Show every point of $J = (-3, 3)$ is an interior point of $J$

Question

Claim: Every point of $J = (-3, 3)$ is an interior point of $J$

Proof: For any $p \in J,$ let $r = \min\{|-3 -p|, |3 - p|\}.$

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I have two questions about the proof above. First one is How do we know this radius works? I tried doing this below. Not sure if it works. Even if what I did is correct, I think I missed points closer to the middle of $J.$

Let $p \in J.$ We want to find a radius $r > 0$ s.t. $N_r(p) \subset J.$ Let $r = \min\{|-3 -p|, |3 - p|\}.$

• $p > 0 \implies r = \min\{|-3 -p|, |3 - p|\} = 3 - p.$

Then $N_r(p) = (p - (3 - p), p + (3 - p) = (2p - 3, 3)$

For the largest positive $p$ which is $p = 3 - \epsilon ,$ we have $(2p - 3, 3) = (3 - 2\epsilon, 3)$ and for the smallest positive $p$ which is $\epsilon,$ we have $(2p - 3, 3) = (2\epsilon - 3, 3)$

• $p < 0 \implies r = \min\{|-3 -p|, |3 - p|\} = 3 + p.$

Then $N_r(p) = (p - (3 + p), p + (3 + p) = (-3 , 3 + 2p)$

For the smallest negative $p$ which is $p = -3 + \epsilon$ we have $(-3 , 3 + 2p) = (-3, -3 + 2\epsilon)$ and the for the largest negative $p$ which is $p =-\epsilon$, we have $(-3 , 3 + 2p) = (-3, 3 - 2\epsilon)$

• $p = 0 \implies r = 3$ and $N_r(0) = (-3, 3)$

Second question is Is there a general method for finding suitable radii in such problems? Thanks.

Answer 1

for any $p\in J$, let $\epsilon=\min\{|-3 -p|, |3 - p|\}$ and then take the radius $\epsilon/2$ around $p$.Then $p$ willbe interior point of $J$.