A union of subsets $ \bigcup_{i \in I} A_i $ is non-redundant if no subset is contained in the union of the others, i.e. for any $ j \in I $, $ A_j \nsubseteq \bigcup_{i \in I - \{j\}} A_i $.

In a metric space, is every open subset a non-redundant union of open balls?

Edit for quality: The open balls form a basis, and it is the canonical basis, it seems interesting to see if they can cover open subsets efficiently. Before asking the question, I had tried to construct it inductively, adding balls as big as possible, unsuccessfully.

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    $\begingroup$ In the separable case a simple inductive construction shows that this is true. I'm not sure about the general case. $\endgroup$ Jul 31, 2021 at 13:22
  • $\begingroup$ One can cover $\mathbb Q$ with an open subset of $\mathbb R$ that has finite measure. $\endgroup$ Jul 31, 2021 at 13:41
  • $\begingroup$ Standard facts imply that every open set has an open cover of refinements of open balls that is minimal in this sense (we cannot omit any member). $\endgroup$ Jul 31, 2021 at 23:18

1 Answer 1


This is true. Let $(X, d)$ be a metric space and $U\subset X$ open.

For each $x\in U$ there is $n\in \mathbb N$ such that $B(x, 2^{-n})\subset U$. Let $n_x$ be the least $n\in\mathbb N$ with that property for each $x\in U$. Let $U_n := \{x\in U~\vert~ n_x = n\}$, then $(U_n)_{n\in \mathbb N}$ is a partition of $U$. Let $<_n$ be a well-order on $U_n$ and define an order $<$ on $U$ via $x<y$ if $n_x = n_y$ and $x<_{n_x} y$ holds or if $n_x < n_y$. This defines a well-order on $U$.

We define balls $B_x$ for $x\in U$ recursively. For the least element $x\in U$ we set $B_x := B(x, 2^{-n_x})$. For each other $x\in U$, assuming $B_y$ is defined for all $y<x$ we define $B_x := B(x,2^{-n_x})$ if $x\notin \bigcup_{y<x} B_y$ and $B_x := \emptyset$ otherwise. Now the union of all nonempty $B_x$ is as desired.

If that union was redundant, let $x\in U$ be any element such that $B_x$ is nonempty and contained in $\bigcup_{y\neq x}B_y$. Then there is $y\in U$ with $x\neq y$ and $x\in B_y$. Then $y>x$ must hold. Otherwise, by definition, $B_x$ would be empty. From $y>x$ follows $n_y \geq n_x$. Therefore from $x\in B_y$ follows $$d(x,y) < 2^{-n_y}\leq 2^{-n_x}$$ hence $y\in B_x$. This is a contradiction because by definition $B_y$ must be empty in this case.

  • $\begingroup$ +1. I didn't think it had anything to do with separability (See the 1st comment to the Q) . $\endgroup$ Aug 1, 2021 at 5:49
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    $\begingroup$ @DanielWainfleet It’s a slight modification of the proof that a metric space is paracompact (the proof by ME Rudin). So that does play a role instead of separability. $\endgroup$ Aug 1, 2021 at 7:58
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    $\begingroup$ @Lukas, please stop answering problem statement and very low quality questions. Please reread the Enforcement of Quality Standards, address to users like you and Henno. $\endgroup$
    – amWhy
    Aug 4, 2021 at 22:06
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    $\begingroup$ @amWhy I am sorry but I fail to see how this is a low quality question. I found this problem interesting myself and thought several hours about it before finding a proof. It wasn't marked as low quality when I answered and also had a score well above 0 when I answered. I therefore don't know how I should have known not to answer. I acknowledge that you know better what doesn't meet the standards but here I fail to see that even in retrospect. $\endgroup$
    – Lukas Betz
    Aug 5, 2021 at 6:59

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