no-cycles condition

Let $X$ be a metric space and let $f:X\to X$ be a homeomorphism.Suppose $ℱ=\{{\mathrm{\Lambda }}_1,\mathrm{\dots },{\mathrm{\Lambda }}_k\}$ is a family of compact invariant sets for $f$. Define a relation $\to$ on $ℱ$ by ${\mathrm{\Lambda }}_i\to {\mathrm{\Lambda }}_j$ if

$$W^u({\mathrm{\Lambda }}_i)\cap W^s({\mathrm{\Lambda }}_j)-\bigcup _{l=1}^k{\mathrm{\Lambda }}_l\ne \mathrm{\varnothing },$$

that is, if the unstable set of ${\mathrm{\Lambda }}_i$ intersects the stable set of ${\mathrm{\Lambda }}_j$ outside the union of the ${\mathrm{\Lambda }}_l$’s.

A cycle for $ℱ$ is a sequence $\{n_i:i=1,\mathrm{\dots },j\}$ such that

$${\mathrm{\Lambda }}_{n_i}\to {\mathrm{\Lambda }}_{n_{i+1}}$$

for and

$${\mathrm{\Lambda }}_{n_j}\to {\mathrm{\Lambda }}_{n_1}.$$

With some abuse of notation, we can write this as

$${\mathrm{\Lambda }}_{n_1}\to {\mathrm{\Lambda }}_{n_2}\to \mathrm{\cdots }\to {\mathrm{\Lambda }}_{n_j}\to {\mathrm{\Lambda }}_{n_1}.$$

If $ℱ$ has no cycles, then we say that it satisfies the no-cycles condition.