I need to check if the set $$F = \{x \in C^{1}[a,b] :|x(a)| \leq B_0, |x'(t)| \leq B_1\} $$ is relatively compact and compact.
The relative compactness of this set can be easily shown using Arzelà-Ascoli Theorem. Then if this set is closed it is also compact. But how to show if this set closed or not closed?
I've tried to use definition of closed set (a closed set is a set whose complement is an open set). The complement of given set is the set $$F^{C} = \{x \in C^{1}[a,b] :|x(a)| \gt B_0 \lor |x'(t)| \gt B_1\} $$
But I don't understand how to choose $\varepsilon \gt 0$ for any $x_0 \in F^C$ such that, for any $y$, $d(x_0, y) < \varepsilon$, $y$ also belongs to $F^C$.
Or maybe there exists an example, which shows that this set is not closed?
Thank you for any help!
In general such a set is not closed (in the uniform metric). Consider, for example, the case $[a,b] = [-1,1]$, $B_0 = 2$, $B_1 = 1$, and the sequence $$ x_n(t) := \sqrt{t^2 + 1/n}, \qquad t\in [-1,1], \ n\in\mathbb{N}. $$ You have that $(x_n)\subset F$ converges uniformly in $[-1,1]$ to the function $x(t) := |t|$, but this function does not belong to $F$.