I try to show that every doubling metric space is separable for these use the doubling condition i try to construct one countable dense set, but i not sure how make these i try using radius like $\frac{1}{2^k}$ around a point but i not sure if these is correct because i cant say that these collection is countable any hint or help i will be very grateful.
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1As we can write $$X=\bigcup_{n\in \mathbb{N}} B(x_0, n)$$ it is enough to show that any fixed ball $B(x_0,n)$ is separable. To deal with a fixed ball, just cover with balls of smaller and smaller radii and pick the centers of those balls as your countable dense set. Use the doubling condition to show that the cover of balls of radius $n/2^k$ is finite and then use the fact that the countable union of countable sets is countable. – Severin Schraven Aug 24 '23 at 00:42
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1Hints: Show an $\epsilon$-separated set must intersect every ball in a finite set. Conclude that a maximal $\epsilon$-separated set must be countable and $\epsilon$-dense. Take countably many $\epsilon$'s converging to $0$. – M W Aug 24 '23 at 00:43
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@SeverinSchraven in order to put $X$ as that union I would have to assume that the distance between two points is always finite? – Nick Weber Aug 24 '23 at 00:53
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1Typically a metric space is assumed to have finite distances, otherwise its called something like an extended metric space. If you try to define the doubling property on an extended metric space, it really only makes sense for balls of finite radius, so separability would not generally be something you could conclude, as you could make a space with uncountably many points and separate them with a distance of infinity and your doubling property would still trivially hold. – M W Aug 24 '23 at 01:05