Continuous functions on locally compact Hausdorff spaces Show every closed Ideal in $C_0(X)$ for a locally compact Hausdorff space X is of the form $K_F=\{f\in C_0(X)\colon f\lvert_F \equiv 0\}$ for some closed subspace $F\subseteq X$.
Here is what I have so far:
Let $I$ be a closed ideal.
I tried to use the Gelfand Transform.
Consider the evaluation map $ev_x: C_0(X) \rightarrow \mathbb{C}$ defined by $ev_x(f)= f(x)$ for all $x \in F$. Then we have $Ker (ev_x)= K_F$
From here, I'm pretty much lost what to do to get the result desired.
Also, I thought about using Urysohn's Lemma here: https://www.math.ksu.edu/~nagy/real-an/1-05-top-loc-comp.pdf . But still didn't anywhere.
Any help will be appreciated.
 A: Given a closed ideal $I\trianglelefteq  C_0(X)$, let
$$
  F=\{x\in X: f(x)=0,  \text{ for all } f\in I\},
  $$
and let us prove that $I=K_F$.
The inclusion  $I\subseteq K_F$ holds for a pretty elementary reason (which nevertheless sounds a bit like a tongue-twister):
every function in $I$ vanishes on any point where all functions in $I$ vanish.
The hard part is to prove the reverse inclusion $K_F\subseteq I$.   For this  pick  $f\in  K_f$.  Fixing  $\varepsilon >0$, recall that the set
$$
  C=\{x\in  X: |f(x)|\geq  \varepsilon \}
  $$
is compact.
Given  $x$ in $C$, we have that $f(x)$ is nonzero, so clearly $x$ is not in $F$.  Consequently not all functions in $I$
vanish at $x$, and hence we may choose  some element $a_x\in  I$, such that $a_x(x)\neq 0$.  The set
$$
  U_x=\{y\in X: a_x(y)\neq 0\}
  $$
is then an open neighborhood of $x$, and hence $\{U_x\}_{x\in C}$ is an open cover of $C$.  By compactness we may pick a
finite subcover $\{U_{x_1},U_{x_2},\ldots ,U_{x_n}\}$.  Next observe that
$$
  g:= \sum_{i=1}^n\overline {a}_{x_i}a_{x_i}
  $$
lies in $I$, and does not vanish on any point of $C$.
There are multiple endings for this story,  and perpahs the slickest way is as follows:  besides an ideal,  $I$ is a
C*-algebra and $g$ is a non-negative element in $I$, so $g^{1/p}$ lies in $I$  for every positive integer $p$.
Moreover the sequence $\{g^{1/p}\}_{p\in {\mathbb N}}$ converges uniformly to 1 on $C$.  It is also easy to see that
$$
  \|g^{1/p}\| =   \|g\|^{1/p} \leq  2,
  $$
for large enough $p$.
Therefore we may find  $p$ such that
$$
  |f(x)-g(x)^{1/p}f(x)| < \varepsilon , \quad\forall x\in  C,
  $$
and
$$
  |f(x)-g(x)^{1/p}f(x)|  \leq  \big (1+|g(x)^{1/p}|\big ) |f(x)|\leq  (1+2)\varepsilon ,
  $$
for every  $x\in  X\setminus C$.
Consequently $\|f - g^{1/p}f\|\leq  3\varepsilon $, so we deduce that $f$ lies in the closure of $I$, and hence also in $I$
because $I$ is closed.

EDIT. Here is another approach,  assuming we already know the answer in case $X$ is compact.
Assuming that $X$ is not compact, let $\hat X=X\cup \{\infty \}$
be the one-point compactification of $X$, and let
$$
  f\in C_0(X)\mapsto \hat f\in  C(\hat X)
  $$
be the natural identification.  One may then prove that $\hat I$ is a closed ideal in $C(\hat X)$, and hence, by the
compact case,
$$
  \hat I=\{g\in C(\hat X) : g|_G=0\},
  $$
where $G$ is a closed subset of $\hat X$.
Notice that necessarily $\infty \in G$, since otherwise one could use Urysohn to
produce some $g\in C(\hat X)$, vanishing on $G$, with $g(\infty )\neq 0$, but this would imply that $g$ lies in
$\hat I\setminus C_0(X)$, which is impossible.
Setting $F=G\cap X$, we have  that $F$ is a closed subset of $X$.   So, upon identifying $C_0(X)$ as the subalgebra of $C(\hat
X)$ formed by the functions vanishing at $\infty $, we have:
$$
  \hat I=\{g\in C(\hat X) : g|_G=0\} = $$$$ =
  \{g\in C(\hat X) : g|_F=0\} \cap  \{g\in C(\hat X) : g(\infty )=0\} = $$$$ =
  \{g\in C(\hat X) : g|_F=0\} \cap  C_0(X) = $$$$ =
  \{g\in C_0(X) : g|_F=0\} = K_F.
  $$
