Prove some more statements about sets and subsets

Proof. Let $x$ be any element in $A$ . Then, $x \in B$ since $A \subset B$ . Further, there is some $y \in B$ such that $y \notin A$ (since $A \subset B$ means that $A$ is a proper subset of $B$ ). Then, since $B \subset C$ we have both $x, y \in C$ . Hence, $A \subset C$ (since every element of $A$ is in $C$ , and there is at least one element of $C$ not in $A$ , so $A \neq C$ ).∎
Proof. Let $x$ be any element of $A$ . Then, $x \in B$ since $A \subseteq B$ . Further, $x \in B$ implies $x \in C$ since $B \subseteq C$ . Thus, $A \subseteq C$ .∎
Proof. This was established in part (a) since we didn’t need $B \neq C$ in that proof. (Since $A$ a proper subset of $B$ guaranteed there was some $y$ in $B$ that wasn’t in $A$ , and since $B$ is a subset, proper or otherwise, of $C$ , this $y$ is in $C$ ; hence, $A$ is a proper subset of $C$ .)∎
Proof. Since $A \subseteq B$ we know every element of $A$ is in $B$ . Hence, if $x \in A$ , then $x \in B$ .∎
If $x \in A$ and $A \in B$ , then we cannot conclude that $x \in B$ . For example, let $A = \{ 1 \}$ and $B = \{ \{ 1 \} \}$ . Then, $A \in B$ (since $A$ is the set $\{ 1 \}$ and $B$ contains this set). However, $1 \in A$ , but $1 \notin B$ (since $B$ contains the set which contains 1, but does not contain the element 1).