Show that every neighborhood is an open set.
Let $E = N_{r}(p)$. Then I want to show that for every $q \in E$, $q$ is an interior point. This means I just have to find a neighborhood of $q$ which is in $E$?
Show that every neighborhood is an open set.
Let $E = N_{r}(p)$. Then I want to show that for every $q \in E$, $q$ is an interior point. This means I just have to find a neighborhood of $q$ which is in $E$?
Dylan's suggestion of drawing a picture is good. Here we have our point $p$, the open ball $E=N_r(p)$, and an arbitrary point $q\in E$. I have labeled the distance from $p$ to $q$ as $d$.
If Dylan and I understand the question correctly, you want to show that there is some $s>0$ for which the open ball $N_s(q)$ is contained in $E$ (i.e. $N_s(q)\subseteq E$)? If that's the case, then can you think of an $s$, depending on $r$ and $d$, for which this is true? Think about drawing the circle of radius $s$ around the point $q$ in the picture.