The subgroup generator function is inclusion monotone

Let $G$ be a group, with $A \subseteq B \subseteq G$. Prove that $\langle A \rangle \leq \langle B \rangle$. Give an example where $A \subseteq B$ with $A \neq B$ but $\langle A \rangle = \langle B \rangle$.

Let $\mathcal{A} = \{ H \leq G \ |\ A \subseteq H \}$ and $\mathcal{B} = \{ H \leq G \ |\ B \subseteq H \}$. Since $A \subseteq B$, we have $A \subseteq H$ whenever $B \subseteq H$; thus $\mathcal{B} \subseteq \mathcal{A}$. By definition, we have $\langle A \rangle = \cap \mathcal{A}$ and $\langle B \rangle = \cap \mathcal{B}$. We know from set theory that $\cap \mathcal{A} \subseteq \cap \mathcal{B}$, so that $\langle A \rangle \subseteq \langle B \rangle$.

Now since $\langle A \rangle$ is itself a subgroup of $G$, we have $\langle A \rangle \leq \langle B \rangle$.

Now suppose $G = \langle x \rangle$ is cyclic. Then $\{x\} \subsetneq G$, but we have $\langle x \rangle = \langle G \rangle$.