Am I right that this statement is false?

Question

For every nonempty set $A$ of real numbers having an upper bound, and for every $d \in \mathbb{R}$, we examine the statement: $$[(\sup A = d, d \notin A) \implies (\exists_{N \in \mathbb{N}}{\forall_{n \ge N}{\exists_{a \in A}{\;d - \tfrac{1}{n} < a \le d - \tfrac{1}{n -1}}}})]$$ I think it's false because consider the set $A = \{1 - \frac{1}{2}, 1 - \frac{1}{3}...\}$. We have $\sup A = 1 = d$ but we cannot find such a number $N$. Am I right?

Answer 1

No, in fact your example satisfies that statment with $N=1$. For any $n\geq 1$, there is an $a\in \{1-\frac{1}{2},1-\frac{1}{3},1-\frac{1}{4},\ldots\}$ such that $$1-\tfrac{1}{n}<a\leq 1-\tfrac{1}{n-1}$$ namely $a=1-\frac{1}{n-1}$.

I recommend considering the set $A=\{1-\frac{1}{3},1-\frac{1}{6},1-\frac{1}{9},\ldots\}$. When $n=3k+2$ for some $k$, $$1-\tfrac{1}{3k+2}<1-\tfrac{1}{3s}\leq 1-\tfrac{1}{3k+1}\iff\tfrac{1}{3k+1}\geq\tfrac{1}{3s}>\tfrac{1}{3k+2}\iff 3k+1\leq 3s<3k+2$$ is impossible for any $a=1-\frac{1}{3s}$ in the set $A$.