Let $A$ be a set of real numbers such that $A \subseteq [0,1]$. I'm having a hard time proving that $C=\left\{\frac{a+1}{n^2} \colon a \in A, n \in \mathbb{N} \right\}$ is not dense in $[0,1]$. How should I approach this? I know that I must find an interval $(x,y) \subseteq [0,1]$ such that $C \cap (x,y) = \phi $. I tried demanding $\frac{a+1}{n^2}<x$ or $\frac{a+1}{n^2}>y$ for every $a \in A$ and $n \in \mathbb{N}$. The problem is that I know nothing about $A$. I'd appreciate any suggestions.
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If the statement is true, I think it would suffice to show the result for $A=[0,1]$.
Regardless, I think the interval $(1/2,1)$ does not intersect $C$.
angryavian
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Suppose that $A=[0,1]$. Then, you can find a set that contains $C$. Namely, for $n=1$, $\frac{A+1}{n^2}=[1,2]$. For $n=2$, $\frac{A+1}{n^1}=[1/4,1/2]$. If you continue in this way, you'll see that all intervals are in the range $[0,1/4]$. More precisely, you know that since $A\subseteq [0,1]$, it follows that $$ C\subseteq \cup_{n=1}^\infty [1/n^2,2/n^2]. $$
Therefore, $C$ does not intersect $(1/2,1)$.
Michael Burr
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If $n=1$ you have $\forall a\in A: c\colon= \frac{a+1}{1} \geq 1$, thus $c\notin C$
If $n\geq2$ it follows $$\forall n\in\mathbb{N}-\{1\} \forall a\in A: \frac{a+1}{n^2}\leq \frac{1+1}{2^2}=\frac{1}{2}$$
Thus your set $C$ does not intersect $\left(0.5, 1\right)$
user127.0.0.1
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