Consider the following theorem:
Let $A$ be a complex non-unital commutative Banach algebra and let $\Omega (A)$ denote its Gelfand spectrum / character space. Then $\Omega (A)$ is locally compact.
I don't understand the following proof:
It is easy to check that $\Omega (A) \cup \{0\}$ is weak star closed in the closed unit ball of $A^\ast$. Since the closed unit ball is weak star compact, $\Omega (A) \cup \{0\}$ is weak star compact and therefore $\Omega (A)$ is locally compact.
First of all, why do we need to consider $\Omega (A) \color{red}{\cup \{0\}}$ if we want to show that $\Omega(A)$ is locally compact? The other step I don't understand is: how does it follow from $\Omega (A) \cup \{0\}$ is weak star compact that $\Omega (A) $ is locally compact? Is it true that a compact space minus one point is locally compact?
We do not need to consider $\Omega(A)\cup \{0\}$, but it makes things easier, since that is a compact (Hausdorff, if that isn't part of your definition of compact) space.
Every open subspace of a compact (Hausdorff) space is locally compact, and since the weak* topology is Hausdorff, singletons are closed, whence
$$\Omega(A) = (\Omega(A)\cup \{0\}) \setminus \{0\}$$
is seen to be locally compact as an open subspace of the compact space $\Omega(A)\cup\{0\}$.