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? 
 A: 
$ϵ<\frac{1}{15}$ is a satisfactory solution.

It's best not to use same letter to mean two different things: the $\epsilon$ that's requested in the problem is somehow also $\frac16+\epsilon$ at the end of proof. You could write $\epsilon = \frac16+\delta$. So, your final answer is 
$$\frac16<\epsilon<\frac{1}{6}+\frac{1}{15}$$
which is correct, but a concrete value would be better. 
Here's a less convoluted approach. Review the three conditions:
$$
\epsilon>\left|\frac13-\frac14\right|,\quad \epsilon>\left|\frac13-\frac12\right|,\quad \epsilon\le \left|\frac13-\frac{17}{30}\right|
$$
(Non-strict inequality in the last condition, because I assume neighborhoods are open. If they are not, adjust.)
They can be condensed into two, because the second implies the first:
$$
\epsilon>\frac16 ,\quad \epsilon\le \frac{7}{30}
$$
So, anything within
$$
\frac{5}{30}<\epsilon\le \frac{7}{30}
$$
works... but there is a natural choice of $\epsilon$ here that is nice and simple and does not depend on strict/non-strict inequalities. 

Generally, in this course it is better to present concrete evidence of existence of epsilons and deltas. Saying: "let $\delta=\epsilon/4$" is better than "pick $\delta$ such that this and that inequalities hold".
A: 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 $1/5$.
