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Sam Nead
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  1. Can you point me out the rigorous definition of "limit of a sequence of arcs" on a surface?
  1. In which topological space are we considering the limit of arcs to obtain a lamination?

Suppose that $S$ is a orientable surface of finite topological type, equipped with a hyperbolic metric of finite volume. Consider the set $A$ of simple geodesic arcs properly embedded in $S$. Consider also the larger set $C$ of closed subsets of $S$. We equip $C$ with the Hausdorff metric.

Suppose that $a_n \in A$ is a sequence of arcs. This also gives a sequence in $C$. If $a_n$ has a limit in $C$, then that is the desired object. It is a theorem that the limit is geodesic laminationa geodesic lamination: a closed subset of $S$ that can be realizedrealised as a disjoint union of simple geodesics.

It is an easy exercise to find a geodesic lamination that is not a Hausdorff limit of arcs.

  1. Where can I find a proof of the statement "every sequence of arcs converges to a lamination on a hyperbolic surface"?

It is not true that every sequence converges. Here is the correct statement, from Section 8.5 of Thurston's notes: "An easy diagonal argument shows that every sequence $\{\gamma_i\}$ has a subsequence which converges geometrically." More detail is given in the proof of Proposition I.4.1.7 of Notes on notes of Thurston by Canary, Epstein, and Green.

  1. Can you point me out the rigorous definition of "limit of a sequence of arcs" on a surface?
  1. In which topological space are we considering the limit of arcs to obtain a lamination?

Suppose that $S$ is a orientable surface of finite topological type, equipped with a hyperbolic metric of finite volume. Consider the set $A$ of simple geodesic arcs properly embedded in $S$. Consider also the larger set $C$ of closed subsets of $S$. We equip $C$ with the Hausdorff metric.

Suppose that $a_n \in A$ is a sequence of arcs. This also gives a sequence in $C$. If $a_n$ has a limit in $C$, then that is the desired object. It is a theorem that the limit is geodesic lamination: a closed subset of $S$ that can be realized as a disjoint union of simple geodesics.

It is an easy exercise to find a geodesic lamination that is not a Hausdorff limit of arcs.

  1. Where can I find a proof of the statement "every sequence of arcs converges to a lamination on a hyperbolic surface"?

It is not true that every sequence converges. Here is the correct statement, from Section 8.5 of Thurston's notes: "An easy diagonal argument shows that every sequence $\{\gamma_i\}$ has a subsequence which converges geometrically." More detail is given in the proof of Proposition I.4.1.7 of Notes on notes of Thurston by Canary, Epstein, and Green.

  1. Can you point me out the rigorous definition of "limit of a sequence of arcs" on a surface?
  1. In which topological space are we considering the limit of arcs to obtain a lamination?

Suppose that $S$ is a orientable surface of finite topological type, equipped with a hyperbolic metric of finite volume. Consider the set $A$ of simple geodesic arcs properly embedded in $S$. Consider also the larger set $C$ of closed subsets of $S$. We equip $C$ with the Hausdorff metric.

Suppose that $a_n \in A$ is a sequence of arcs. This also gives a sequence in $C$. If $a_n$ has a limit in $C$, then that is the desired object. It is a theorem that the limit is a geodesic lamination: a closed subset of $S$ that can be realised as a disjoint union of simple geodesics.

It is an easy exercise to find a geodesic lamination that is not a Hausdorff limit of arcs.

  1. Where can I find a proof of the statement "every sequence of arcs converges to a lamination on a hyperbolic surface"?

It is not true that every sequence converges. Here is the correct statement, from Section 8.5 of Thurston's notes: "An easy diagonal argument shows that every sequence $\{\gamma_i\}$ has a subsequence which converges geometrically." More detail is given in the proof of Proposition I.4.1.7 of Notes on notes of Thurston by Canary, Epstein, and Green.

fix brain-o's.
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Sam Nead
  • 28.2k
  • 5
  • 72
  • 131
  1. Can you point me out the rigorous definition of "limit of a sequence of arcs" on a surface?
  1. In which topological space are we considering the limit of arcs to obtain a lamination?

Suppose that $S$ is a orientable surface of finite topological type, equipped with a hyperbolic metric of finite volume. Consider the set $A$ of simple geodesic arcs properly embedded in $S$. Consider also the larger set $C$ of closed subsets of $S$. We equip $C$ with the Hausdorff topology, to make it a metric space.

Suppose that $a_n \in A$ is a sequence of arcs. This also gives a sequence in $C$. If $a_n$ has a limit in $C$, then that is the desired object. It is a theorem that the limit is geodesic lamination: a closed subset of $S$ that can be realized as a disjoint union of simple geodesics.

It is an easy exercise to find a geodesic lamination that is not a Hausdorff limit of arcs.

  1. Where can I find a proof of the statement "every sequence of arcs converges to a lamination on a hyperbolic surface"?

Surely thisIt is discussed in the reference you are reading? If you give a link to it, perhaps we can find the location of this resultnot true that every sequence converges.

This Here is also discussed (very tersely) inthe correct statement, from Section 8.5 of Thurston's notes. The exact quote is: "An easy diagonal argument shows that every sequence $\{\gamma_i\}$ has a subsequence which converges geometrically." More detail is given in the proof of Proposition I.4.1.7 of Notes on notes of Thurston by Canary, Epstein, and Green.

  1. Can you point me out the rigorous definition of "limit of a sequence of arcs" on a surface?
  1. In which topological space are we considering the limit of arcs to obtain a lamination?

Suppose that $S$ is a orientable surface of finite topological type, equipped with a hyperbolic metric of finite volume. Consider the set $A$ of simple geodesic arcs properly embedded in $S$. Consider also the larger set $C$ of closed subsets of $S$. We equip $C$ with the Hausdorff topology, to make it a metric space.

Suppose that $a_n \in A$ is a sequence of arcs. This also gives a sequence in $C$. If $a_n$ has a limit in $C$, then that is the desired object. It is a theorem that the limit is geodesic lamination: a closed subset of $S$ that can be realized as a disjoint union of simple geodesics.

It is an easy exercise to find a geodesic lamination that is not a Hausdorff limit of arcs.

  1. Where can I find a proof of the statement "every sequence of arcs converges to a lamination on a hyperbolic surface"?

Surely this is discussed in the reference you are reading? If you give a link to it, perhaps we can find the location of this result.

This is also discussed (very tersely) in Section 8.5 of Thurston's notes. The exact quote is "An easy diagonal argument shows that every sequence $\{\gamma_i\}$ has a subsequence which converges geometrically." More detail is given in the proof of Proposition I.4.1.7 of Notes on notes of Thurston by Canary, Epstein, and Green.

  1. Can you point me out the rigorous definition of "limit of a sequence of arcs" on a surface?
  1. In which topological space are we considering the limit of arcs to obtain a lamination?

Suppose that $S$ is a orientable surface of finite topological type, equipped with a hyperbolic metric of finite volume. Consider the set $A$ of simple geodesic arcs properly embedded in $S$. Consider also the larger set $C$ of closed subsets of $S$. We equip $C$ with the Hausdorff metric.

Suppose that $a_n \in A$ is a sequence of arcs. This also gives a sequence in $C$. If $a_n$ has a limit in $C$, then that is the desired object. It is a theorem that the limit is geodesic lamination: a closed subset of $S$ that can be realized as a disjoint union of simple geodesics.

It is an easy exercise to find a geodesic lamination that is not a Hausdorff limit of arcs.

  1. Where can I find a proof of the statement "every sequence of arcs converges to a lamination on a hyperbolic surface"?

It is not true that every sequence converges. Here is the correct statement, from Section 8.5 of Thurston's notes: "An easy diagonal argument shows that every sequence $\{\gamma_i\}$ has a subsequence which converges geometrically." More detail is given in the proof of Proposition I.4.1.7 of Notes on notes of Thurston by Canary, Epstein, and Green.

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Sam Nead
  • 28.2k
  • 5
  • 72
  • 131
  1. Can you point me out the rigorous definition of "limit of a sequence of arcs" on a surface?
  1. In which topological space are we considering the limit of arcs to obtain a lamination?

Suppose that $S$ is a orientable surface of finite topological type, equipped with a hyperbolic metric of finite volume. Consider the set $A$ of simple geodesic arcs properly embedded in $S$. Consider also the larger set $C$ of closed subsets of $S$. We equip $C$ with the Hausdorff topology, to make it a metric space.

Suppose that $a_n \in A$ is a sequence of arcs. This also gives a sequence in $C$. If $a_n$ has a limit in $C$, then that is the desired object. It is a theorem that the limit is geodesic lamination: a closed subset of $S$ that can be realized as a disjoint union of simple geodesics.

It is an easy exercise to find a geodesic lamination that is not a Hausdorff limit of arcs.

  1. Where can I find a proof of the statement "every sequence of arcs converges to a lamination on a hyperbolic surface"?

Surely this is discussed in the reference you are reading.? If you give a link to it, perhaps we can find the location of this result.

This is also discussed (very tersely) in Thurston's notesSection 8.5 of Thurston's notes. The exact quote is "An easy diagonal argument shows that every sequence $\{\gamma_i\}$ has a subsequence which converges geometrically." More detail is given in the proof of Proposition I.4.1.7 of Notes on notes of Thurston by Canary, Epstein, and Green.

  1. Can you point me out the rigorous definition of "limit of a sequence of arcs" on a surface?
  1. In which topological space are we considering the limit of arcs to obtain a lamination?

Suppose that $S$ is a orientable surface of finite topological type, equipped with a hyperbolic metric of finite volume. Consider the set $A$ of simple geodesic arcs properly embedded in $S$. Consider also the larger set $C$ of closed subsets of $S$. We equip $C$ with the Hausdorff topology, to make it a metric space.

Suppose that $a_n \in A$ is a sequence of arcs. This also gives a sequence in $C$. If $a_n$ has a limit in $C$, then that is the desired object. It is a theorem that the limit is geodesic lamination: a closed subset of $S$ that can be realized as a disjoint union of simple geodesics.

It is an easy exercise to find a geodesic lamination that is not a Hausdorff limit of arcs.

  1. Where can I find a proof of the statement "every sequence of arcs converges to a lamination on a hyperbolic surface"?

Surely this is discussed in the reference you are reading. If you give a link to it, perhaps we can find the location of this result.

This is also discussed in Thurston's notes.

  1. Can you point me out the rigorous definition of "limit of a sequence of arcs" on a surface?
  1. In which topological space are we considering the limit of arcs to obtain a lamination?

Suppose that $S$ is a orientable surface of finite topological type, equipped with a hyperbolic metric of finite volume. Consider the set $A$ of simple geodesic arcs properly embedded in $S$. Consider also the larger set $C$ of closed subsets of $S$. We equip $C$ with the Hausdorff topology, to make it a metric space.

Suppose that $a_n \in A$ is a sequence of arcs. This also gives a sequence in $C$. If $a_n$ has a limit in $C$, then that is the desired object. It is a theorem that the limit is geodesic lamination: a closed subset of $S$ that can be realized as a disjoint union of simple geodesics.

It is an easy exercise to find a geodesic lamination that is not a Hausdorff limit of arcs.

  1. Where can I find a proof of the statement "every sequence of arcs converges to a lamination on a hyperbolic surface"?

Surely this is discussed in the reference you are reading? If you give a link to it, perhaps we can find the location of this result.

This is also discussed (very tersely) in Section 8.5 of Thurston's notes. The exact quote is "An easy diagonal argument shows that every sequence $\{\gamma_i\}$ has a subsequence which converges geometrically." More detail is given in the proof of Proposition I.4.1.7 of Notes on notes of Thurston by Canary, Epstein, and Green.

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Sam Nead
  • 28.2k
  • 5
  • 72
  • 131
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