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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.

Bhaskar Vashishth
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Salamo
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  • $h(A\cup B\cup C)=max{ h(A),h(B),h(C)}$. – san May 30 '15 at 19:51
  • But only if A,B, C are T-invariant... – Salamo May 30 '15 at 19:54
  • I would guess that it is always true – san May 30 '15 at 22:14
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    if, say, $A$ is not $T$-invariant, what do you mean by $h(A,T_{|A})$? You need $A$ to be $T$-invariant for $T_{|A}$ to define a self-map of $A$. So, I think the question makes sense just under the assumption that your three spaces are invariant, so that you have well-defined restrictions. – Simone Jun 05 '15 at 09:19
  • $h(A\cup B\cup C,T_{|A\cup B\cup C})=h(A\cup B\cup C,T_{|A})+h(A\cup B\cup C,T_{|B})+h(A\cup B\cup C,T_{|C})=h(A,T_{|A})+h(B,T_{|B})+h(C,T_{|C})$ – JMP Jun 06 '15 at 15:45
  • Then maybe, I wrote that down wrong. I think, I mean $h(A\cup B\cup C,T)$. Then the assumption is not true, isn't it? Anyhow, I do not see why $A,B,C$ should be invariant under T. – Salamo Jun 07 '15 at 13:53

1 Answers1

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Using this definition of a restricted topology, we can state that:

$$h(A∪B∪C,T|_{A∪B∪C})$$

$$=\;h(A∪B∪C,T|_A)\;+\;h(A∪B∪C,T|_B)\;+h\;(A∪B∪C,T|_C)$$

$$=\;h(A,T|_A)\;+\;h(B,T|_B)\;+\;h(C,T|_C)$$

JMP
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