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

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.

• In the separable case a simple inductive construction shows that this is true. I'm not sure about the general case. Jul 31, 2021 at 13:22
• One can cover $\mathbb Q$ with an open subset of $\mathbb R$ that has finite measure. Jul 31, 2021 at 13:41
• 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). Jul 31, 2021 at 23:18

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 $$x if $$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 we define $$B_x := B(x,2^{-n_x})$$ if $$x\notin \bigcup_{y and $$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.