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Let $(X,d)$ be a metric space. Does the following property have a name?

For every $\epsilon>0$ a countable collection of subsets $(A_i)_{i=1}^\infty$ exists such that $X=\bigcup_{i=1}^\infty A_i$ and $diam(A_i)<\epsilon$.

EDIT 2: Is separability equivalent?

Martin Sleziak
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Carucel
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  • I believe this is a metric version of the notion of a second countable space from topology: https://en.wikipedia.org/wiki/Second-countable_space – Justin Benfield Aug 18 '16 at 13:42
  • Thank you! I don't see why, but if that is correct this is if and only if it is a seperable metric space, by this question. Could you explain why this corresponds to the notion of a second countable space? – Carucel Aug 18 '16 at 13:47
  • Not second countable, but totally bounded instead (took me some more hopping around wikipedia to find the exact same property as you describe here): https://en.wikipedia.org/wiki/Totally_bounded_space – Justin Benfield Aug 18 '16 at 13:52
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    It's also not "totally bounded", since then $X$ has to be covered by finitely many sets of diameter $\epsilon$. In my question I asked for a countable such collection. Perhaps there is no name for this property... – Carucel Aug 18 '16 at 13:59
  • I suspect that any space satisfying your property would be a Lindelöf space (take any given cover of the space, by choosing an epsilon small enough, you should be able to find a countable collection such that every set of the cover is itself covered by elements of the countable collection). – Justin Benfield Aug 18 '16 at 14:25
  • Seperable, second countable, and Lindelof are all the same for metric spaces. However, I don't see why the property in my question is equivalent to any of them. – Carucel Aug 18 '16 at 14:34

2 Answers2

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The property you stated is equivalent to separability of $X$ (which is equivalent to $X$ being second countable, since it is metric). Let's denote $B_\epsilon(x)$ the open ball of radius $\epsilon$ around $x$.

If $X$ is separable, take a countable dense subset $D=\left\{x_1,x_2,\ldots\right\}$. Given $\epsilon>0$, take $A_n=B_{\epsilon/2}(x_n)$. The collection $\left\{A_n\right\}$ has the desired properties.

Conversely, suppose the condition you wrote is satisfies. For a given $n$, choose a family $\left\{A^n_i\right\}_i$, with $diam A^n_i<1/n$ and $X=\bigcup_i A^n_i$. For each $n$ and each $i$ choose $x_{n,i}\in A^n_i$. Then $D=\left\{x_{n,i}:n,i=1,2,\ldots\right\}$ is a countable dense subset of $X$.

Luiz Cordeiro
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I believe a space satisfying your property is separable: Consider a for some fixed $\delta>0$, any point $x$ in your space. Now let $\epsilon<\delta$ and consider the centers of the given countable cover by $A_i$'s of your space. Since $x$ is in your space, the $A_i$'s cover the space, and $\epsilon<\delta$, it follows that there is an $i_0\in\mathbb{N}$ such that $x\in A_{i_0}$, moreover, we have that the center of $A_{i_0}$, $a_{i_0}$ must be within the $\delta$-neighborhood of $x$.

Note: There is one tricky point I haven't dealt with, and that's that nothing in the property you've given requires that the centers of the sequence of centers of the $A_i$'s for a smaller $\epsilon$ will contain the centers of a sequence for a larger $\epsilon$, but this can be addressed by 'appending' them to the sequence (we take add in the $A_i^\epsilon$'s to the sequence for countably many epsilons (e.g. use rational epsilons), and hence obtain a countable dense subset; it's still countable because countable union of countable sets is countable).

Justin Benfield
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