# $\epsilon$-cover of a compact metric space

Let $(W,d)$ be a compact metric space. For a subset $A \subset W$, define its diameter by $\delta(A):=\sup\{d(x,y)\mid x,y \in A\}$. A cover ${\cal U}=\{A_\lambda\}_{\lambda\in \Lambda}$ of $W$ is said to be an $\epsilon$-cover ($\epsilon>0$) if $\delta(A_\lambda)<\epsilon$ for any $\lambda\in \Lambda$. Set $$k(\epsilon):=\inf\{\# \Lambda \mid \mbox{ there exists an \epsilon-cover of W with the index set \Lambda} \},$$ where $\# \Lambda$ means the cardinality of $\Lambda$. Note that, $\# \Lambda=+\infty$ if is an infinite set.

(1) Show that $k(\epsilon)$ is finite for any $\epsilon>0$, and is non-strictly decreasing.

It seems that the set $A_\lambda$ in the definition of $\epsilon$-cover above needs not be open. If $A_\lambda$ is open for all $\lambda\in \Lambda$ then it's somewhat easy to prove $k(\epsilon)$ is finite. Because if there exists an $\epsilon$-cover of $W$ with the index set $\Lambda$ then there exists a finite sub $\epsilon$-cover of $W$ since $W$ is compact. I don't know how to deal with the other cases.

(2) Show that $\delta(W)<+\infty$.

To prove the diameter of some metric space is finite, do we often prove the metric is bounded? It seems to me that the definition of $\epsilon$-cover above somehow relates to that of $\epsilon$-net. Then the metric $W$ should be called totally bounded? But any compact metric space is undoubtedly totally bounded. I'm afraid I'm messing the definitions up.

(3) Show that an isometry (a map preserving distance) $\tau:W \to W$ is always a bijection.

I guess $k(\epsilon)$ will play some role here but cannot figure that out.

Any help would be much appreciated.

• "But any metric space is undoubtedly totally bounded." This is incorrect, take e.g. $\mathbb{R}$. – Matias Heikkilä Oct 11 '16 at 13:28
• There are also bounded metric spaces that are not totally bounded (e.g. the unit ball in $l^2$). – Matias Heikkilä Oct 11 '16 at 13:30
• Sorry I have missed the word "compact". Edited! – user Oct 11 '16 at 13:52

1) To show that $k(\epsilon)$ is finite, it is enough to show that there is some finite covering. But it is easy enough to find an open covering, so you should use that (non-open coverings are not that useful).

The existance-proof works like that: For each point $x\in W$ take the (open) $\epsilon/2$-ball around $x$. These (infinitely many) balls surely cover the whole space. As the space is compact, there is is a finite subset of these balls which still cover the whole space. Note that $\epsilon/2$-balls have diameter $\le \epsilon$.

And for non-strictly decreasing, you just have to note that if $\epsilon_1 \le \epsilon_2$, than any $\epsilon_1$-cover will also be a $\epsilon_2$-cover. That means that by increasing $\epsilon$, you can only increase the set of allowed coverings, which means the infimum can only decrease.

2) By (1) we know there is a finite set of $x_1, ..., x_n\in W$ such that the unit-balls around these points cover the whole space. Furthermore, let

$M = \max_{i\neq j}d(x_i,x_j)$

which is some finite constant. Now take two arbitrary points $x,y\in W$. They both lie it (potentially differet) unit-balls. I.e. there are $i,j$ such that $d(x,x_i), d(y,x_j) < 1$. By the triangle inequality you get $d(x,y) \le 2 + M$, which implies $\delta(W) \le 2 + M < \infty$.

3) Injectivity: Let $\tau(x) = \tau(y)$. Then $d(x,y) = d(\tau(x),\tau(y)) = 0$, which implies $x=y$.

Surjectivity (by contradiction): Let $y\not\in \tau(W)$. Then there is an $\epsilon>0$ such that $B_\epsilon(y) \cap \tau(W) = \emptyset$. Now take a minimal (open) $\epsilon$-cover of $W$. i.e. $A_1,...,A_{k(\epsilon)}$ such that they cover the whole space and $\delta(A_i) \le \epsilon$. $y$ must be inside at least one of the $A_i$, w.l.o.g. assume that $y\in A_1$. Then $A_1\subset B_\epsilon(y)$, so $A_1\cap\tau(W) = \emptyset$. Now $\tau^{-1}(A_2),..., \tau^{-1}(A_{k(\epsilon)})$ form an open $\epsilon$-cover of $W$ which cannot be, because $k(\epsilon)$ is minimal.

• For the proof of part (1), in the first line your the second paragraph should we take the $\epsilon/2$-ball around $x$ instead of $\epsilon$-ball? Because the diameter of any set in the defined covering is less than $\epsilon$. – user Oct 11 '16 at 14:59
• you are right. I corrected the answer. – Simon Oct 11 '16 at 15:12