Is $S= \{x:x(\omega)\geq 1,x\in C(\Omega)\}$ closed or compact where $\Omega$ is a compact metric space?

Question

Suppose we equipped the Banach space $(C(\Omega),\|\cdot\|_\infty)$ with the norm topology(uniform topology) where $\Omega$ is a compact metric space. According to the definition of topology, $C(\Omega)$ is both a closed and open set at the same time. Moreover, $C(\Omega)$ is not compact because it is not even bounded, i.e. $d(f,0)$ is not bounded. (Am I right on the compactness?). I further consider a subset such that $S= \{x:x(\omega)\geq 1,x\in C(\Omega)\}$ where $\omega\in \Omega$ is a given element. Is the set closed or even compact? How about $S= \{x:x(\omega)\leq 1,x\in C(\Omega)\}$ and $S= \{x:x(\omega)= 1,x\in C(\Omega)\}$

Answer 1

$S = \{x: x(\omega) \ge 1\}$ is indeed closed in $C(\Omega)$ because the norm topology is finer than the pointwise (product) topology.

Note that $e_x: C(\Omega) \to \mathbb{R}$ defined by $e_x(f) = f(x)$, is a continuous map for any $x \in \Omega$, because again $|e_x(f) - e_x(g)| \le \|f-g\|$ and so $e_x$ is even uniformly continuous, and $S = e_\omega^{-1}[[1,\infty)]$, the inverse image of a closed set of the reals under a continuous function, hence closed.

Similarly $\{f: f(\omega) = 1\} = e_\omega^{-1}[\{1\}]$ is also closed.

The first is never bounded, as it contains all constant maps with values $c \ge 1$ which is an unbounded subset of $C(\Omega)$. The second one is never bounded if $\Omega$ has more than 1 point. (If $p \neq \omega$, let for each $c \in \mathbb{R}$, $f_c$ be a continuous function on $\Omega$ with $f_c(p) = c$ and $f(\omega) = 1$ (Urysohn functions will do, but it's even easier on a metric space domain), the $f_c$ are in $S$ and unbounded). In both cases $S$ is not compact.