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Notations and definitions

Let $E$ be a finite dimensional vector space with norm $||\;||$.

Let $B$ denote the closed unit ball in $E$ and $B_r[a]$ the closed ball centered at $a$ with radius $r$.

I will say $B$ has a finite cover with closed balls of radius $r>0$, if for a given $r >0$, $$\exists N\in \mathbb N, \exists (a_1,\ldots,a_N)\in E^N, B \subset \cup_{i=1}^N B_r[a_i]$$

With the help of sequential compactness, I was able to prove the existence for any $r>0$ of a finite cover of $B$ with closed balls of radius $r$.

Now let $m(r) \in \mathbb N$ be the least number of closed balls of radius $r$ needed to cover $B$.

Question

I am asked to prove that the well-defined function $$\begin{array}{ccccc} m & : & \mathbb R_+ & \to & \mathbb N \\ & & r & \mapsto & m(r) \\ \end{array}$$

is decreasing.

This result is very intuitive : you need less bigger balls to cover $B$, but I'm unable to prove it...

How can I link two covers of $B$ with different radii? Should I go for contradiction ?

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Any cover with $n$ small balls implies that a cover with larger balls and the same $n$ exists. Just use the same centers. –  Thomas Mar 15 at 16:16
    
@Thomas I don't think this is valid. The definition of cover requires an equality, not just an inclusion. If I take bigger balls with same centers, $B$ will be included in the union of the balls, but not necessarily equal this union. –  G.T.R Mar 15 at 16:26
    
I missed the $=$ sign. You cannot cover a $2$ dimensional ball with finitely many (!) smaller (!) $2$- dimensional balls using this definition, though (and it won't work in higher dimensions, either), at least if you use the usual Euclidean norm. Some norms where unit balls are rectangular may allow that. –  Thomas Mar 15 at 16:32
    
Ok, I'll switch to inclusions (it makes more sense!). I agree with your reasoning. Do you want to write it as answer ? –  G.T.R Mar 15 at 16:44

1 Answer 1

up vote 1 down vote accepted

This is obvious. Note that $m(r)$ is not only an infimum, but a minimum. This means that you can cover $B$ with $m:=m(r)$ suitably placed balls of radius $r$. When $r'>r$ then $m$ balls of radius $r'$ with the same centers will obviously cover $B$, but there might be a better covering with balls of radius $r'$. So $m(r')\leq m(r)$. (You cannot hope for strict monotonicity.)

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