Prove that the following statements are logically equivalent:
- $A \subseteq B$
- $A \cap B = A$
- $A \cup B = B$
- $B^c \subseteq A^c$
Here is what I have so far. I am not sure how much of it is correct or if there is an easier way to go about it.
Assume $A \subseteq B$ to prove $A \cap B = A$:
Suppose $x\in A$ since $A \subseteq B$ then $x\in B$. $\therefore x\in A\cap B$ $\therefore A\subseteq A\cap B$.
Assume $A \cap B = A$ to prove $A\cup B=B$:
Suppose $x\in A$ then $x\in A \cap B$ $\therefore$ $x\in B$ from the definition of intersection. If $x\in B$ then $x\in A\cup B$. This proves $A\cap B = A \subseteq A\cup B=B$.
Assume $A\cup B=B$ to prove $B^c \subseteq A^c$:
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Assume $B^c \subseteq A^c$ to prove $A \subseteq B$:
$x\in A$ implies $x\notin B^c$ because $B^c\subseteq A^c$ hence $x\in B$ and $A\subseteq B$
From this we can see that $x \Leftrightarrow y$ for any $(x,y) \in \{A \subseteq B, A \cap B = A, A\cup B=B, A \subseteq B\}$ because we can get to any statement from any other statement.
Best Answer
My suggestion for proving is to begin with what you want to prove and in the middle, try to use the assumption. I mean if you want to show $A \subset B$, begin with that $x \in A$ and try to reach $x \in B$.
I write the proof for some cases:
Assume $B^c \subseteq A^c$ to prove $A \subseteq B$:
$x \in A \Rightarrow x \not \in A^c \Rightarrow x \not \in B^c$ (because $ B^c \subseteq A^c) \Rightarrow x \in B$. Therefore $A \subseteq B$.
Assume $A\cup B=B$ to prove $B^c \subseteq A^c$:
$x \in B^c \Rightarrow x \not \in B \Rightarrow x \not \in A\cup B \Rightarrow x \not \in A \Rightarrow x \in A^c$. Therefore $B^c \subseteq A^c$.