Before I come to my actual questions, I want to give some context (e.g. definitions, …).
(1) Let $X$ be a normed space over a field $\mathbb{K} \in \{\mathbb{R}, \mathbb{C}\}$. Then we define $ X' := \{ \varphi: X \to \mathbb{K} \ \text{is linear and continuous} \} $ which is called the topological dual space of $X$.
(2) On $X'$ we can define the weak* topology which is characterized as the weakest topology on $X'$ which makes the functionals $ x': X' \to \mathbb{K},\ x'(\varphi):=\varphi(x) $ continuous (for all $x \in X$).
(3) Alaoglu's theorem shows that the closed unit ball $B_{X'} := \{\varphi \in X' : \lVert \varphi \rVert \leq 1 \}$ is compact regarding the weak* topology. This is one of the great advantages of the weak* topology over the norm topology because otherwise the closed unit ball would be compact only in the finite-dimensional case.
(4) Now let $A$ be a commutative unital Banach algebra over $\mathbb{C}$. Then we define
$$ \Gamma_A := \{\varphi:A\to \mathbb{C},\ \varphi \ \text{is linear and multiplicative with}\ \varphi(1_A) = 1 \} $$
which is called the Gelfand space of $A$. It is easy to show that $\Gamma_A \subset B_{A'}$.
My goal: I want to show that $\Gamma_A \subset A'$ is a (weak*-)compact Hausdorff space. (Later I want to show the Gelfand-Naimark theorem but for now, this is enough).
I was able to show the Hausdorff part. For the compact part, I can use Alaoglu's theorem which leaves me with the task to show that $\Gamma_A$ is a (weak*-)closed subset of $A'$.
Attempt 1: My first naive approach to showing that $\Gamma_A$ is closed w.r.t. the weak* topology was to show that for a sequence $(\varphi_n) \subset \Gamma_A$ which converges to some $\varphi \in A'$ we already have $\varphi \in \Gamma_A$. But then I figured out that this approach which works for metric spaces does not work for topological spaces.
Attempt 2: Since I only have very rudimentary knowledge of topology, I tried to do this without sequences and in some textbook I found that you can write
$$
\Gamma_A := \bigcap_{x,y\in A} \{\varphi \in A' : \varphi(xy) = \varphi(x) \varphi(y) \} \cap \{\varphi: \varphi(1_A) = 1 \}.
$$
However, I can't figure out why those sets are closed under the weak* topology. Maybe I can use that the functions $x'$, $y'$ and $(xy)'$ are continuous? But then I would need that sums and scalar products of continuous functions are also continuous but I am not sure if this is true in this topological setting.
Question: Why is the Gelfand space $\Gamma_A$ closed with respect to the weak* topology?
It would be nice if the answer would need as little understanding of topological spaces as possible.
Best Answer
I will go with the following solution:
We can write \begin{align*} \Gamma_A &= \bigcap_{x,y\in A} \{ \varphi \in A' : \varphi(xy) = \varphi(x) \varphi(y) \} \cap \{\varphi: \varphi(1_A) =1 \} \\ &= \bigcap_{x,y \in A} \{ \varphi \in A': (xy)'(\varphi) - x'(\varphi) y'(\varphi) = 0 \} \cap \{\varphi: 1_A'(\varphi)=1 \}. \end{align*}
Now, notice that $x'$, $y'$, $(xy)'$ and $1_A'$ are (by definition) continuous functions from $A'$ to $\mathbb{K}$. Since $$ + : \mathbb{K} \times \mathbb{K} \to \mathbb{K},\ (x,y)\mapsto x+y \quad \text{and} \quad \cdot: \mathbb{K} \times \mathbb{K} \to \mathbb{K},\ (x,y)\mapsto x \cdot y $$ are continuous functions, we know that $(xy)' - x' \cdot y'$ is also a continuous function.
Thus, $\Gamma_A$ is closed as the intersection of closed sets.