My teacher was explaining fractals to me and he said "clearly $H_{\delta}^{s}(F)$ is non-increasing with $s$". I do not understand how it is that he has come to this. Could somebody tell me please why it is that it is non-increasing?
I know this might be a silly question because he said clearly, but I do not see it.
First off, the phrases "clearly" and "it's obvious that..." should be banned from mathematics. These are basically mathematician for one of the following:
Never believe a mathematician who says that something is clear. It almost never is, and it is insulting to the student when it happens.
Recall that $$ \mathcal{H}^{s}_{\delta}(E) := \inf\left\{ \sum_{j=1}^{\infty} |E_j|^s : E\subseteq \bigcup_{j=1}^{\infty} E_j \text{ and } |E_j| < \delta \right\},$$ where $|E_j|$ is the diameter of $E_j$. Observe that for $s\ge 0$ and $\alpha < 1$, the function $s \mapsto \alpha^s$ is a decreasing function of $s$. From this, it follows that if $s < t$, then $$ \sum_{j=1}^{\infty} |E_j|^s \ge \sum_{j=1}^{\infty} |E_j|^t, $$ as long as $|E_j| < \delta < 1$. But then we have \begin{align} \mathcal{H}^{s}_{\delta}(E) &= \inf\left\{ \sum_{j=1}^{\infty} |E_j|^s : E\subseteq \bigcup_{j=1}^{\infty} E_j \text{ and } |E_j| < \delta \right\} \\ &\ge \inf\left\{ \sum_{j=1}^{\infty} |E_j|^t : E\subseteq \bigcup_{j=1}^{\infty} E_j \text{ and } |E_j| < \delta \right\} \\ &= \mathcal{H}^{t}_{\delta}(E) \end{align} whenever $s < t$ and $\delta < 1$. Hence $\mathcal{H}^{s}_{\delta}(E)$ is non-increasing in $s$ so long as $\delta < 1$. We can actually deal with $\delta \ge 1$ in most cases, but the arguments are slightly delicate, and rely on properties of the underlying space (I think that separability may be good enough, but don't quote me on that---if we assume $E\subseteq \mathbb{R}^n$, we should be okay).
On the positive side, we are often concerned with the asymptotic behaviour, i.e. with the limit $$\lim_{\delta\to 0} \mathcal{H}^s_{\delta}(E),$$ hence we can assume $\delta < 1$ with no loss of generality.