Inexistence of a finite $\frac{1}{2}$-net

Question

Let $C\left( [0;1] \right)$ be the set of all continuous functions $f \colon [0;1] \to \mathbb{R}$, and $\| f \|_\infty := \sup\limits_{0 \leq x \leq 1} |f(x)| $ for all $f \in C\left( [0;1] \right)$. Show that in the normed space $\left( C\left( [0;1] \right), \| \cdot \|_\infty \right)$ , the closed ball $\overline{B}_1(0)$ (with center $g \equiv 0$, radius $1$) has no finite $\dfrac{1}{2}$-net!

Let $(X,d)$ be a metric space, $\varepsilon > 0$ and $A \subset X$. Then $A$ is said to be an $\varepsilon$-net if $ \bigcup\limits_{a \in A} B_{\varepsilon}(a) = X $.

I don't know how to prove this, can anyone help, please?

Answer 1

It's worth considering carefully what the ball you're trying to cover looks like. This ball $\overline{B_1} (0)$ is all the continuous functions $[0,1]\rightarrow \mathbb{R}$ that only take values within distance $1$ of $0$. In other words, your ball is all continuous functions $[0,1]\rightarrow [-1,1]$.

To show it has no finite $\frac{1}{2}$ net, you need to show that given any finite set $A = \{f_1, f_2, \dots, f_n\}$ of continuous functions $[0,1]\rightarrow [-1,1]$, there exists some continuous function $g: [0,1]\rightarrow [-1,1]$ such that $||g-f_i||_\infty > \frac{1}{2}$ for all $f_i$ (i.e., the function $g$ only has to take one single value far away from each $f_i$, of which there are finitely many).

For one possible way of doing this, click the spoiler below:

Pick $n$ distinct points $0 < x_1 < x_2 < \dots < x_n < 1$ in $(0,1)$, and let $y_i = f_i(x_i)$. We construct a function $g$ such that $|g(x_i)-f_i(x_i)| > \frac{1}{2}$ as follows: if $y_i \geq 0$, let $g(x_i) = -1$, and if $y_i < 0$, let $g(x_i) = 1$. We can extend this to a continuous function by just making $g$ a straight line from $g(x_i)$ to $g(x_{i+1})$ on the intervals $[x_i, x_{i+1}]$. The function $g$ does not lie in any $\frac{1}{2}$ net of $A$, and we are done. For a picture example, click the next spoiler.