We denote by $C[0, 1]$ the space of continuous functions on $[0, 1]$ under the supremum norm, equipped with the Borel sigma algebra.


A *covering* of $C[0, 1]$ is a (possibly countably infinite) collection of Borel sets $E\_i$ such that $\bigcup\_i E\_i = C[0, 1]$.


A covering is said to have *finite eccentricity* if there exist constants $0 < c \leq C$ such that each $E\_i$ contains a open ball of radius $c$ and is contained within an open ball of radius $C$. We call $c$ and $C$ the *eccentricity constants* of the cover.


A covering is said to be *point finite* if every $f \in C[0, 1]$ lies in finitely many of the $E\_i$.


**Question:** Let $\epsilon > 0$ and $0< \delta < 1$ be arbitrary. Does there exist a point finite covering of $C[0, 1]$ of finite eccentricity with eccentricity constants $(1-\delta)\epsilon$ and $(1+\delta)\epsilon$?



Does $C[0, 1]$ admit a covering by sets of arbitrarily small eccentricity?
To give a positive answer to the question it is enough to, for a fixed $\varepsilon$, give a collection of disjoint balls in $C[0,1]$ of radius $\varepsilon$ which is dense in $C[0,1]$. Indeed, then for $\delta$ as small as you want you can take the sets $E\_i$ to be the balls and adjoin every point $f$ outside the balls to one ball at distance $<\delta$ of $f$ (there are countably many balls so you can do this in such a way that the resulting sets are Borel), obtaining a covering such that each point is in just one set, and with eccentricity constants $\varepsilon,\varepsilon+\delta$.


To build this collection of balls we first consider a sequence of distinct points $0,1,a\_3,a\_4,\dots$ dense in $[0,1]$. Now for $n\geq2$, let $A\_n$ be the set of functions which take some arbitrary values in the points $a\_1,a\_2,\dots,a\_n$ and are defined by linear interpolation in the rest of the interval. Then $A\_n$ is isometric to $\mathbb{R}^n$ in the square metric, with the coordinates given by the values of the function in $a\_1,\dots,a\_n$. Moreover, in this coordinates, the inclusion of $A\_n$ in $A\_{n+1}$ is given by a linear inclusion of $\mathbb{R}^n$ into $\mathbb{R}^{n+1}$ which preserves the first $n$ coordinates.


Now we can consider a tiling of $A\_2$ by balls of radius $\varepsilon$ (in the square metric of $\mathbb{R}^2$, squares of side $2\varepsilon$), and we call $C\_2$ the collection of their centers. As the inclusion $A\_2\to A\_3$ is linear and preserves the first $2$ coordinates, we can extend this tiling to a tiling of $A\_3$ by cubes of radius $\varepsilon$. More precisely, the set $C\_3$ of centers of cubes in $A\_3$ will be given in coordinates by $\{p+k(0,0,\dots,0,2\varepsilon);p\in C\_2,k\in\mathbb{Z}\}$. Concretely, that implies $C\_2\subseteq C\_3$. This way we can extend this tilings of $A\_n$ to tilings of $A\_{n+1}$ by balls (hypercubes) of radius $\varepsilon$, with centers in a set $C\_n$, with $C\_n\subseteq C\_{n+1}\forall n$. Note that by construction, the points of the $C\_n$ are at distance $\geq 2\varepsilon$ from each other.


So, the collection of balls of radius $\varepsilon$ and center in $\cup\_{n=2}^\infty C\_n$ will be disjoint, and it is dense in $\cup\_{n=2}^\infty A\_n$. As $\cup\_{n=2}^\infty A\_n$ is dense in $C[0,1]$, the collection of balls is dense in $C[0,1]$ too. So now we can just use the construction of the first paragraph to get the cover we want.


