Does $C[0, 1]$ admit a covering by sets of arbitrarily small eccentricity?

Question

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 $(1+\delta)\epsilon$?

Answer 1

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.

Answer 2

We may take $\epsilon=1$ as if $\bigcup E_i$ has eccentricity constants $1-\delta$ and $1+\delta$, then $\bigcup rE_i$ has eccentricity constants $r(1\pm\delta)$. The thing that you are really trying to control is the ratio of the eccentricity constants. Here is a construction with the ratio of the eccentricity constants arbitrarily close to 2. I don’t see how you could get the ratio to be arbitrarily close to 1 though.

Take a countable dense collection of functions. Then choose a maximal sub-collection of functions that are at least 1 unit apart and let $\eta>0$. Any continuous function is now within $1+\eta$ of this collection- otherwise it can be approximated by something in the dense collection, contradicting the maximality.

Now call your functions $(f_n)$. Define $S_n=\{g:\|g-f_n\|\le (1+\eta)\inf_m\{\|g-f_m\|\}\}$. Finally, take $B_n=S_n\setminus\bigcup_{j<n}S_j$.

If $\|g-f_n\|<\frac 1{2+\eta}$, then since the $f_j$'s are 1-separated, $\|g-f_m\|>\frac{1+\eta}{2+\eta}$ for any $m\ne n$. Hence $g\in S_n$, so that we have shown $B(f_n,\frac 1{2+\eta})\subset B_n$ On the other hand, if $\|g-f_n\|\ge(1+\eta)^2$, then we have already shown that there exists $m$ with $\|g-f_m\|<1+\eta$, so that $g\not\in S_n$. It follows that $B(f_n,\frac 1{2+\eta})\subset B_n\subset B(f_n,1+\eta)$. Since $\eta$ can be chosen arbitrarily close to 0, we can get a ratio of eccentricity constants as close to 2 as desired.