Find an $\epsilon$ such that the $\epsilon$ neighborhood of $\frac{1}{3}$ contains $\frac{1}{4}$ and $\frac{1}{2}$ but not $\frac{17}{30}$

Question

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.

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

Answer 1

$ϵ<\frac{1}{15}$ 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.

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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".

Answer 2

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 $1/5$.