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🚧 Omitted last few questions of 6.2

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tomit4 2026-07-21 17:25:46 -07:00
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chapter_6/exercises.md
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@ -2665,29 +2665,43 @@ Prove each statement in 39-44.
a. $(A - B) \cup (B - A) \cup (A \cap B) = A \cup B$ a. $(A - B) \cup (B - A) \cup (A \cap B) = A \cup B$
Omitted.
b. The sets $(A - B)$, $(B - A)$, and $(A \cap B)$ are mutually disjoint. b. The sets $(A - B)$, $(B - A)$, and $(A \cap B)$ are mutually disjoint.
Omitted.
40. For every positive integer $n$, if $A$ and $B_1, B_2, B_3, \dots$ are any 40. For every positive integer $n$, if $A$ and $B_1, B_2, B_3, \dots$ are any
sets, then sets, then
$$ A \cap \left(\bigcup_{i = 1}^{n}B_i\right) = \bigcup_{i = 1}^{n}(A \cap B_i) $$ $$ A \cap \left(\bigcup_{i = 1}^{n}B_i\right) = \bigcup_{i = 1}^{n}(A \cap B_i) $$
Omitted.
41. For every positive integer $n$, if $A_1, A_2, A_3, \dots$ and $B$ are any 41. For every positive integer $n$, if $A_1, A_2, A_3, \dots$ and $B$ are any
sets, then sets, then
$$ \bigcap_{i = 1}^{n}(A_i - B) = \left(\bigcup_{i = 1}^{n}A_i\right) - B $$ $$ \bigcap_{i = 1}^{n}(A_i - B) = \left(\bigcup_{i = 1}^{n}A_i\right) - B $$
Omitted.
42. For every positive integer $n$, if $A_1, A_2, A_3, \dots$ and $B$ are any 42. For every positive integer $n$, if $A_1, A_2, A_3, \dots$ and $B$ are any
sets, then sets, then
$$ \bigcap_{i = 1}^{n}(A_i - B) = \left(\bigcap_{i = 1}^{n}A_i\right) - B $$ $$ \bigcap_{i = 1}^{n}(A_i - B) = \left(\bigcap_{i = 1}^{n}A_i\right) - B $$
Omitted.
43. For every positive integer $n$, if $A$ and $B_1, B_2, B_3, \dots$ are any 43. For every positive integer $n$, if $A$ and $B_1, B_2, B_3, \dots$ are any
sets, then sets, then
$$ \bigcup_{i = 1}^{n}(A \times B_i) = A \times \left(\bigcup_{i = 1}^{n}B_i\right) $$ $$ \bigcup_{i = 1}^{n}(A \times B_i) = A \times \left(\bigcup_{i = 1}^{n}B_i\right) $$
Omitted.
44. For every positive integer $n$, if $A$ and $B_1, B_2, B_3, \dots$ are any 44. For every positive integer $n$, if $A$ and $B_1, B_2, B_3, \dots$ are any
sets, then sets, then
$$ \bigcap_{i = 1}^{n}(A \times B_i) = A \times \left(\bigcap_{i = 1}^{n}B_i\right) $$ $$ \bigcap_{i = 1}^{n}(A \times B_i) = A \times \left(\bigcap_{i = 1}^{n}B_i\right) $$
Omitted.