I am thinking about the following question of covering number and maximal separated subset:
Let $X$ be a Banach space and let $B_X(y,r)$ denote the ball of $X$ whose center is at $y$ with radius $r$.
(Definition of covering number)
Denote $N_{\epsilon}(A,X):=\min\{n: \exists y^1, y^2, ... ,y^n, where\ \ each \ \ y^j\in X: A\subset \cup_{j=1}^n B_X(y^j,\epsilon)\}$.
(Definition of maximal separated subset)
Denote $F\subset A$ be the maximal $2\epsilon$-separated subset of $A$ in the metric of $X$, i.e. $||x-y ||\geq2\epsilon$ for any two distinct $x,y\in F$, and $A\subset \cup_{x\in F}B_X(x,2\epsilon)$.
Prove that: $$|F|\leq N_{\epsilon}(A,X) $$ where $|F|$ is the cardinality of the finite set $F$.
My idea is that for the maximal $2\epsilon$-separated subset of $A$, the set $A$ is covered by the balls with radius at least $2\epsilon$. While for the cover number, we use a set of balls with a smaller radius $\epsilon$ to cover $A$. My intuition is that if the radius becomes smaller, the number of balls we used could be larger. But I don't know how to give a good proof since the cover number is considering the minimum of the cover.
Any suggestions would be welcome and thank you in advance!!
We can assume w.l.g that $N_\epsilon(A,X)<\infty$ (otherwise the inequality is obvious). Note that for a non totally bounded set $A$ then $N_\epsilon(A,X)=\infty$ it will happen for some $\epsilon's$. First we claim that if $N_\epsilon(A,X)<\infty$ then $2\epsilon-$separated subsets of $A$ exist. Indeed, pick an $\epsilon-$cover of $A$, say $\{y^1,...,y^n\}$ (we can assume, probably by adding more points that $\{y^1,...,y^n\}\subseteq A)$ such that $$A\subseteq \bigcup_{j=1}^{n}B_X(y^j,\epsilon)$$ If the above balls are disjoint, then $||y^j-y^i||\geq \epsilon$ for every $i\neq j$. Since, $$A\subseteq \bigcup_{j=1}^{n}B_X(y^j,2\epsilon)$$ Then, $\{y^1,...,y^n\}$ is an $2\epsilon-$separated subset of $A$. If the above balls are not disjoint, then let $$A_1=\{y^j:\, B_X(y^j,\epsilon)\cap B_X(y^1,\epsilon)\neq \emptyset,\ j=1,...,n\}$$ If $A_1$ contains only $y^1$ we keep the ball $B_X(y^1,\epsilon)$ which is disjoint from all the others and we continuing with the point $y^2$ and the set $A_2$ defined in similar fashion as with $A_1$. So, suppose that $A_1$ contains also other points than $y^1$. We claim that $$\tag{1}\bigcup_{y\in A_1}B_X(y,\epsilon)\subseteq B_X(y^1,2\epsilon)$$ Indeed, if $z\in B_X(y,\epsilon)$ for some $y\in A_1$ then, since $B_X(y^1,\epsilon)\cap B_X(y,\epsilon)\neq \emptyset$ we can pick an elememnt $w$ in the intersection of the two balls. Then, by triangle inequality, $$|z-y^1|\leq |z-w|+|w-y^1|<2\epsilon$$ Hence, $z\in B_X(y^1,2\epsilon)$ and $(1)$ is justified. Now, from the set $A_1$ we keep only $y^1$ and we throw all the other points. Now, by considering the set $\{y^1,...,y^n\}\setminus A_1$ if this is empty we will have $A\subseteq B_X(y^1,2\epsilon)$ and $\{y^1\}$ is an $2\epsilon-$separated subset of $A$. Otherwise, we pick a point $y\in \{y^1,...,y^n\}\setminus A_1$ and we define the set $$A_2=\{y^j:\, B_X(y^j,\epsilon)\cap B_X(y,\epsilon)\neq \emptyset\}$$ and apply the same procedure as before. Continuing this way, with a recursive argument we end up with a subset $F'\subseteq \{y^1,...,y^n\}$ such that $$A\subseteq \bigcup_{y\in F'}B_X(y,2\epsilon)$$ with the above union being disjoint, meaning that $|y-y'|\geq \epsilon$ for every $y\neq y'$ in $F'$. Hence, $F'$ is an $2\epsilon-$separated subset of $A$.
Now, that we justified that $2\epsilon-$separated subsets of $A$ exists, we can pick a random $F'=\{x^1,...,x^m\}\subseteq A,\ 2\epsilon-$separated subset of $A$, and a random $\{y^1,...,y^n\},\ \epsilon-$cover of $A$ (this time we dont assume that $\{y^1,...,y^n\}\subseteq A$ otherwise this restriction will violate the generality of the argument). We claim that $m\leq n$. If we prove this, we would have prove that $|F|\leq N_\epsilon(A,X)$. Now, $$A\subseteq \bigcup_{j=1}^{n}B_X(y^j,\epsilon)$$ Now, since every $x^k$ belongs to $A$ there will exists $y^k$ such that $x^k\in B_X(y^k,\epsilon)$. Define the function $\phi:\{x^1,...,x^m\}\to \{y^1,...,y^n\}$ by $\phi(x^k)=y^k$ for every $k=1,...,m$. Our claim is that $\phi$ is one to one. Indeed, if $\phi(x^k)=\phi(x^l)$ then, $x_k,x_l\in B_X(y_k,\epsilon)$. This implies that $$|x^k-x^l| \leq |x^k-y^k|+|x^l-y^k|<2\epsilon$$ and now using the fact that $|x^k-x^l|\geq 2\epsilon$ for every $k\neq l$ we end up with $x_k=x_l$. Hence, $\phi$ is one to one which means that $m\leq n$ as desired.