# Show that the space of increasing, bounded function is not totally bounded w.r.t. $\sup$-norm

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 $$\pi/2$$.