Here is the exercise that I got.

Verify that the class $\mathcal{G} = \left\{ g: \mathbb{R} \to [0,1], g \text{ is increasing}\right\}$ is not totally bounded for the supremum norm on $\mathbb{R}$.

I am trying to prove this by constructing a counterexample (for a given $\epsilon > 0$ and some functions $g_1,\cdots,g_n$), but I don't know how. The rough idea I have is that, to construct $f$ such that $\|f - g_i\|_\infty > \epsilon$ for all $g_i$'s. So maybe only at one point $x_i$, $f$ and $g_i$ are far away. For example, maybe at point $x_1$, $f$ is only close to $\max_i g_i$ and far away from $\min_i g_i$, but at another point $x_2$, $f$ is only close to $\min_i g_i$ but far away from $\max_i g_i$, but I don't know how to formalize this, partly because I don't know how far is $\max_i g_i$ from $\min_i g_i$.

The picture in my head is that, if $g_i$ is like a increasing straight line, then I can consider $f = \frac{1}{2}$, so $f$ and $g$ will be far away when $x$ is big or small. If $g$ is quite flat, I can take $f$ to be an increasing line.

I am not sure if I am thinking correctly and how to proceed. Could someone give me a hint?


The below will work for functions into $[-\pi/2,\pi/2]$. You would just need to shift and stretch a little for your case. Recall that $\arctan(x)$ is strictly increasing and ranges in $(-\pi/2,\pi/2)$. Replacing $x$ by $\alpha x$ for $\alpha>1$ does a horizontal compression. So for large $\alpha$, $\arctan(\alpha x)$ will approach its asymptotes faster despite still starting off at $(0,0)$.

The sequence $f_n(x) = \arctan(nx)$, $n\geq 1$ is contained in $\mathcal{G}$ but can't be contained in finitely many balls of radius $\frac{\pi}{6}$, $B(g,\frac{\pi}{6})$. If they were, then all $f_n$'s would be within distance $\frac{\pi}{3}$ of the set $\{f_{n_1},\ldots,f_{n_k}\}$ for some finite indices $n_1 < \ldots < n_k$. But (feel free to check) that if you let $n >> n_k$, then the distance will approach $\pi/2$.


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