Currently I'm trying to prove that for : An infinite linear space X with norm, there exist an infinite family of disjoint balls $B(x_{n},\epsilon)$ that are in $B(0,1)$ .
I've already checked this link
Fitting an infinite collection of balls in an infinite dimensional unit ball
but trully I can't understand the last sentence, why we changed the centers and how we are sure that these balls are disjoint ??And also this proof can work for any $\epsilon$??
Thanks in advance!!
If U is open and not empty, then there is some r > 0, x in U
with B(x,r) subset U. Thus K = closure B(x,r/2) subset U.
U - K is open and not empty.
Use that to construct a series of pairwise disjoint closed sets inside of U.
In the proof in the link there was no changing of centers. Open balls and centers are not mentioned until the last sentence. Before that, there was just a sequence $(x_n)_n$ of unit vectors. And of course unit vectors do not belong to $B(0,1).$
Let $0<\epsilon <1.$ Let $S_n$ be the linear span of $\{x_j:1\leq j\leq n\}.$ Let the distance $d(x_{n+1},S_n)$, from $x_{n+1}$ to $S_n,$ be greater than $1-\epsilon.$
Let $r=\frac {1}{3}(1-\epsilon).$ Let For $m<n,$ we have $$\|\frac {2}{3}x_n - \frac {2}{3}x_m\|\geq \frac {2}{3}d(x_n,S_{n-1})>\frac {2}{3}(1-\epsilon)=2r.$$ $$\text {Therefore }\quad B(\frac {2}{3}x_m,r)\cap B(\frac {2}{3}x_n,r)=\phi.$$ And each $B(\frac {2}{3}x_m,r)\subset B(0,1)$ because $\|\frac {2}{3}x_m\|+r=\frac {2}{3}+r=1-\frac {1}{3}\epsilon <1.$
Since $\epsilon$ can be arbitrarily small, $r$ can be arbitrarily close to $1/3.$
I don't know whether $r<1/3$ is the best possible result in general. It seems so, if we restrict ourselves to this type of construction.
Because, with $0<K<1,$ when $m\ne n$: To ensure that $B(Kx_m,r)\cap B(Kx_n,r)=\phi$ we must have $r\leq \|Kx_m-Kx_n\|/2,$ and to ensure that $B(Kx_n,r)\subset B(0,1)$ we must have $r\leq 1-K.$ Together these imply $3r\leq \|Kx_m-Kx_n\|+1-K,$ so if the best upper bound that we have for $\|Kx_m-Kx_n\|$ is less than $K$, then $r < 1/3.$
