Is $C[a,b]$ isomorphic to $C([a,b]\cup[c,d])$ It is very simple to prove that $C[a,b]$ is isomorphic to subspace of $C([a,b]\cup[c,d])$ and vice versa $C([a,b]\cup[c,d])$ is isomorphic to subspace of $C[a,b]$.
Is it true for $\bigl(C[a,b], C([a,b]\cup[c,d])\bigr)$ pair?
If it is true, can one find/prove existence of isometrical isomorphism?
 A: If isomorphic means isomorphic as rings then no, they're not isomorphic. Keen fact:


Suppose $K_1$ and $K_2$ are compact Hausdorff spaces. Then $C(K_1)$ is isomorphic to $C(K_2)$ if and only if $K_1$ and $K_2$ are homeomorphic.


(Sketch: $C(K)$ is a Banach algebra with maximal ideal space $K$.)
This is a keen thing, because it shows that the topological structure of $K$ is determined by the purely algebraic structure of $C(K)$. (Note that "isomorphism" here means just an algebraic isomorphism as rings, nothing about "isometric" or  even "continuous".)
(An ad hoc argument for the $K_1$ and $K_2$ you have in mind: Does there exist a non-constant $f\in C(K)$ with $f^2=f$?)

Come to think of it, you mention "subspace"; maybe you're just asking about isomorphism as Banach spaces. I believe they are, may know for sure after lunch. They're not isometrically isomorphic, however: If $K$ is a compact Hausdorff space then $K\times\Bbb T$ is homeomorphic to the set of extreme points of the closed unit ball of $C(K)^*$.
