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I'm trying to solve the following problem:

Say whether this statement is true for every $A\subset\Bbb{R}$ bounded from above ($A\not=\varnothing$) and for every $d\in\Bbb{R}$:

$$[(\sup{A} = d\notin A)\implies(\exists_{N\in\Bbb{N}}\forall_{n\ge N, n\in\Bbb{N}}\exists_{a\in A}\space\space d-\frac{1}{n}<a\le d-\frac{1}{n+1})]$$

Intuition tells me that if we prove that $\exists_{\epsilon>0}$ such that $A=(d-\epsilon, d)$ then we've practically proven the statement above. But is that really the case? Or rather it's something that is simply not true and there is other way to solve this problem? Can someone point me in the right direction?

qiubit
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1 Answers1

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No, this is not true. For example, take $A= \{ -\frac{1}{2n} : n \in \mathbb{N} \}$. Clearly the supremum is $0$, but there is no chance to get some $a \in A$ in the intervals $\left( -\frac{1}{2n} , -\frac{1}{2n+1} \right)$...

Crostul
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