This question is motivated by
Does excision imply that these inclusions are isomorphisms in homology?
There are two variants of the excision axiom for a homology theory:
(E) Let $U, A \subset X$. If $\overline{U} \subset int(A)$, then the inclusion map $i : (X \backslash U, A \backslash U) \to (X,A)$ induces isomorphisms $i_\ast : H_n(X \backslash U, A \backslash U) \to H_n(X,A)$ for each $n$.
(E-open) As above, but restricted to open $U$.
When axiomatic homology theory was introduced by Eilenberg and Steenrod, they used axiom (E-open). However, they also showed that singular homology theory satisfies (E).
In the literature (E) prevails, but it is certainly not a universally accepted standard.
It seems that (E) is strictly stronger than (E-open). To verify this we would need an example of a homology theory satisfying (E-open) but not (E), but I could not find any.
Does anybody know such an example?
Edit:
Eilenberg and Steenrod define their axioms for homology theories living on an admissible category $\mathfrak{A}$ of pairs of spaces and maps of such pairs. This means that $\mathfrak{A}$ contains a single point space and is closed with respect to a few elementary operations. Some well-known theories live on the category $\mathfrak{T}^2$ of all pairs of topological spaces (for example singular homology), other theories are only defined on smaller categories.
The answer to my question depends on $\mathfrak{A}$. For example, if $\mathfrak{A} = \mathfrak{C}^2$ = category of all compact pairs, then (E) and (E-open) are equivalent simply because only an open $U$ gives us a compact pair $(X \backslash U, A \backslash U)$.
This means that the desired example must live on an admissible $\mathfrak{A}$ having the following property ($\ast$):
Call a triple $(X,A,U)$ of spaces excision enabled if both $(X,A)$ and $(X \backslash U, A \backslash U)$ belong to $\mathfrak{A}$ and $\overline{U} \subset int(A)$.
($\ast$) The class of excision enabled triples is strictly larger than class of excision enabled triples with an open $U$.
This is certainly true for $\mathfrak{T}^2$.