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Gerry Myerson
Gerry Myerson
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Refining ladders and orbit segments - with a pictuirepicture

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Daniel Mansfield
Daniel Mansfield
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Refining ladders and orbit segments. - with a pictuire

Refining ladders and orbit segments. - with a pictuire

I am trying to understand the following paragraph from The Classification of Non-Singular Actions, Revisited, page 5 paragraph 2.

Remember that $S \in [T]$ so that for every $x\in X, S(x) = T^{n(x)}$. If $M_1$ is large compared with $n(x)$ for all but $\epsilon_1$ of the space, and compared with $M$ as well (and $\epsilon_1$ is taken small enough), the orbit segment $\{T^jx\}_{j=0}^M$ is in fact contained in some orbit segment of $\mathcal{L}_0$ for most $x$.

So, choosing $M_1$ and $\epsilon_1$ apply Rohlin's lemma to the induced transformation on $C(0)$. This gives us that $\{{T \vert_{C(0)}}^jx\}_{j=0}^{M_1}$ is an orbit segment for all but $\epsilon_1$ of $C(0)$.

But how does this imply that we have orbit segments outside of $C(0)$? I do not understand the significance of choosing $M_1$ to be large compared with $n(x)$ or $M$.

Refining ladders and orbit segments.

I am trying to understand the following paragraph from The Classification of Non-Singular Actions, Revisited, page 5 paragraph 2.

Remember that $S \in [T]$ so that for every $x\in X, S(x) = T^{n(x)}$. If $M_1$ is large compared with $n(x)$ for all but $\epsilon_1$ of the space, and compared with $M$ as well (and $\epsilon_1$ is taken small enough), the orbit segment $\{T^jx\}_{j=0}^M$ is in fact contained in some orbit segment of $\mathcal{L}_0$ for most $x$.

So, choosing $M_1$ and $\epsilon_1$ apply Rohlin's lemma to the induced transformation on $C(0)$. This gives us that $\{{T \vert_{C(0)}}^jx\}_{j=0}^{M_1}$ is an orbit segment for all but $\epsilon_1$ of $C(0)$.

But how does this imply that we have orbit segments outside of $C(0)$? I do not understand the significance of choosing $M_1$ to be large compared with $n(x)$ or $M$.

Refining ladders and orbit segments - with a pictuire

I am trying to understand the following paragraph from The Classification of Non-Singular Actions, Revisited, page 5 paragraph 2.

Remember that $S \in [T]$ so that for every $x\in X, S(x) = T^{n(x)}$. If $M_1$ is large compared with $n(x)$ for all but $\epsilon_1$ of the space, and compared with $M$ as well (and $\epsilon_1$ is taken small enough), the orbit segment $\{T^jx\}_{j=0}^M$ is in fact contained in some orbit segment of $\mathcal{L}_0$ for most $x$.

So, choosing $M_1$ and $\epsilon_1$ apply Rohlin's lemma to the induced transformation on $C(0)$. This gives us that $\{{T \vert_{C(0)}}^jx\}_{j=0}^{M_1}$ is an orbit segment for all but $\epsilon_1$ of $C(0)$.

But how does this imply that we have orbit segments outside of $C(0)$? I do not understand the significance of choosing $M_1$ to be large compared with $n(x)$ or $M$.

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Daniel Mansfield
Daniel Mansfield
  • 718
  • 3
  • 11

Refining ladders and orbit segments.

I am trying to understand the following paragraph from The Classification of Non-Singular Actions, Revisited, page 5 paragraph 2.

Remember that $S \in [T]$ so that for every $x\in X, S(x) = T^{n(x)}$. If $M_1$ is large compared with $n(x)$ for all but $\epsilon_1$ of the space, and compared with $M$ as well (and $\epsilon_1$ is taken small enough), the orbit segment $\{T^jx\}_{j=0}^M$ is in fact contained in some orbit segment of $\mathcal{L}_0$ for most $x$.

So, choosing $M_1$ and $\epsilon_1$ apply Rohlin's lemma to the induced transformation on $C(0)$. This gives us that $\{{T \vert_{C(0)}}^jx\}_{j=0}^{M_1}$ is an orbit segment for all but $\epsilon_1$ of $C(0)$.

But how does this imply that we have orbit segments outside of $C(0)$? I do not understand the significance of choosing $M_1$ to be large compared with $n(x)$ or $M$.