In a calculus of variations problem, I am given $V = \mathcal C([0,1], \mathbb R)$ to be a normed function space. I want to show that the set $$A = \{x\in V \quad \vert \quad x(0) = x(1) = 0, \quad \max_{s\in [0,1]} ||x(s)|| \leq 1\}$$ is closed (and also be able to show it for different sets).
Unfortunately my background in real analysis is quite weak. I have read about how to show this, particularly showing that $A$ is closed iff it contains all its limit points. But I still do not really understand the technical procedure.
Can someone guide me through it?
Since i can not comment, i will write is as a post.
One of the ways to prove a set is closed is to show that the set contains its limit points. Assuming (because you did not mention it) that the norm is $\|x\|=\max_{s \in [0,1]} |x(s)|$
Let $\{x_n\}$ be a sequence in $A$ that converges (w.r.t $\|.\|$) to $x \in V$. We want to show that $x \in A$. Since $x_n \in A$ for every $n \in \mathbb N$ then we have $x_n(0)=x_n(1)=0$ and $\|x_n\|=\max_{s\in[0,1]} | x_n(s)| \leq 1$.
Fix $\epsilon >0$. Then there is $N>0$ such that for $ n >N$
$$|x_n(0) -x(0)| \leq \max_{s \in [0,1]}|x_n(0) -x(0)| \leq \|x_n - x\| \leq \epsilon, $$
i.e $\lim_{n \to \infty} |x_n(0) -x(0)|=0$, implies $\lim_{n \to \infty} x_n(0)=x(0)=0$.
Similarly, we get $\lim_{n \to \infty} x_n(1)=x(1)=0$.
The show that $x$ is in the unit ball, we argue as follows: for $n >N$ we get
$$\|x\| \leq \|x-x_n\| + \|x_n\| < \epsilon +1$$
since $\epsilon$ was chosen arbitrarily small, then we get $\|x\| \leq 1$. Therefore, $x \in A$, thus $A$ is closed in $V$.