I've started to learn about spectral theory and I'm looking at some examples. The spectrum is defined as $$ \sigma(a) = \{\lambda \in \mathbb C | a-1\lambda \text{ is not invertible} \}$$
If $A= C(X)$ is the algebra of continuous functions on a compact Hausdorff space then $\sigma(f) = f(X)$. That's clear to me. But why is the spectrum of $a$ if $A = \ell^\infty (X)$ the space of bounded sequences $\sigma(a) = \overline{a(X)}$ (closure in $\mathbb C$)? Why is it the closure?
Because the spectrum is necessarily a closed set, and so $\overline{a(X)} \subset \sigma(a)$. Furthermore, if $\lambda \notin \overline{a(X)}$, then $\exists \epsilon > 0$ such that $$ |a(x) - \lambda| > \epsilon $$ Hence, $(a-\lambda)^{-1} \in \ell^{\infty}(X)$ (since it is bounded by $1/\epsilon$). Hence, $\lambda \notin \sigma(a)$. Hence, $\sigma(a) = \overline{a(X)}$.