Finding the Spectrum of an element of $\ell^\infty$ I have done Q.1.2.1 already and it is quite clear.But I am not sure about the next one.

How does the closure come into the picture. I have a feeling that it should be $f(S)$ only.
Am I missing something? Kindly Help!!!
Thanks & Regards in advance
 A: The spectrum of $f\in\ell^\infty(S)$ definitely has to include (at least some) limit points of $f(S)$.  Consider for example $S=\mathbb N$, and $f\in\ell^\infty(\mathbb N)$ given by $f(n)=\frac{1}{n}$.  Then $0\notin f(\mathbb N)$, but $f$ is not invertible in $\ell^\infty(\mathbb N)$, so $0$ must be in $\sigma(f)$. 
As for the proof of the claim, here is a guideline for which you can fill in the details:
Fix $f\in\ell^\infty(S)$.  If $\lambda\notin\overline{f(S)}$,  then there is some $\varepsilon>0$ such that $|f(s)-\lambda|\geq\varepsilon$ for all  $s\in S$.  Then show that $g:S\to\mathbb C$ defined by $g(s)=\frac{1}{f(s)-\lambda}$ is in $\ell^\infty(S)$, and an inverse to $f-\lambda$.  
For the other direction, suppose $\lambda\in \overline{f(S)}$.  Then for each $\varepsilon>0$, there is some $s\in S$ such that $|f(s)-\lambda|<\varepsilon$.  Then either $g$ as given above, is not well-defined, or has norm $>\frac{1}{\varepsilon}$, and since $\varepsilon>0$ was arbitrary, we have $g\notin\ell^\infty(S)$.
