# Why locally compact in the Gelfand representation?

I’m missing something in the Gelfand representation. Let’s just say $\mathfrak{A}$ is a Banach algebra. Then it’s a Banach space, and so we have $\mathfrak{A}^\ast$. The multiplicative linear functionals on $\mathfrak{A}$ are continuous and have norm $\leq 1$, and so still sit inside the unit ball of $\mathfrak{A}$, which is weak-$\ast$ compact. The set of multiplicative linear functionals are weak-$\ast$ closed, so the set of mulitplicative linear functionals is then weak-$\ast$ compact.

Where does not having a unit make the above not work? I always see that the multiplicative linear functionals are locally compact in the non-unital case, but I cannot spot where the unit is needed in the above.

#### Solutions Collecting From Web of "Why locally compact in the Gelfand representation?"

Assume $A$ is non-unital. Denote by $X(A)$ the space of its characters and denote
$$X(A)_+=X(A)\cup\{0\}$$
One can show that $X(A)_+$ is weak-$^*$ closed subset of weak-$^*$ compact unit ball of $A^*$. So $X(A)_+$ is weak-$^*$ compact.

One can show that $X(A)_+$ is a one point compactification of $X(A)$, so $X(A)$ is locally compact.