Let $A$ be a nonempty upper bounded set. So I know that for all $\varepsilon>0$, $\sup A<a+\varepsilon$, where $a$ is an arbitrary element of $A$. But since $x<y+\varepsilon$ (for all $\varepsilon>0$) implies $x\leq y$, $\sup A\leq a$ which is a contradiction. What's wrong?
2026-04-09 13:29:05.1775741345
About approximation property of supremum
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The $a$ you mention depends on the $\varepsilon$. The property is : for all $\varepsilon>0$, there exists $a(\varepsilon)\in A$ such that $\sup A<\varepsilon +a(\varepsilon)$.