Totally bounded space Suppose $M=\left \{ f\in L^1([0,1])\, |\, 0<f(x)<\frac 1{\sqrt x}   \text{almost everywhere on} \, (0,1) \right\}$.
Is it true or not, that $M$ is totally bounded?
 A: It is not totally bounded.
For an abstract reason why not, note that $M$ contains a ball of the $L^\infty$ norm.  If $M$ were totally bounded then the inclusion $L^\infty([0,1]) \hookrightarrow L^1([0,1])$ would be compact.  But it isn't.
We can use the example given there to make a concrete proof.  Let $e_n(x) = \sin(2 \pi n x)$.  It's well known, and simple to check, that the $e_n$ are orthogonal in $L^2([0,1])$,  i.e. $\int_0^1 e_n(x) e_m(x)\,dx = 0$ for $n \ne m$.  We also have $\int_0^1 e_n(x)^2\,dx = \frac{1}{2}$.  We can then observe that, for $n \ne m$, $\int_0^1 (e_n(x) - e_m(x))^2\,dx = 1$.
Now, noting that $\frac{1}{2}|e_n - e_m| \le 1$, we have $\frac{1}{2}|e_n - e_m| \ge \frac{1}{4}|e_n - e_m|^2$.  Thus, for $n \ne m$, we have
$$\frac{1}{2} \int_0^1 |e_n(x) - e_m(x)|\,dx \ge \frac{1}{4} \int_0^1 (e_n(x)-e_m(x))^2 \,dx = \frac{1}{4}.$$
So we have shown $\lVert e_n - e_m \rVert_1 \ge 1/2$.
Set $f_n = \frac{1}{2} + \frac{1}{4} e_n$.  Then $\frac{1}{4} \le f_n \le \frac{3}{4}$ so $f_n \in M$, and we have $\lVert f_n - f_m \rVert_1 \ge \frac{1}{8}$ for all $n \ne m$.  Thus if we have any cover of $M$ by balls of some radius $\epsilon < \frac{1}{16}$, each ball can contain at most one of the $f_n$, and hence the cover is not finite.
A: Another useful set of functions in $M$ are the 'square waves'. Let $f_1$ be the indicator function of $\cup_{i=0}^{\infty} [i, i+\frac{1}{2}]$, and define $f_{n+1}(x) = f(2^n x)$. It is simple to show that $||f_n - f_m|| = \frac{1}{4}$ as long as $m\neq n$.
