# Exterior Covering Number of $\epsilon /2$ is greater of equal than Covering Number of $\epsilon$

Definition $$(\epsilon -Net)$$ :let $$(T,d)$$ be a metric space .Consider a subset $$K \subset T$$ and let $$\epsilon >0$$, A subset $$N \subset K$$ is called $$\epsilon -Net$$ of $$K$$ if every point in $$K$$ is within a distance $$\epsilon$$ if every point in K is within a distance $$\epsilon$$ of some point of $$N$$,i.e.

$$\forall x \in K \ \ \ \exists x_0 \in N \ : d (x,x_0) \leq \ \epsilon$$

Definition(Covering Number): For metric space $$(T,d)$$ The covering number of $$K \subset T$$ respect to a given $$\epsilon \geq 0$$ ,denotes as $$N(K,d,\epsilon )$$, is the smallest possible cardinarity an $$\epsilon -Net$$ of K , or equivalently ,is the smallest number of closed balls with centers in $$K$$ and radii $$\epsilon$$ whose union covers $$K$$

Definition(Exterior Covering Number): For metric space $$(T,d)$$ The exterior covering number of $$K \subset T$$ respect to a given $$\epsilon \geq 0$$ ,denotes as $$N^{ext}(K,d,\epsilon )$$, is the smallest number of closed balls with centers not necessary in $$K$$ and radii $$\epsilon$$ whose union covers $$K$$

then I was asked to prove that:

$$N(K,d,\epsilon) \leq N^{ext} (K,d,\epsilon /2)$$

how to see that ?

here is my attempt: since each $$\epsilon -ball$$ in $$N(K,d, \epsilon)$$ should intersect at least one $$\epsilon /2 -ball$$ in $$N^{ext}(N,d,\epsilon/2)$$ ,thus I am trying to show that by contradiction: for each two distinct $$\epsilon -ball$$ in $$N(K,d, \epsilon)$$ ,the $$\epsilon /2 -balls$$ they intersect must contain a distinct one.

Suppose that $$\bar B(x_1, \epsilon / 2), \ \dots, \ \bar B(x_{N^{\rm ext}}, \epsilon / 2)$$ is an external covering of $$K$$ of minimal size.
For each $$i \in \{ 1, \dots, N^{\rm ext}\}$$, there exists a $$k_i \in K$$ that is contained in $$\bar B(x_i, \epsilon / 2)$$. (Otherwise $$\bar B(x_i, \epsilon / 2)$$ would be redundant, contradicting the minimality of size of the covering).
By the triangle inequality, $$\bar B(x_i , \epsilon / 2) \subseteq \bar B(k_i, \epsilon),$$ for each $$i$$, which means that $$\bar B(k_1, \epsilon ), \ \dots, \ \bar B(k_{N^{\rm ext}}, \epsilon)$$ is an internal covering of size $$N^{\rm ext}$$. Hence the smallest internal covering has size at most $$N^{\rm ext}$$.