Show that the following four conditions are equivalent:
$(i) A \subset B, \quad (ii) A - B = \phi, \quad (iii) A \cup B = B, \quad (iv) A \cap B = A$

(A) We need to show that $(i) \Leftrightarrow (ii) \Leftrightarrow (iii) \Leftrightarrow (iv)$.
$(i) \Leftrightarrow (ii)$:
Assume $A \subset B$. If $x \in A - B$,then $x \in A$ and $x \notin B$. But $A \subset B$ implies $x \in A \Rightarrow x \in B$,which contradicts $x \notin B$. Thus,$A - B = \phi$.
Conversely,assume $A - B = \phi$. If $x \in A$,then $x$ must be in $B$ (otherwise $x \in A - B$,which is empty). Thus,$A \subset B$.
$(i) \Leftrightarrow (iii)$:
Assume $A \subset B$. Since $B \subset A \cup B$ is always true,we only need to show $A \cup B \subset B$. If $x \in A \cup B$,then $x \in A$ or $x \in B$. If $x \in A$,then $x \in B$ (as $A \subset B$). If $x \in B$,then $x \in B$. Thus,$A \cup B \subset B$,so $A \cup B = B$.
Conversely,assume $A \cup B = B$. If $x \in A$,then $x \in A \cup B$. Since $A \cup B = B$,$x \in B$. Thus,$A \subset B$.
$(i) \Leftrightarrow (iv)$:
Assume $A \subset B$. Since $A \cap B \subset A$ is always true,we show $A \subset A \cap B$. If $x \in A$,then $x \in B$ (as $A \subset B$). Thus,$x \in A$ and $x \in B$,so $x \in A \cap B$. Thus,$A \cap B = A$.
Conversely,assume $A \cap B = A$. If $x \in A$,then $x \in A \cap B$,which implies $x \in A$ and $x \in B$. Thus,$x \in B$,so $A \subset B$.