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An open ball is an open set

How to prove that in any metric space an epsilon-neighborhood is an open set? solution: Suppose x belongs to V(p). Then d(x, p)

Not every open set can be written as a union of countably many epsilon neighborhoods. For example, take R with the discrete topology. Then any epsilon neighborhood is either R or a singleton set and so no proper uncountable set can be written as a countable union of these. However, R with its usual metric does have this property

is that correct?

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    Here we go: http://math.stackexchange.com/questions/104083/an-open-ball-is-an-open-set2012-09-06

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In a metric space $(X, d)$, a set $U$ is open if and only if for all $x \in U$, there exists a $r \in \mathbb{R}$, $r > 0$, such that $B_r(x) \subset U$.

Now let $\epsilon > 0$. You would like to show that $B_\epsilon(z)$ is open for any $z \in X$. Now you just need to verify the above definition. Given any $x \in B_\epsilon(z)$, let $r = \min\{d(z,x), \epsilon - d(z,x)\}$. Then $B_r(x) \subset B_\epsilon(z)$. Hence $B_\epsilon(z)$ is open.


Alternatively, sometimes the metric topology is also defined to be topology generated by basis of balls. In this definition, it is immediately clear that any ball is open.