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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 $E$?

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Not every neighborhood is an open set. Wkipedia states this ("Note that the neighbourhood V need not be an open set itself"). Also, what's $N_r(p)$? – George Lowther Aug 22 '11 at 23:19
Maybe you're trying to show that every ball (something of the form $N_r(p)$, in your notation) in a metric space is open? A neighborhood of a point $p$ is usually an open set containing $p$, or a set containing an open set containing $p$, depending on the author. – Dylan Moreland Aug 22 '11 at 23:22
Anyway, if I'm right about the nature of your actual question then I think it would be good to draw a picture (so we're in the setting of $\mathbf R^2$ with the usual metric). – Dylan Moreland Aug 22 '11 at 23:25
How do you define neighborhood? – lhf Aug 22 '11 at 23:32
up vote 3 down vote accepted

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 $d$.

enter image description here

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.

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$s = r-d$ would work? – James Aug 23 '11 at 0:05
@James: Yup! This follows from the triangle inequality, which is part of the axioms of a metric space. – Zev Chonoles Aug 23 '11 at 0:08

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