Let $X$ be a compact metric space and $T\colon X\to X$ continuous.
By $h(A\cup B\cup C,T_{|A\cup B\cup C})$ denote the toplogical entropy of $T$, restricted on $A\cup B\cup C$, where $A,B,C\subset X$ are disjoint.
Is then $$ h(A\cup B\cup C,T_{|A\cup B\cup C})=h(A,T_{|A})+h(B,T_{|B})+h(C,T_{|C})? $$ that is can the topological entropy of $T$ be "splitted up"?
I tried to find an answer but failed. Intuitively, I would think that this is true.
Using this definition of a restricted topology, we can state that:
$$h(A∪B∪C,T|_{A∪B∪C})$$
$$=\;h(A∪B∪C,T|_A)\;+\;h(A∪B∪C,T|_B)\;+h\;(A∪B∪C,T|_C)$$
$$=\;h(A,T|_A)\;+\;h(B,T|_B)\;+\;h(C,T|_C)$$