Let $X$ be locally compact Hausdorff space and $A=C_0(X),$ the space of continuous functions vanishing at infinity and $f \in A$. I'm trying to verify the following:
$\sigma_A(f),$ the spectrum of $f$ is compact.
My attempt:
I've shown that $\sigma_A(f)=f(X) \cup \{0\}.$ Now, it suffices to show that $f(X)$ is compact. Since, $f\in A,$ therefore $f(X)$ is bounded.
I'm having trouble showing that $f(X)$ is closed.
I took $\lambda \not \in f(X).$ I need to show that there exists a neighbourhood of $\lambda$ such that no element in that neighbourhood is in the range.
Any hints are appreciated.
Let $\lambda \in \overline{f(X)}.$ Then, there exists a sequence $(f(x_n))_n$ such that $f(x_n) \to \lambda.$ Let $\epsilon >0$ be given. Since $f \in A,$ there exists a compact set $K$ in $X$ such that $$|f(x)|\leq \epsilon$$ for all $x \in X\setminus K.$
Case 1: $K$ has infinite points of the sequence $(x_n)_n.$ Then there is a subsequence $(x_{n_k})_k$ of $(x_n)_n$ in $K.$ Hence, $f(x_{n_k})\in f(K)$ and $f(K)$ is closed(since, compact). Now, $f(x_{n_k}) \to \lambda.$ Therefore, $$\lambda \in \overline{f(K)}=f(K)\subseteq f(X)$$
Case 2: $K$ has only finite points of the sequence $(x_n)_n.$ Then there exists $n_0 \in \mathbb N,$ such that $x_n \in X \setminus K$ for all $n\geq n_0.$ Hence, $|f(x_n)|\leq \epsilon$ for all $n\geq n_0.$ It follows that $\lambda =0.$
We have thus obtained $$\overline{f(X)} \subseteq f(X) \cup \{0\}.$$
Lastly, for each $n \in \mathbb N,$ there exists a compact set $K_n$ in $X$ such that $$|f(x)|\leq \frac 1n$$ for all $x \in X\setminus K_n.$ Let $x_n \in X \setminus K_n.$ Then $|f(x_n)| \leq \frac 1n$. Hence, $|f(x_n)| \to 0.$ Therefore, $0 \in \overline{f(X)}.$
So,$$\overline{f(X)} = f(X) \cup \{0\}.$$ Hence, $\sigma_A(f)$ is closed. Since, it is also bounded, we may conclude that it is compact.