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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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Miss Independent
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  • What is your definition of open set in a metric space? It is not that difficult to prove that, under one definition, every open $\epsilon$-ball is a nbhd of each of its points. – Pedro Sep 06 '12 at 18:56
  • I'm using the following definition of an open set in a metric space (X,d): a set A is open if ∀a∈A,∃r>0 s.t. Br(a)⊆A. – Miss Independent Sep 06 '12 at 19:09
  • Why are we worrying about countability? Or is that for an entirely different question? At any rate, the first part has been asked before and there were some nice pictures. Will try to find the threads. – Dylan Moreland Sep 06 '12 at 19:25
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    Here we go: http://math.stackexchange.com/questions/104083/an-open-ball-is-an-open-set – Dylan Moreland Sep 06 '12 at 19:28

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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.

William
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