Boundary of complex bounded domain












0












$begingroup$


Assume we have a bounded domain in the complex plane with smooth boundary. Is it possible to write the boundary as disjoint union of finitely/countably many closed (smooth) curves?



And furthermore, can you always obtain an “outer” closed curve whose interior contains the whole domain?



In general, if the boundary is not smooth, really strange things can happen. But is this case nice enough? I'm grateful for counterexamples or sketches of proofs.






complex-analysis differential-geometry algebraic-topology differential-topology curves







share|cite|improve this question











$endgroup$












  • $begingroup$
    Imagine the disc with a slit through it. Do you count this? Your curves will not be injective.
    $endgroup$
    – Mike Miller
    Jan 5 at 15:09










  • $begingroup$
    But the boundary is not smooth, not even $C^1$, is it?
    $endgroup$
    – YoungMath
    Jan 5 at 15:14












  • $begingroup$
    I missed that sentence. What does "smooth boundary" mean to you? With any definition I can think of the result is then obvious (that precisely means the boundary is a smooth curve).
    $endgroup$
    – Mike Miller
    Jan 5 at 15:23










  • $begingroup$
    I want to know if they are all closed and how many there are.
    $endgroup$
    – YoungMath
    Jan 5 at 15:37
















0












$begingroup$


Assume we have a bounded domain in the complex plane with smooth boundary. Is it possible to write the boundary as disjoint union of finitely/countably many closed (smooth) curves?



And furthermore, can you always obtain an “outer” closed curve whose interior contains the whole domain?



In general, if the boundary is not smooth, really strange things can happen. But is this case nice enough? I'm grateful for counterexamples or sketches of proofs.






complex-analysis differential-geometry algebraic-topology differential-topology curves







share|cite|improve this question











$endgroup$












  • $begingroup$
    Imagine the disc with a slit through it. Do you count this? Your curves will not be injective.
    $endgroup$
    – Mike Miller
    Jan 5 at 15:09










  • $begingroup$
    But the boundary is not smooth, not even $C^1$, is it?
    $endgroup$
    – YoungMath
    Jan 5 at 15:14












  • $begingroup$
    I missed that sentence. What does "smooth boundary" mean to you? With any definition I can think of the result is then obvious (that precisely means the boundary is a smooth curve).
    $endgroup$
    – Mike Miller
    Jan 5 at 15:23










  • $begingroup$
    I want to know if they are all closed and how many there are.
    $endgroup$
    – YoungMath
    Jan 5 at 15:37














0












0








0





$begingroup$


Assume we have a bounded domain in the complex plane with smooth boundary. Is it possible to write the boundary as disjoint union of finitely/countably many closed (smooth) curves?



And furthermore, can you always obtain an “outer” closed curve whose interior contains the whole domain?



In general, if the boundary is not smooth, really strange things can happen. But is this case nice enough? I'm grateful for counterexamples or sketches of proofs.






complex-analysis differential-geometry algebraic-topology differential-topology curves







share|cite|improve this question











$endgroup$




Assume we have a bounded domain in the complex plane with smooth boundary. Is it possible to write the boundary as disjoint union of finitely/countably many closed (smooth) curves?



And furthermore, can you always obtain an “outer” closed curve whose interior contains the whole domain?



In general, if the boundary is not smooth, really strange things can happen. But is this case nice enough? I'm grateful for counterexamples or sketches of proofs.






complex-analysis differential-geometry algebraic-topology differential-topology curves




complex-analysis differential-geometry algebraic-topology differential-topology curves






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Jan 5 at 11:51







YoungMath

















asked Jan 5 at 11:37









YoungMathYoungMath

190110




190110












  • $begingroup$
    Imagine the disc with a slit through it. Do you count this? Your curves will not be injective.
    $endgroup$
    – Mike Miller
    Jan 5 at 15:09










  • $begingroup$
    But the boundary is not smooth, not even $C^1$, is it?
    $endgroup$
    – YoungMath
    Jan 5 at 15:14












  • $begingroup$
    I missed that sentence. What does "smooth boundary" mean to you? With any definition I can think of the result is then obvious (that precisely means the boundary is a smooth curve).
    $endgroup$
    – Mike Miller
    Jan 5 at 15:23










  • $begingroup$
    I want to know if they are all closed and how many there are.
    $endgroup$
    – YoungMath
    Jan 5 at 15:37


















  • $begingroup$
    Imagine the disc with a slit through it. Do you count this? Your curves will not be injective.
    $endgroup$
    – Mike Miller
    Jan 5 at 15:09










  • $begingroup$
    But the boundary is not smooth, not even $C^1$, is it?
    $endgroup$
    – YoungMath
    Jan 5 at 15:14












  • $begingroup$
    I missed that sentence. What does "smooth boundary" mean to you? With any definition I can think of the result is then obvious (that precisely means the boundary is a smooth curve).
    $endgroup$
    – Mike Miller
    Jan 5 at 15:23










  • $begingroup$
    I want to know if they are all closed and how many there are.
    $endgroup$
    – YoungMath
    Jan 5 at 15:37
















$begingroup$
Imagine the disc with a slit through it. Do you count this? Your curves will not be injective.
$endgroup$
– Mike Miller
Jan 5 at 15:09




$begingroup$
Imagine the disc with a slit through it. Do you count this? Your curves will not be injective.
$endgroup$
– Mike Miller
Jan 5 at 15:09












$begingroup$
But the boundary is not smooth, not even $C^1$, is it?
$endgroup$
– YoungMath
Jan 5 at 15:14






$begingroup$
But the boundary is not smooth, not even $C^1$, is it?
$endgroup$
– YoungMath
Jan 5 at 15:14














$begingroup$
I missed that sentence. What does "smooth boundary" mean to you? With any definition I can think of the result is then obvious (that precisely means the boundary is a smooth curve).
$endgroup$
– Mike Miller
Jan 5 at 15:23




$begingroup$
I missed that sentence. What does "smooth boundary" mean to you? With any definition I can think of the result is then obvious (that precisely means the boundary is a smooth curve).
$endgroup$
– Mike Miller
Jan 5 at 15:23












$begingroup$
I want to know if they are all closed and how many there are.
$endgroup$
– YoungMath
Jan 5 at 15:37




$begingroup$
I want to know if they are all closed and how many there are.
$endgroup$
– YoungMath
Jan 5 at 15:37










1 Answer
1






active

oldest

votes


















2












$begingroup$

By assumption $partial U = overline U - U$ is a smooth manifold (without boundary). Because $partial U = overline U cap (Bbb C - U)$, this is a closed set; you have assumed it is bounded, so it is compact. A compact connected curve without boundary is diffeomorphic to $S^1$, and a compact 1-manifold without boundary is diffeomorphic to some finite disjoint union thereof.



There is some difficulty in counting the number of components, which has to do with the topology of the domain. There is one of these per "end" of $U$. A simply connected domain has one end; an annulus has two. (In general, suppose $U$ is the interior of a compact manifold with boundary $overline U$. Then the number of ends agrees with the number of boundary components, which furthermore agrees with the rank of $H_1(U)$ via what is called Poincare-Lefschetz duality.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    “A compact connected curve without boundary is diffeomorphic to $S^1$, and a compact 1-manifold without boundary is diffeomorphic to some finite disjoint union thereof.” That sounds quite remarkable, though quite intuitve. Do you have any reference for that? Moreover, does any connected boundary component has no boundary itself, so that the statement is applicable?
    $endgroup$
    – YoungMath
    Jan 5 at 16:12










  • $begingroup$
    @YoungMath It's in any book on smooth manifolds you choose. You build a nonvanishing vector field and flow along it; if you are compact, either you eventually exit some boundary point or end up back where you started. And yes, "smooth boundary" literally means that every point has a neighborhood so that the curve looks like $Bbb R$ inside of $Bbb R^2$, and this local model has no boundary. The second statement follows immediately from the first statement because any manifold is a disjoint union of its path components.
    $endgroup$
    – Mike Miller
    Jan 5 at 16:13












  • $begingroup$
    Perfect. That's what I needed, thank you! :)
    $endgroup$
    – YoungMath
    Jan 5 at 16:15











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1 Answer
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1 Answer
1






active

oldest

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active

oldest

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active

oldest

votes









2












$begingroup$

By assumption $partial U = overline U - U$ is a smooth manifold (without boundary). Because $partial U = overline U cap (Bbb C - U)$, this is a closed set; you have assumed it is bounded, so it is compact. A compact connected curve without boundary is diffeomorphic to $S^1$, and a compact 1-manifold without boundary is diffeomorphic to some finite disjoint union thereof.



There is some difficulty in counting the number of components, which has to do with the topology of the domain. There is one of these per "end" of $U$. A simply connected domain has one end; an annulus has two. (In general, suppose $U$ is the interior of a compact manifold with boundary $overline U$. Then the number of ends agrees with the number of boundary components, which furthermore agrees with the rank of $H_1(U)$ via what is called Poincare-Lefschetz duality.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    “A compact connected curve without boundary is diffeomorphic to $S^1$, and a compact 1-manifold without boundary is diffeomorphic to some finite disjoint union thereof.” That sounds quite remarkable, though quite intuitve. Do you have any reference for that? Moreover, does any connected boundary component has no boundary itself, so that the statement is applicable?
    $endgroup$
    – YoungMath
    Jan 5 at 16:12










  • $begingroup$
    @YoungMath It's in any book on smooth manifolds you choose. You build a nonvanishing vector field and flow along it; if you are compact, either you eventually exit some boundary point or end up back where you started. And yes, "smooth boundary" literally means that every point has a neighborhood so that the curve looks like $Bbb R$ inside of $Bbb R^2$, and this local model has no boundary. The second statement follows immediately from the first statement because any manifold is a disjoint union of its path components.
    $endgroup$
    – Mike Miller
    Jan 5 at 16:13












  • $begingroup$
    Perfect. That's what I needed, thank you! :)
    $endgroup$
    – YoungMath
    Jan 5 at 16:15
















2












$begingroup$

By assumption $partial U = overline U - U$ is a smooth manifold (without boundary). Because $partial U = overline U cap (Bbb C - U)$, this is a closed set; you have assumed it is bounded, so it is compact. A compact connected curve without boundary is diffeomorphic to $S^1$, and a compact 1-manifold without boundary is diffeomorphic to some finite disjoint union thereof.



There is some difficulty in counting the number of components, which has to do with the topology of the domain. There is one of these per "end" of $U$. A simply connected domain has one end; an annulus has two. (In general, suppose $U$ is the interior of a compact manifold with boundary $overline U$. Then the number of ends agrees with the number of boundary components, which furthermore agrees with the rank of $H_1(U)$ via what is called Poincare-Lefschetz duality.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    “A compact connected curve without boundary is diffeomorphic to $S^1$, and a compact 1-manifold without boundary is diffeomorphic to some finite disjoint union thereof.” That sounds quite remarkable, though quite intuitve. Do you have any reference for that? Moreover, does any connected boundary component has no boundary itself, so that the statement is applicable?
    $endgroup$
    – YoungMath
    Jan 5 at 16:12










  • $begingroup$
    @YoungMath It's in any book on smooth manifolds you choose. You build a nonvanishing vector field and flow along it; if you are compact, either you eventually exit some boundary point or end up back where you started. And yes, "smooth boundary" literally means that every point has a neighborhood so that the curve looks like $Bbb R$ inside of $Bbb R^2$, and this local model has no boundary. The second statement follows immediately from the first statement because any manifold is a disjoint union of its path components.
    $endgroup$
    – Mike Miller
    Jan 5 at 16:13












  • $begingroup$
    Perfect. That's what I needed, thank you! :)
    $endgroup$
    – YoungMath
    Jan 5 at 16:15














2












2








2





$begingroup$

By assumption $partial U = overline U - U$ is a smooth manifold (without boundary). Because $partial U = overline U cap (Bbb C - U)$, this is a closed set; you have assumed it is bounded, so it is compact. A compact connected curve without boundary is diffeomorphic to $S^1$, and a compact 1-manifold without boundary is diffeomorphic to some finite disjoint union thereof.



There is some difficulty in counting the number of components, which has to do with the topology of the domain. There is one of these per "end" of $U$. A simply connected domain has one end; an annulus has two. (In general, suppose $U$ is the interior of a compact manifold with boundary $overline U$. Then the number of ends agrees with the number of boundary components, which furthermore agrees with the rank of $H_1(U)$ via what is called Poincare-Lefschetz duality.






share|cite|improve this answer









$endgroup$



By assumption $partial U = overline U - U$ is a smooth manifold (without boundary). Because $partial U = overline U cap (Bbb C - U)$, this is a closed set; you have assumed it is bounded, so it is compact. A compact connected curve without boundary is diffeomorphic to $S^1$, and a compact 1-manifold without boundary is diffeomorphic to some finite disjoint union thereof.



There is some difficulty in counting the number of components, which has to do with the topology of the domain. There is one of these per "end" of $U$. A simply connected domain has one end; an annulus has two. (In general, suppose $U$ is the interior of a compact manifold with boundary $overline U$. Then the number of ends agrees with the number of boundary components, which furthermore agrees with the rank of $H_1(U)$ via what is called Poincare-Lefschetz duality.







share|cite|improve this answer












share|cite|improve this answer



share|cite|improve this answer










answered Jan 5 at 15:49









Mike MillerMike Miller

37.2k472139




37.2k472139












  • $begingroup$
    “A compact connected curve without boundary is diffeomorphic to $S^1$, and a compact 1-manifold without boundary is diffeomorphic to some finite disjoint union thereof.” That sounds quite remarkable, though quite intuitve. Do you have any reference for that? Moreover, does any connected boundary component has no boundary itself, so that the statement is applicable?
    $endgroup$
    – YoungMath
    Jan 5 at 16:12










  • $begingroup$
    @YoungMath It's in any book on smooth manifolds you choose. You build a nonvanishing vector field and flow along it; if you are compact, either you eventually exit some boundary point or end up back where you started. And yes, "smooth boundary" literally means that every point has a neighborhood so that the curve looks like $Bbb R$ inside of $Bbb R^2$, and this local model has no boundary. The second statement follows immediately from the first statement because any manifold is a disjoint union of its path components.
    $endgroup$
    – Mike Miller
    Jan 5 at 16:13












  • $begingroup$
    Perfect. That's what I needed, thank you! :)
    $endgroup$
    – YoungMath
    Jan 5 at 16:15


















  • $begingroup$
    “A compact connected curve without boundary is diffeomorphic to $S^1$, and a compact 1-manifold without boundary is diffeomorphic to some finite disjoint union thereof.” That sounds quite remarkable, though quite intuitve. Do you have any reference for that? Moreover, does any connected boundary component has no boundary itself, so that the statement is applicable?
    $endgroup$
    – YoungMath
    Jan 5 at 16:12










  • $begingroup$
    @YoungMath It's in any book on smooth manifolds you choose. You build a nonvanishing vector field and flow along it; if you are compact, either you eventually exit some boundary point or end up back where you started. And yes, "smooth boundary" literally means that every point has a neighborhood so that the curve looks like $Bbb R$ inside of $Bbb R^2$, and this local model has no boundary. The second statement follows immediately from the first statement because any manifold is a disjoint union of its path components.
    $endgroup$
    – Mike Miller
    Jan 5 at 16:13












  • $begingroup$
    Perfect. That's what I needed, thank you! :)
    $endgroup$
    – YoungMath
    Jan 5 at 16:15
















$begingroup$
“A compact connected curve without boundary is diffeomorphic to $S^1$, and a compact 1-manifold without boundary is diffeomorphic to some finite disjoint union thereof.” That sounds quite remarkable, though quite intuitve. Do you have any reference for that? Moreover, does any connected boundary component has no boundary itself, so that the statement is applicable?
$endgroup$
– YoungMath
Jan 5 at 16:12




$begingroup$
“A compact connected curve without boundary is diffeomorphic to $S^1$, and a compact 1-manifold without boundary is diffeomorphic to some finite disjoint union thereof.” That sounds quite remarkable, though quite intuitve. Do you have any reference for that? Moreover, does any connected boundary component has no boundary itself, so that the statement is applicable?
$endgroup$
– YoungMath
Jan 5 at 16:12












$begingroup$
@YoungMath It's in any book on smooth manifolds you choose. You build a nonvanishing vector field and flow along it; if you are compact, either you eventually exit some boundary point or end up back where you started. And yes, "smooth boundary" literally means that every point has a neighborhood so that the curve looks like $Bbb R$ inside of $Bbb R^2$, and this local model has no boundary. The second statement follows immediately from the first statement because any manifold is a disjoint union of its path components.
$endgroup$
– Mike Miller
Jan 5 at 16:13






$begingroup$
@YoungMath It's in any book on smooth manifolds you choose. You build a nonvanishing vector field and flow along it; if you are compact, either you eventually exit some boundary point or end up back where you started. And yes, "smooth boundary" literally means that every point has a neighborhood so that the curve looks like $Bbb R$ inside of $Bbb R^2$, and this local model has no boundary. The second statement follows immediately from the first statement because any manifold is a disjoint union of its path components.
$endgroup$
– Mike Miller
Jan 5 at 16:13














$begingroup$
Perfect. That's what I needed, thank you! :)
$endgroup$
– YoungMath
Jan 5 at 16:15




$begingroup$
Perfect. That's what I needed, thank you! :)
$endgroup$
– YoungMath
Jan 5 at 16:15


















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