Using Papa Rudin as a reference, specifically starting at the middle of page 38.
For any locally compact Hausdorff space $X$ and topological field $K$ define $K_c(X)$ to be the space of all compactly-supported $K$-valued functions on $X$. By the same arguments in the text, we should have a vector space $K_c(X)$ here. Since $K$ is a topological field we can add functions and scalar multiply.
This raises the question: is $K_c(X)$ also a ring either under pointwise multiplication or function composition?
Attempt. $\text{supp} (f \cdot g) = \overline{\{ x \in X : f(x) g(x) \neq 0 \} }= \overline{\text{supp}(f) \cap \text{supp}(g)}$ as the values are in a field which is an integral domain in particular. Not sure about the closure of the intersection though.
If $A, B$ are two subsets of a Hausdorff topo space $X$ and $\overline{A}, \overline{B}$ are both compact, then they are closed since compact subsets of a Hausdorff space are closed. Then $\overline{A \cap B} \subset \overline{A} \cap \overline{B}$ as the left is the intersection of all closed sets containing $A \cap B$ and certainly $\overline{A} \cap \overline{B}$ contains $A \cap B$ and is closed.
The intersection of a closed set with a compact set is compact in a Hausdorff space by Corollaries on page 36. And by Theorem 2.4 a closed subset of a compact set is comppact. So that I think I'm done with the proof as $\overline{A \cap B} \subset \overline{A} \cap \overline{B}$ which is compact.
Thus can I assume that $K_c(X)$ is a $K$-algebra for any topological field $K$? Better yet, for any topological integral domain $K$?
Best Answer
Your proof is correct. However, it is easier to argue that $\{ x \in X : f(x) g(x) \neq 0 \} \subset \{ x \in X : f(x) \neq 0 \}$ which implies $\text{supp} (f \cdot g) \subset \text{supp} (f)$: Hence $\text{supp} (f \cdot g)$ is compact because closed subsets of compact sets are compact.
Therefore $K_c(X)$ is a $K$-algebra for any topological field $K$.