \cup D)$
Solución
Parta de $(A \cup D) \Delta (B \cup D)$
$\Rightarrow [(A \cup D) - (B \cup D)] \cup [(B \cup D) - (A \cup D)]$
$\Rightarrow [(A \cup D) \cap \complement_{E}(B \cup D)] \cup [(B \cup D) \cap \complement_{E}(A \cup D)] $
Por De Morgan
$ \Rightarrow [(A \cup D) \cap (\complement_{E}B \cap \complement_{E}D)] \cup [(B \cup D) \cap (\complement_{E}A \cap \complement_{E}D)]$
Por distributiva
$\Rightarrow [((\complement_{E}D \cap D) \cup (\complement_{E}D \cap A)) \cap \complement_{E}B] \cup [((\complement_{E}D \cap D) \cup (\complement_{E}D \cap B)) \cap \complement_{E}A]$
$\Rightarrow [\complement_{E}D \cap A \cap \complement_{E}B] \cup [\complement_{E}D \cap B \cap \complement_{E}A]$
Por distributiva
$\Rightarrow \complement_{E}D \cap [(A - B) \cup (B - A)]$
$\Rightarrow A \Delta B -D$
17) Sean $A, B$ y $C$ tres subconjuntos, no vacíos, de un conjunto E.
Muestre que si $(A \times B) \cup (B \times A) = C \times C \Rightarrow A = B = C$
Solución
Asuma que $\exists a, b, c (a \in A, b \in B, c \in C : a \not = b \not = c)$
$\Rightarrow A \times B = \{(a, b)\} \wedge B \times A = \{(b, a)\} \wedge C \times C = \{(c, c)\}$
$\Rightarrow (A \times B) \cup (B \times A) = \{(a, b), (b, a)\}$
Por hipótesis
$\{(a, b), (b, a)\} = \{(c, c)\}$
Por definición de producto cartesiano
$\Rightarrow a = b = c$
$ \Rightarrow \Leftarrow$
$\Rightarrow \forall a, b, c (a \in A, b \in B, c \in C : a = b = c)$
$\therefore (A \times B) \cup (B \times A) = C \times C \Rightarrow A = B = C$
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