Lebl-Basic-Analysis-I · Problem 1.1.4

The key insight is that $A$ is a "smaller" subset nested inside $B$, so its bounds must be squeezed between $B$'s bounds. Since $\inf B$ is the greatest lower bound of all elements in $B$, and every element of $A$ is also in $B$, the infimum of $A$ cannot go below $B$'s infimum. By the same logic, $\sup A$ cannot exceed $\sup B$. Finally, for any nonempty set, its infimum is always at most its supremum by definition. Chain these three inequalities together to get the result.

The key insight is that $A$ is a "smaller" subset nested inside $B$, so its bounds must be squeezed between $B$'s bounds. Since $\inf B$ is the greatest lower bound of all elements in $B$, and every element of $A$ is also in $B$, the infimum of $A$ cannot go below $B$'s infimum. By the same logic, $\sup A$ cannot exceed $\sup B$. Finally, for any nonempty set, its infimum is always at most its supremum by definition. Chain these three inequalities together to get the result.