- #1
Someone2841
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Below is a attempt at a proof of a distributive property of union/intersections of sets. A critique would be very much appreciated. Thanks in advance!
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For sets A, B, and C, the formula,
##A \cup (B \cap C) = (A \cup B) \cap (A \cup C)##
is true
Two sets are equal if they contain the same elements. Let x denote an element of set ##A \cup (B \cap C)## and y an element of ##(A \cup B) \cap (A \cup C)##. Where X and Y are sets, ##X \subset Y## and ##e \in Y##, define the function ##f_X(e)## as
##f_X(e) =
\begin{cases}
1, & e \in X \\
0, & e \notin X
\end{cases}##
We want to show that ##\forall e \in A \cup B \cup C, f_{A \cup (B \cap C)}(e) = f_{(A \cup B) \cap (A \cup C)}(e)##, since this would imply ##e \in A \cup (B \cap C) \iff e \in (A \cup B) \cap (A \cup C)##. First, two basic and easily verifiable properties of ##f_X(e)## are:##\begin{alignat}{0}
f_{X \cap Y}(e) = f_X(e)f_Y(e) & \text{(1) - since e must be in X and Y} \\
f_{X \cup Y}(e) = f_X(e) + f_Y(e) - f_X(e)f_Y(e) & \text{(2) - since e may be in X or Y}
\end{alignat}##
And now:
##f_{A \cup (B \cap C)}(e)##
##
\begin{alignat}{0}
=&f_A(e) + f_{B \cap C}(e) - f_A(e)f_{B \cap C}(e) & \text{(1)}\\
=&f_A(e) + f_B(e) f_C(e) - f_A(e)f_B(e)f_C(e) & \text{(0)}\\
=&f_{(A \cup B) \cap (A \cup C)}& \text{(1)}
\end{alignat}
##
And therefore ##e \in A \cup (B \cap C) \iff e \in (A \cup B) \cap (A \cup C)## and ##A \cup (B \cap C) = (A \cup B) \cap (A \cup C) \text{ } \Box##
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P.S.: This is not a homework assignment; however, it is very much like a homework assignment. Would this type of problem go better in the homework section?
-----------------
For sets A, B, and C, the formula,
##A \cup (B \cap C) = (A \cup B) \cap (A \cup C)##
is true
Two sets are equal if they contain the same elements. Let x denote an element of set ##A \cup (B \cap C)## and y an element of ##(A \cup B) \cap (A \cup C)##. Where X and Y are sets, ##X \subset Y## and ##e \in Y##, define the function ##f_X(e)## as
##f_X(e) =
\begin{cases}
1, & e \in X \\
0, & e \notin X
\end{cases}##
We want to show that ##\forall e \in A \cup B \cup C, f_{A \cup (B \cap C)}(e) = f_{(A \cup B) \cap (A \cup C)}(e)##, since this would imply ##e \in A \cup (B \cap C) \iff e \in (A \cup B) \cap (A \cup C)##. First, two basic and easily verifiable properties of ##f_X(e)## are:##\begin{alignat}{0}
f_{X \cap Y}(e) = f_X(e)f_Y(e) & \text{(1) - since e must be in X and Y} \\
f_{X \cup Y}(e) = f_X(e) + f_Y(e) - f_X(e)f_Y(e) & \text{(2) - since e may be in X or Y}
\end{alignat}##
And now:
##f_{A \cup (B \cap C)}(e)##
##
\begin{alignat}{0}
=&f_A(e) + f_{B \cap C}(e) - f_A(e)f_{B \cap C}(e) & \text{(1)}\\
=&f_A(e) + f_B(e) f_C(e) - f_A(e)f_B(e)f_C(e) & \text{(0)}\\
=&f_{(A \cup B) \cap (A \cup C)}& \text{(1)}
\end{alignat}
##
And therefore ##e \in A \cup (B \cap C) \iff e \in (A \cup B) \cap (A \cup C)## and ##A \cup (B \cap C) = (A \cup B) \cap (A \cup C) \text{ } \Box##
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P.S.: This is not a homework assignment; however, it is very much like a homework assignment. Would this type of problem go better in the homework section?