- #1
evinda
Gold Member
MHB
- 3,836
- 0
Hello! (Wave)
Let $U$ be a set and $A,B$ subsets of $U$.
I want to prove the following identities:
That's what I have tried:
Let $U$ be a set and $A,B$ subsets of $U$.
I want to prove the following identities:
- $A \cap A^c= \varnothing, A \cup A^c=U$
- $(A^c)^c=A$
- $(A \cap B)^c=A^c \cup B^c$
- $(A \cup B)^c=A^c \cap B^c$
- $A \setminus B=A \cap B^c$
That's what I have tried:
-
Let $x \in A \cap A^c$.
Then $x \in A \wedge x \in A^c \leftrightarrow x \in A \wedge x \in U \setminus A \leftrightarrow x \in A \wedge (x \in U \wedge x \notin A)$, that is a contradiction.
So, there is no element $x$, such that $x \in A \cap A^c$, therefore $A \cap A^c=\varnothing$.
$$$$- Let $x \in A \cup A^c$.
Then $x \in A \lor x \in A^c \leftrightarrow x \in A \lor x \in U \setminus A \leftrightarrow x \in A \lor (x \in U \wedge x \notin A)$
How can I continue?
$$$$ - Let $x \in (A^c)^c \leftrightarrow x \in U \setminus A^c \leftrightarrow x \in U \wedge x \notin A^c \leftrightarrow x \in A \cup A^c \wedge x \notin A^c \leftrightarrow (x \in A \lor x \in A^c) \wedge x \notin A^c \leftrightarrow x \in A$
$$$$ - Let $x \in (A \cap B)^c \leftrightarrow x \in U \setminus A \cap B \leftrightarrow x \in U \wedge x \notin A \cap B \leftrightarrow x \in U \wedge ((x \in A \wedge x \notin B) \lor (x \notin A \wedge x \in B)) \leftrightarrow (x \in U \wedge (x \in A \wedge x \notin B)) \lor (x \in U \wedge (x \notin A \wedge x \in B))$
How can I continue?
$$$$ - Let $x \in (A \cup B)^c \leftrightarrow x \in U \setminus (A \cup B) \leftrightarrow x \in U \wedge x \notin A \cup B \leftrightarrow x \in U \wedge (x \notin A \wedge x \notin B) \leftrightarrow (x \in U \wedge x \notin A) \wedge (x \in U \wedge x \notin B) \leftrightarrow x \in A^c \wedge x \in B^c \leftrightarrow x \in A^c \cap B^c$
$$$$ - Let $x \in A \setminus B \leftrightarrow x \in A \wedge x \notin B \leftrightarrow x \in A \wedge x \in U \setminus B \leftrightarrow x \in A \wedge x \in B^c \leftrightarrow x \in A \cap B^c$