Show $A = \{f \in C([0,1], R) \ | \max f(x) \le 1 \}$ is closed.
I tried to show $A$ is compact, which is not compact; and I tried to show by definition of closed set but failed.
I am thinking of showing the limit is exist in the set but I am not sure how to show the limit exist first and then show it is in the set.
$\endgroup$ 33 Answers
$\begingroup$Let $f$ be a limit point of $A$. There exists a sequence $f_n$ in $A$ such that $f_n \to f$ uniformly on $[0,1]$. Since $\max f_n(x) \le 1$, $f_n(x) \le 1$ for all $x$ and $n$; taking the limit yields $f(x) \le 1$ for all $x$. Hence $\max f(x) \le 1$. Furthermore, since $f$ is the uniform limit of a sequence of continuous functions, $f$ is continuous. Therefore $f\in A$. This shows that $A$ is closed.
$\endgroup$ 4 $\begingroup$Let $f \in \bar A$. Then, there's a sequence $\{f_n\}_{n\ge0} \subset A$ such that: $f_n \rightarrow f$.
As $\{f_n\}_{n\ge0} \subset A$: $f_n(x) \le 1$, $\forall$ $n \ge 0$ and $x \in [0,1]$.
This gives: $$\lim_{n\to\infty}f_n(x) \le 1, \ \forall \ x \in [0,1]$$
That's: $$f(x) \le 1, \ \forall \ x \in [0,1]$$
Therefore $f \in A$. This shows $\bar A \subset A$.
$\endgroup$ 1 $\begingroup$Here is another approach:
Let $\phi(f) = \max \{ f(x) : x \in [0,1] \}$.
Prove that $\phi$ is a continuous functional $C([0,1], \mathbb R) \to \mathbb R$.
Note that $A=\phi^{-1}(-\infty,1]$ and conclude that $A$ is closed.
