In trying to understand the definition of Hausdorff dimension I'm trying to get a good understanding of Hausdorff measure. On wikipedia they first make the following definition (found here):
Let $(X,d)$ be a metric space, $S$ be any subset of $X$ and $\delta>0$ be a real number, then define
$H_{\delta }^{d}(S)=\inf {\Bigl \{}\sum _{{i=1}}^{\infty }(\operatorname {diam}\;U_{i})^{d}:\bigcup _{{i=1}}^{\infty }U_{i}\supseteq S,\,\operatorname {diam}\;U_{i}<\delta {\Bigr \}}$.
Then we take $\lim_{\delta\to 0} H_{\delta }^{d}(S)$ as a definition of $d$-dimensional Hausdorff measure.
What I am stuck on right now is the motivation for this limit that they give saying:
"Note that $H_{\delta }^{d}(S)$ is monotone decreasing in $\delta$, since the larger $\delta$ is, the more collections of sets are permitted, making the infimum smaller."
As I understand it we are taking the infimum over a set of sums of all the diameters of each set in a cover. Now say we increase $\delta$ so that we can cover $S$ with a $\delta$-ball. Then how is increasing $\delta$ even more going to permit a cover with an even smaller "total diameter" than before? I might have forgotten or misunderstood something trivial but I don't see why the infimum would necessarily change by allowing "bigger" sets in the covers.
In conclusion my question is:
Q: Will someone give an explanation to why the statement made in the above quotation is true (and/or preferably give some intuition for it)?
(My level is undergraduate.)
Let $\delta_1 < \delta_2$.
The family $\mathcal{U}_{\delta_2}=\left\{ U \mid \bigcup _{i=1}^{\infty }U_{i}\supseteq S,\,\operatorname{diam}(U_{i})<\delta_2 \right\}$ contains the family $\mathcal{U}_{\delta_1}=\left\{ U \mid \bigcup _{i=1}^{\infty }U_{i}\supseteq S,\,\operatorname{diam}(U_{i})<\delta_1 \right\}$. Indeed, is $U \in \mathcal{U}_{\delta_1}$, then $\operatorname{diam}(U) < \delta_1 < \delta_2$, so $U \in \mathcal{U}_{\delta_2}$.
Then, taking the infimum over $\mathcal{U}_{\delta_2}$ leads to something possibly smaller than taking it over $\mathcal{U}_{\delta_1}$ (this is just set theory).
The last sentence can be reread as $H_{\delta_2 }^{d}(S) \leq H_{\delta_1 }^{d}(S)$, so we have proved that, given $\delta_1 < \delta_2$, we have $H_{\delta_1 }^{d}(S) \geq H_{\delta_2 }^{d}(S)$, so $H_{\delta}^{d}(S)$ is decreasing in $\delta$.