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

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.


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 $x=1,\frac12,\ldots,\frac1m$.

  • $\begingroup$ Can you please elaborate on the part of how to use the continuity of $f$ at $0$ to show that the given cover is actually open? $\endgroup$ – Silver moon Feb 11 '15 at 6:49
  • $\begingroup$ @Martin: Continuity at $0$ is used to show that it’s a cover, not to show that each $U_n$ is open. To show that each $U_n$ is open, you need to specify the topology on $S$. Are you using the topology of pointwise convergence? $\endgroup$ – Brian M. Scott Feb 11 '15 at 6:53
  • $\begingroup$ $S$ is a metric space with induced metric of sup norm. However, can you please suggest how to use continuity argument to show U is an open cover? $\endgroup$ – Silver moon Feb 11 '15 at 6:56
  • $\begingroup$ @Martin: For any $f\in S$, $\lim_{n\to\infty}f\left(\frac1n\right)=f(0)$, so $$\left|f(0)-f\left(\frac1n\right)\right|<1$$ for all sufficently large $n$. $\endgroup$ – Brian M. Scott Feb 11 '15 at 6:59
  • $\begingroup$ ok Brian, so by continuity at $0$, we can pass to the limit of $f(1/n)$, I understand this part, but my question is how do we say that this is an open cover? $\endgroup$ – Silver moon Feb 11 '15 at 7:02

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.

  • $\begingroup$ Thanks Oxeimon, your answer is same as Brian's $\endgroup$ – Silver moon Feb 11 '15 at 8:32
  • $\begingroup$ we literally finished writing them like 5 seconds apart. $\endgroup$ – oxeimon Feb 11 '15 at 17:37

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