Find an $\epsilon$ such that the $\epsilon$ neighborhood of $\frac{1}{3}$ contains $\frac{1}{4}$ and $\frac{1}{2}$ but not $\frac{17}{30}$ I am self studying analysis and wrote a proof that is not confirmed by the text I am using to guide my study.  I am hoping someone might help me comfirm/fix/improve this.  
The problem asks: 

Find an $\epsilon$ such that $J_{\epsilon}(\frac{1}{3})$ contains $\frac{1}{4}$ and $\frac{1}{2}$ but not $\frac{17}{30}$

Here $J_\epsilon(x)$ means the $\epsilon$-neighborhood of $x$.
I know that $d\left(\frac{1}{4},\frac{1}{3}\right)<d\left(\frac{1}{2},\frac{1}{3}\right)$ 
$$\left|\frac{1}{2} - \frac{1}{3}\right|  =  \frac{1}{6}$$ 
I know that:
$$J_{\frac{1}{6}}(\frac{1}{3})  = \left(\frac{1}{6},\frac{1}{2}\right)$$ 
and so $$J_{\frac{1}{6}+\epsilon}\left(\frac{1}{3}\right)  =\left(\frac{1}{6}-\epsilon,\frac{1}{2}+\epsilon\right)$$ where $\epsilon<\frac{1}{15}$ is a satisfactory solution. 

Am I allowed to generalize this way with epsilon? Would the answer be better If provide some concrete value of epsilon? 

Compute the distances from $1/3$,
$$\Big|\frac14-\frac13\Big|=\frac1{12}=\frac5{60},$$
$$\Big|\frac12-\frac13\Big|=\frac1{6}=\frac{10}{60},$$
$$\Big|\frac{17}{30}-\frac13\Big|=\frac{7}{30}=\frac{14}{60}.$$
So any $$\frac{10}{60}<\epsilon\le\frac{14}{60}$$will do, say $1/5$.
