If $(X,d\_X)$ is a compact metric space, and $(Y,d)$ is another metric space. Moreover, suppose that the metric capacity of $(Y,d)$ is at-least that of $(X,d\_X)$, that is
$$
\kappa\_X(\epsilon)\leq \kappa\_Y(\epsilon) ; (\forall \epsilon \in (0,1]
.
$$


Here the metric capacity of a metric space $(X,d\_X)$ is defined by
$$
\kappa\_X(\epsilon)\triangleq\sup\left\{
k : \exists x\_0,\dots,x\_k \in X : \exists r>0, \sqcup\_{i=1}^k B(x\_i;\epsilon r)\subseteq B(x\_0;r)
\right\}
$$


Does there necessarily exist a *continuous* injection of $(X,d\_X)$ into a metric ball in $(Y,d)$ of some (finite) radius? 


**Note:**


* I'm not looking for Lipschitz, just continuous inclusion.
* $X,Y\neq \emptyset$.



Continuous inclusion of metric spaces of smaller capacity
The answer is **no**. You can have $X$ path connected and $Y$ totally disconnected satisfying your condition. Then any continuous map from $X$ to $Y$ is necessarily constant so there are no injective maps from $X$ to $Y$. 


Here is another example: $X=S^1$ is a unit circle of with the geodesic metric and $Y=\mathbb{R}$. Since $X$ and $Y$ are locally isometric, it follows that
$$
\kappa\_X(\epsilon)\leq \kappa\_Y(\epsilon) ; (\forall \epsilon \in (0,1]).
$$
Indeed, if $r\le\pi$ in the definition of $\kappa$ the corresponding quantities are equal (balls of radius $r$ are isometric balls). If $r>\pi$, then the quantity in the definition of $\kappa\_X(\epsilon)$ is less than the corresponding one in the definition of $\kappa\_Y(\epsilon)$ (ball in $Y$ is bigger - segment of length $2r$ than the ball in $S^1$ - the unit circle and they both are locally isometric).


However there is no injective map from $X=S^1$ to $Y=\mathbb{R}$. 


