Prove that set is closed and ideal of another set

Question

Let $(X; d)$ be a compact metric space. For a given $x_0 \in X$, define $C_{x_0}(X,\mathbb{R})$ by

${C}_{x_0} = \{ f \in \mathcal{C}(X,d): f(x_0) = 0 \}$.

I need to prove that $C_{x_0}(X,\mathbb{R})$ is closed in $\mathcal{C}(X,\mathbb{R})$ and that $C_{x_0}(X,\mathbb{R})$ is an ideal of $\mathcal{C}(X,R)$. How do I prove this?

Answer 1

Consider the linear functional $\ell$ given by $\ell(f) = f(x_0)$. Then $\ell$ is continuous and $C_{x_0} = \ell^{-1}(\{0\})$ is closed as the preimage of a closed set under a continuous map.

For the ideal property first note that $C_{x_0}$ is a subspace of $C(X, \mathbb{R})$ (by the point-wise definition of addition and scalar multiplication). Furthermore, we have $(fg)(x_0) = (gf)(x_0) = f(x_0)g(x_0) = 0$ for all $f \in C_{x_0}$ and $g \in C(X, \mathbb{R}$).

Answer 2

Hint for the ideal property use the pointwise ring structure of $C(X,\Bbb R)$ to deduce that $C_{x_0}(X,\Bbb R)$ is closed under addition and closed under multiplication with arbitrary elements of $C(X,\Bbb R)$.

Hint I guess the topology on $C(X,\Bbb R)$ is given by supnorm $\Vert \cdot \Vert_\infty$. Then the map $\operatorname{ev}_{x_0}: C(X,\Bbb R) \rightarrow \Bbb R, f \mapsto f(x_0)$ is continuous, since the preimage of some interval $(t-\xi,t+\xi)$ is given by $$\operatorname{ev}_{x_0}^{-1}((t-\xi,t+\xi))=\{f \in C(X,\Bbb R) \mid f(x_0) \in (t-\xi,t+\xi)\}$$ and every such $f$ satisfies that every $g$ with $\Vert f-g\Vert_\infty < \min(\vert f(x_0)-(t-\xi)\vert, \vert f(x_0)-(t+\xi)\vert)$ is also in $\operatorname{ev}_{x_0}^{-1}((t-\xi,t+\xi))$. So any such $f$ has an open neighborhood in $\operatorname{ev}_{x_0}^{-1}((t-\xi,t+\xi))$, which shows that $\operatorname{ev}_{x_0}$ is continuous.