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.
Best Answer
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.