Showing a set is closed I have to show that the set $ A \subset C[0,1]$ defined by $A = [f \colon 0 \leq f(x) \leq 1 \forall x \in [0,1] ]$ is closed in the $||.||_\infty$ norm.
Now i know the definition for open/closed set etc, but im not very good (or confident) in applying them. (especially when it comes to closed sets, open sets i find easier). How would i go about doing this?
 A: Let $(f_n)\subset A:f_n\to f$. Then for every $n$, $x\in[0,1]$ we have that $0\leq f_n(x)\leq 1$,what do you think of $supf(x)$, for $x\in[0,1]$? will it be $\leq$ than $1$?
A: I shall prove that $A':=C[0,1]\setminus A$ is open. Consider an $f_0\in A'$. There is a point $\xi\in[0,1]$ where the condition $0\leq f(x)\leq1$ is violated. So there is an $h>0$ with $f_0(\xi)=-h$ or $f_0(\xi)=1+h$. Whenever $\|f-f_0\|_\infty<{h\over2}$ we know that $$f(\xi)<f_0(\xi)+{h\over2}=-{h\over2}<0$$ in the first case, and that $$f(\xi)>f_0(\xi)-{h\over2}=1+{h\over2}>1$$ in the second case. It follows that $f\in A'$.
A: Show that for $x\in[0,1]$ the evaluation map $e_x\colon C[0,1]\to\mathbb R$, $e_x(f)=f(x)$ is continuous. Then write $A=\bigcap_x e_x^{-1}[[0,1]]$.
A: First note that since $f \in C[0,1]$ it follows from the Extreme value theorem that $f$ achieves its maximum on $[0,1]$ at least once.
Take $\{f_{n}\}_{n}^{\infty} \subset A$ such that $f_{n} \rightarrow f$ with respect to the norm $||\frac{}{}||_{\infty}$. We can prove by contradiction, so assume $f \notin A$. It follows then that either $||f||_{\infty} > 1$ or $||f||_{\infty} < 0$. 
Case 1:
There exists a $x \in [0,1]$ such that $f(x) = c$ where $c > 1$. We can take $||f||_{\infty} = c$ since we know that $f$ achieves its maximum on $[0,1]$. we also know that $0 \leq ||f_{n}||_{\infty} \leq 1$ since $\{f_{n}\}_{n}^{\infty} \subset A$.
Using the reverse triangle inequality we get: $c-1 \leq | ||f_{n}||_{\infty}-||f||_{\infty}| \leq ||f_{n}-f||_{\infty}$. This contradicts assumption that $f_{n} \rightarrow f$ in $||\frac{}{}||_{\infty}$.
Case 2:
$||f||_{\infty} = d < 0$. It follows then that $f(x) < d\text{    }$ for all $x \in [0,1]$. We once again note that $0 \leq ||f_{n}||_{\infty} \leq 1$.
Using the definition of convergence we can take $\epsilon := d$ and it the follows that for any $N \in \mathbb{N}$ and $n \geq N$ that $\text{sup}\{|f_{n}(x)-f(x)|: x \in [0,1]\} \geq \epsilon$. This contradicts the assumption that $f_{n} \rightarrow f$ in $||\frac{}{}||_{\infty}$ 
$\therefore f \in A$
Conclusion: $A$ is a closed subset.   
A: Maybe this will help you understand this, and similar, questions: The $||\cdot||_\infty$ norm is also called the "uniform convergence" norm, for the following reason: A set $A \subset C[0, 1]$ is closed in the $||\cdot||_\infty$ norm if and only if for any uniformly convergent sequence $f_n \in A$, we have $\lim f_n \in A$.
Using this, your question reduces to the following simple fact: A uniformly convergent sequence of bounded functions converges to a function with the same bounds. (In fact, pointwise convergence would be enough for this. Think about it.)
Proof: (Mathematicians would regard this fact as a tautological truth, evident from the definitions, but I am writing it explicitly so you can see what's going on.)
First direction: Suppose $A$ is closed. Let $f_n \in A$ be a uniformly convergent sequence and denote its limit by $f = \lim f_n$.
Then for any $\epsilon > 0$ there is some $N$ such that for all $n > N$, for all $x \in [0, 1]$, $|f_n(x) - f(x)| < \epsilon$.
This means for any $\epsilon > 0$ there is some $N$ such that for all $n > N$, $\sup_{x\in[0,1]} |f_n(x) - f(x)| < \epsilon$.
Therefore for any $\epsilon > 0$ there is some $N$ such that for all $n > N$, $||f_n -f||_\infty < \epsilon$.
Finally, this means $\lim_{n\to\infty} ||f_n - f|| = 0$, which by the closedness of $A$ under the $||\cdot||_\infty$ norm, means $f \in A$.
Second direction: Suppose any $f_n \in A$ uniformly convergent sequence yields $\lim f_n \in A$; we need to show $A$ is closed.
To show $A$ is closed we need to show that for any $f_n \in A$ that has a limit in the $||\cdot||_\infty$ norm, the limit is in $A$.
Let $f$ be a $||\cdot||_\infty$ limit of $f_n \in A$. This means $||f_n -f||_\infty \to 0$.
Therefore [argument omitted... just read the statements of "first direction" in reverse order...]
Therefore $f_n$ uniformly converges to $f$, and by the assumption $f \in A$. So, $A$ is closed.
