Show that a space of functions is not totally bounded.

Let $F = \{f\colon [0,1] \to \Bbb R\mid f \text{ is bounded }\}$; $d(f,g) = \sup|f(x) - g(x)|$ for $x \in [0,1]$. $F_1 = \{f\colon [0,1] \to \Bbb R:\mid |f(x)| \le 1, \forall x \in [0,1]\}$. Show that $F_1$ is not totally bounded.

I'm given the definition of a totally bounded space as:

A metric space $(X,d)$ is said to be totally bounded provided the following condition is satisfied. For each $\epsilon > 0$, there exists a finite set $F_\epsilon$ (called an $\epsilon$-net) contained in $X$ such that $X = \bigcup_{z \in F_{\epsilon}} N(z;\epsilon)$.

This is a homework question, so I would like the work to be my own, however I'm having a tough time knowing where to start. I've begun by assuming that $F_1$ is totally bounded, and hoping to find a contradiction but thus far have made little progress.

Any help is appreciated!

• Whats is the definition of $N$ if I may ask? – MoebiusCorzer Nov 4 '15 at 22:59
• @MoebiusCorzer presumably an open $\epsilon$-neighborhood of the point $z$. – Kevin Carlson Nov 4 '15 at 23:00
• @KevinCarlson Thank you. – MoebiusCorzer Nov 4 '15 at 23:00
• Yes, that is the epsilon neighborhood of a point z. – Iff Nov 4 '15 at 23:10

Some hints: you want to find an $\epsilon$ so that there's no such net, so pick your favorite $\epsilon$ (smaller than $1$!) Then given finitely many functions $f_1,...,f_n$ on the interval bounded between $-1$ and $1$, you need to produce $f_{n+1}$ not within $\epsilon$ of any of them, and still bounded between $-1$ and $1$. There aren't any continuity conditions on members of $F$, so you can construct $f_{n+1}$ arbitrarily: try a "diagonalization" process, where you make it differ from each of the other $f_i$ at a different point.