Prove that there is a bounded separable metric space such that for any $0 < ε < \text{diam }X$ and any $s > 0$, $H^s_ε(X) = ∞.$
Here ${\displaystyle H_{\delta }^{d}(S)=\inf \left\{\sum _{i=1}^{\infty }(\operatorname {diam} U_{i})^{d}:\bigcup _{i=1}^{\infty }U_{i}\supseteq S,\operatorname {diam} U_{i}<\delta \right\}}$ if we take $\delta \to 0$ then we get $ H^{d}(S)$ which is the Hausdorff measure of $S$ in the Hausdorff dimension $d$(definition).
Now my question probably doesn't need any knowledge of Hausdorff measure. Let me include my try:
I am trying to create a separable metric subspace in $l^{\infty}$[ observe that $l^{\infty}$ is not seperable] so that that whenever I take an open cover there will be basis under each open set of the cover and I will try to show that their sum would be infinity.
Now the easiest construction of the separable set is any countable subset of $l^{\infty}$. Initially I was thinking to take $S=\{(a_n)| \text{ $a_n$ is zero at all co-ordinates except $n$ where it is 1}\}_{n=1}^{\infty}$. What I observed that $S$ is countable but it is discrete so the open covers will eventually give me diameter zero. Then I tried to improve my set $T=\left\{B(x,r_k)|r_k=\frac 1k, x\in S\right\}$. $T$ is separable and all covers of $T$ will be discrete except( and hence of diameter zero) the open sets that will contain the points of $S$ let us name those open sets by $U_n$ i.e $a_n\in U_n$. Now can I set a lower bound of $\text{diam }(U_n)$ so that the countable sum would be $\infty$? If not can you help me answer the question with an example?