In a compact metric space, every pair of homeomorphic open subsets has isomorphic basis?

Question

Let X be a compact metric space, V and W be open subsets. Suppose there is an homemorphism from V to W. Let B be a countable basis for X and B(V), B(W) the relativized basis for V and W, respectively. Are B(V) and B(W) order isomorphic (under subset order)? For instance, for X the unit interval, and B the usual basis of intervals with rational endpoints, The problem is that an homeomorphism between V and W does not need to send a basic open set into a basic open set. I do not know how to define an order isomorphism between B(V) and B(W).

Answer 1

Not in general: for instance, $B(V)$ and $B(W)$ do not even have to have the same cardinality. For a really simple example, consider $X=\{0,1,2\}$, $V=\{0,1\}$, $W=\{1,2\}$, and the basis $B$ that contains all the singletons as well as $\{0,1\}$. Then $B(V)$ has three elements but $B(W)$ has only the two singletons.

More generally, $B$ might contain $V$ (or a superset of $V$) but not contain any superset of $W$, and then $B(V)$ will have a greatest element but $B(W)$ will not.