If $\emptyset\neq A\subseteq\mathbb{R}$ is a set bound from above, and $A$ does not have a maximal value, and $B\subseteq A$ and $B$ has a maximal value, how can I prove that $\sup(A)=\sup(A\setminus B)$?
Doesn't this seem impossible?
2026-04-11 20:20:03.1775938803
How can I prove the equality of $\sup(A)$ and $sup(A\setminus B)?$
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Since $B$ has a maximum there exists some $a \in A$ such that $b \leq a$ for all $b \in B$. Since $A$ does not have a maximal value there exists some $x > a$ with $x \in A$, otherwise $a$ would be maximal. Since all $x > a$ are not in $B$ we have $\sup (A \setminus B) = \sup A$.