Proving noncompactnes by showing open cover with no finite subcover

Question

Define:

$$S = \{f \in C([0, 1],\Bbb R) : |f(x)| \le 1 \; \forall x \in [0, 1]\}$$

I have an open cover for the set $S$:

$$U_{n} := \{f \in C([0, 1],\Bbb R): |f(0) − f(1/n)| < 1\}$$

for each $n \in \Bbb N$.

Please help me out with showing that this doesn't have a finite subcover. I have the idea that I need to find a sequence of functions which are not covered by this open cover.

Answer 1

HINT: First you have to verify that $\{U_n:n\in\Bbb Z^+\}$ actually is an open cover of $S$, if you’ve not done so already; showing that it’s a cover requires using the fact that each $f\in S$ is continuous at $0$. To show that it has no finite subcover, it suffices to show that for each $m\in\Bbb Z^+$ there is a function

$$f_m\in S\setminus\bigcup_{n\le m}U_m\;.$$

In other words, $f_m$ must be in $S$, and it must satisfy

$$\left|f_m(0)-f_m\left(\frac1n\right)\right|\ge 1$$

for $n=1,2,\ldots,m$. I suggest trying to construct such an $f_m$ by setting $f_m(0)=0$ and making $f_m$ piecewise linear, with the value $1$ at $x=1,\frac12,\ldots,\frac1m$.

Answer 2

Take the piecewise functions $f_m$ whose graphs on $[0,1]$ is given by a line segment of slope $m$ from $(0,0)$ to $(1/m,1)$ and then a flat line from $(1/m,1)$ to $(1,1)$.

No matter how many (finitely many) $U_n$'s you take, simply pick an $m$ bigger than all of the chosen $n$'s, then $f_m$ won't be in your cover.