# 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.

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?

• Assuming your notation is $J_\epsilon$ is the $\epsilon$ neighborhood of a point, then $\frac 1 6$ is a perfectly acceptable answer to this question, they just wanted you to find a particular $\epsilon$ that works – Alan Dec 15 '14 at 16:07
• Yes, that is the correct definition of that notation for this instance. – 123 Dec 15 '14 at 16:13
• @Alan: But $J_{1/6}(\frac13)$ doesn't contain $\frac12$. So $\epsilon$ needs to be strictly greater than $\frac16$. – TonyK Dec 15 '14 at 17:29
• @tonyk Ahh, yes. So a smidge more, but not enough to get to $\frac {17} {30}$. I blame sleep dep :) – Alan Dec 15 '14 at 18:49

$ϵ<\frac{1}{15}$ 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".

• I think the OP meant to say, "$\frac16 < \epsilon < \frac16 + \frac{1}{15}$ is a satisfactory solution." – TonyK Dec 15 '14 at 17:31

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 $1/5$.

• Likely to cause confusion, since OP is using $\epsilon$ differently here. – Thomas Andrews Dec 15 '14 at 18:21
• I cannot fix the confusion introduced by the OP, I am answering the question. – Yves Daoust Dec 15 '14 at 18:41
• That assumes the point of answering this question is to spell out an answer. He already has an answer, and he was asking for some validation, clarification, simplification. You appear to only have read the title of the question. – Thomas Andrews Dec 15 '14 at 18:55