How to prove that $\Delta( A)$ with the Gelfand topology is compact and Hausdorff?

Question

How do I prove:

$\Delta( A)$ with the Gelfand topology is compact and Hausdorff.

I've tried proving it closed, but I'm having difficulties with how to begin writing a proof. And I have no idea how to prove the Hausdorff condition.

Answer 1

I take it $A$ is a Banach algebra with identity and $\Delta(A)$ is the maximal ideal space?

The point is that $$\Delta(A)\subset K=\prod_{x\in A}\overline {D(0,||x||)}.$$ The Gelfand topology is the relative topology inherited from $K$. So it's Hausdorff, just because each of those disks is Hausdorff.

And $K$ is compact, so yes to show $\Delta(A)$ is compact you only need to show it's a closed subset of $K$. There's a slight subtlety here. An element of $\Delta(A)$ is by definition a map $\phi:A\to C$ such that

1. $\phi$ is linear.

2. $\phi(xy)=\phi(x)\phi(y)$ for all $x,y\in A$

3. There exists $x\in A$ with $\phi(x)\ne 0$.

(And then it's a not quite trivial fact that in fact $\Delta(A)\subset K$.)

Now, it's clear from the definition of the product topology that the set of elements of $K$ satisfying (1) and (2) is closed. But not so for condition (3)! And in fact if $A$ has no identity then (3) does not define a closed subset of $K$, and in fact $\Delta(A)$ is not compact, at least not in general.

But if $A$ has an identity then (3) is equivalent to

4. $\phi(e)=1$

and the set of elements of $K$ satisfying (4) is closed, qed.

Note: If you prefer you can talk about the closed unit ball of the Banach-space dual, with the weak* topology, instead of $K$ with the product topology.