Let $B \subset [1,2]$. Define the set $E \subset C[1,2]$ by

$$E = \{x \in C[1,2]: |x(t)| < 1 \text{ if } t\in B\}$$

Prove that if $B$ is closed in $[1,2]$, then $E$ is an open set in $C[1,2]$.

I'm completely stumped on this. My instinct is to show that for every $\epsilon > 0$, there exists a d $\delta_\epsilon > 0$, such that when we have $t\in V_1(t_o; \delta)$ — a ball centered at $t_o$ radius $\delta$ — then we must have $x(t) \in V_C(t_o; \epsilon$). Here I denote $V_1$ to be the normal ball, or $\delta$-neighborhood, in $\mathbb{R}^1$ and $V_C$ to be the ball with radius defined by the metric

$$d(x,y) = \max_{t\in [1,2]}|x(t) - y(t)|$$

Roughly speaking, I think this means that when this metric is less than $\epsilon$ like

$$d(x,y) = \max_{t\in [1,2]}|x(t) - y(t)| < \epsilon \tag{A}$$

we can select an open set from $B$ that somehow shows that (A) is satisfied. I'm not really sure how to connect the dots on this. Especially since $B$ could be a set of discrete points, based on the given, even a Cantor-type set.

Here's my intuition. No matter how close we make two continuous functions in $E$, we can select from the subset of $B$ on which they are forced to abide by, an open set to produce a function that's "in between" the two continuous functions under consideration. Thus, there is an infinite number of such "in between" functions and there is no function that's the "farthest function", just like there's no largest element of $[0,1)$. Sorry for the ramble. Am I thinking about this correctly? And can I get some hints? Thank you!

  • 1
    $\begingroup$ Hint: $|u|_{|B}$ is continuous on a compact set so reaches a maximum. If $u \in E$, said maximum is lower than $1$, so any small perturbation of $u$ will be in $E$. $\endgroup$ – Mindlack Feb 13 at 13:42

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