Connected Partitions of Spheres
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Let $U,V$ be disjoint non-empty connected open subsets of the sphere $S^2$ such that $partial U=partial V$ and $operatorname{cl}(Ucup V)=S^2$. Must $U$ and $V$ be simply connected?
This seems intuitively obvious, but I'm not sure how to best show it. I have a painstaking proof where one basically "rasterizes" the problem and reduces it to some discrete statement about any finite grid, but this feels like a poor way to go about it. Is there a better way to show this statement?
algebraic-topology
asked Jan 7 at 20:19
Milo BrandtMilo Brandt
39.8k476140
$endgroup$
|
show 1 more comment
$begingroup$
Let $U,V$ be disjoint non-empty connected open subsets of the sphere $S^2$ such that $partial U=partial V$ and $operatorname{cl}(Ucup V)=S^2$. Must $U$ and $V$ be simply connected?
This seems intuitively obvious, but I'm not sure how to best show it. I have a painstaking proof where one basically "rasterizes" the problem and reduces it to some discrete statement about any finite grid, but this feels like a poor way to go about it. Is there a better way to show this statement?
algebraic-topology
asked Jan 7 at 20:19
Milo BrandtMilo Brandt
39.8k476140
$endgroup$
$begingroup$
In $S^3$ it is not true. Consider the Alexander horned sphere (see en.wikipedia.org/wiki/Alexander_horned_sphere). It separates $S^3$ in two non-empty components $U, V$ as in your question, but one of these is not simply connected.
$endgroup$
– Paul Frost
Jan 8 at 17:22
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I wonder if it is possible to show that the common boundary of $U, V$ is connected?
$endgroup$
– Paul Frost
Jan 8 at 17:29
$begingroup$
@PaulFrost It's certainly true that the boundary is connected; the discrete method I know about for this problem also establishes that the boundary is connected, but it's not a very satisfying proof.
$endgroup$
– Milo Brandt
Jan 8 at 17:35
$begingroup$
I asked because one can show that the complement of compact connected $B subset S^2$ splits into the disjoint union of at most countably many open disks.
$endgroup$
– Paul Frost
Jan 8 at 17:42
$begingroup$
@PaulFrost Ah, that's interesting, I didn't know that. That seems like it would be most of the way to a proof - though I don't have a quick way to see that the boundary is indeed connected. (But that might be easier; it might even be doable via homology)
$endgroup$
– Milo Brandt
Jan 8 at 20:35
|
show 1 more comment
$begingroup$
Let $U,V$ be disjoint non-empty connected open subsets of the sphere $S^2$ such that $partial U=partial V$ and $operatorname{cl}(Ucup V)=S^2$. Must $U$ and $V$ be simply connected?
This seems intuitively obvious, but I'm not sure how to best show it. I have a painstaking proof where one basically "rasterizes" the problem and reduces it to some discrete statement about any finite grid, but this feels like a poor way to go about it. Is there a better way to show this statement?
algebraic-topology
asked Jan 7 at 20:19
Milo BrandtMilo Brandt
39.8k476140
$endgroup$
Let $U,V$ be disjoint non-empty connected open subsets of the sphere $S^2$ such that $partial U=partial V$ and $operatorname{cl}(Ucup V)=S^2$. Must $U$ and $V$ be simply connected?
This seems intuitively obvious, but I'm not sure how to best show it. I have a painstaking proof where one basically "rasterizes" the problem and reduces it to some discrete statement about any finite grid, but this feels like a poor way to go about it. Is there a better way to show this statement?
algebraic-topology
algebraic-topology
asked Jan 7 at 20:19
Milo BrandtMilo Brandt
39.8k476140
asked Jan 7 at 20:19
Milo BrandtMilo Brandt
39.8k476140
edited Jan 8 at 16:05
Milo Brandt
asked Jan 7 at 20:19
Milo BrandtMilo Brandt
39.8k476140
asked Jan 7 at 20:19
Milo BrandtMilo Brandt
39.8k476140
asked Jan 7 at 20:19
Milo BrandtMilo Brandt
39.8k476140
39.8k476140
$begingroup$
In $S^3$ it is not true. Consider the Alexander horned sphere (see en.wikipedia.org/wiki/Alexander_horned_sphere). It separates $S^3$ in two non-empty components $U, V$ as in your question, but one of these is not simply connected.
$endgroup$
– Paul Frost
Jan 8 at 17:22
$begingroup$
I wonder if it is possible to show that the common boundary of $U, V$ is connected?
$endgroup$
– Paul Frost
Jan 8 at 17:29
$begingroup$
@PaulFrost It's certainly true that the boundary is connected; the discrete method I know about for this problem also establishes that the boundary is connected, but it's not a very satisfying proof.
$endgroup$
– Milo Brandt
Jan 8 at 17:35
$begingroup$
I asked because one can show that the complement of compact connected $B subset S^2$ splits into the disjoint union of at most countably many open disks.
$endgroup$
– Paul Frost
Jan 8 at 17:42
$begingroup$
@PaulFrost Ah, that's interesting, I didn't know that. That seems like it would be most of the way to a proof - though I don't have a quick way to see that the boundary is indeed connected. (But that might be easier; it might even be doable via homology)
$endgroup$
– Milo Brandt
Jan 8 at 20:35
|
show 1 more comment
$begingroup$
In $S^3$ it is not true. Consider the Alexander horned sphere (see en.wikipedia.org/wiki/Alexander_horned_sphere). It separates $S^3$ in two non-empty components $U, V$ as in your question, but one of these is not simply connected.
$endgroup$
– Paul Frost
Jan 8 at 17:22
$begingroup$
I wonder if it is possible to show that the common boundary of $U, V$ is connected?
$endgroup$
– Paul Frost
Jan 8 at 17:29
$begingroup$
@PaulFrost It's certainly true that the boundary is connected; the discrete method I know about for this problem also establishes that the boundary is connected, but it's not a very satisfying proof.
$endgroup$
– Milo Brandt
Jan 8 at 17:35
$begingroup$
I asked because one can show that the complement of compact connected $B subset S^2$ splits into the disjoint union of at most countably many open disks.
$endgroup$
– Paul Frost
Jan 8 at 17:42
$begingroup$
@PaulFrost Ah, that's interesting, I didn't know that. That seems like it would be most of the way to a proof - though I don't have a quick way to see that the boundary is indeed connected. (But that might be easier; it might even be doable via homology)
$endgroup$
– Milo Brandt
Jan 8 at 20:35
$begingroup$
In $S^3$ it is not true. Consider the Alexander horned sphere (see en.wikipedia.org/wiki/Alexander_horned_sphere). It separates $S^3$ in two non-empty components $U, V$ as in your question, but one of these is not simply connected.
$endgroup$
– Paul Frost
Jan 8 at 17:22
$begingroup$
In $S^3$ it is not true. Consider the Alexander horned sphere (see en.wikipedia.org/wiki/Alexander_horned_sphere). It separates $S^3$ in two non-empty components $U, V$ as in your question, but one of these is not simply connected.
$endgroup$
– Paul Frost
Jan 8 at 17:22
$begingroup$
I wonder if it is possible to show that the common boundary of $U, V$ is connected?
$endgroup$
– Paul Frost
Jan 8 at 17:29
$begingroup$
I wonder if it is possible to show that the common boundary of $U, V$ is connected?
$endgroup$
– Paul Frost
Jan 8 at 17:29
$begingroup$
@PaulFrost It's certainly true that the boundary is connected; the discrete method I know about for this problem also establishes that the boundary is connected, but it's not a very satisfying proof.
$endgroup$
– Milo Brandt
Jan 8 at 17:35
$begingroup$
@PaulFrost It's certainly true that the boundary is connected; the discrete method I know about for this problem also establishes that the boundary is connected, but it's not a very satisfying proof.
$endgroup$
– Milo Brandt
Jan 8 at 17:35
$begingroup$
I asked because one can show that the complement of compact connected $B subset S^2$ splits into the disjoint union of at most countably many open disks.
$endgroup$
– Paul Frost
Jan 8 at 17:42
$begingroup$
I asked because one can show that the complement of compact connected $B subset S^2$ splits into the disjoint union of at most countably many open disks.
$endgroup$
– Paul Frost
Jan 8 at 17:42
$begingroup$
@PaulFrost Ah, that's interesting, I didn't know that. That seems like it would be most of the way to a proof - though I don't have a quick way to see that the boundary is indeed connected. (But that might be easier; it might even be doable via homology)
$endgroup$
– Milo Brandt
Jan 8 at 20:35
$begingroup$
@PaulFrost Ah, that's interesting, I didn't know that. That seems like it would be most of the way to a proof - though I don't have a quick way to see that the boundary is indeed connected. (But that might be easier; it might even be doable via homology)
$endgroup$
– Milo Brandt
Jan 8 at 20:35
|
show 1 more comment
1 Answer
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This follows from the fact that a connected open subset of $S^2$ is simply connected iff its complement is connected. This fact itself is proved in many complex analysis textbooks, see also Complement is connected iff Connected components are Simply Connected
answered Jan 9 at 6:00
Lukas GeyerLukas Geyer
13.8k1555
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$begingroup$
This follows from the fact that a connected open subset of $S^2$ is simply connected iff its complement is connected. This fact itself is proved in many complex analysis textbooks, see also Complement is connected iff Connected components are Simply Connected
answered Jan 9 at 6:00
Lukas GeyerLukas Geyer
13.8k1555
$endgroup$
add a comment |
$begingroup$
This follows from the fact that a connected open subset of $S^2$ is simply connected iff its complement is connected. This fact itself is proved in many complex analysis textbooks, see also Complement is connected iff Connected components are Simply Connected
answered Jan 9 at 6:00
Lukas GeyerLukas Geyer
13.8k1555
$endgroup$
add a comment |
$begingroup$
This follows from the fact that a connected open subset of $S^2$ is simply connected iff its complement is connected. This fact itself is proved in many complex analysis textbooks, see also Complement is connected iff Connected components are Simply Connected
answered Jan 9 at 6:00
Lukas GeyerLukas Geyer
13.8k1555
$endgroup$
This follows from the fact that a connected open subset of $S^2$ is simply connected iff its complement is connected. This fact itself is proved in many complex analysis textbooks, see also Complement is connected iff Connected components are Simply Connected
answered Jan 9 at 6:00
Lukas GeyerLukas Geyer
13.8k1555
answered Jan 9 at 6:00
Lukas GeyerLukas Geyer
13.8k1555
answered Jan 9 at 6:00
Lukas GeyerLukas Geyer
13.8k1555
answered Jan 9 at 6:00
Lukas GeyerLukas Geyer
13.8k1555
13.8k1555
add a comment |
add a comment |
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$begingroup$
In $S^3$ it is not true. Consider the Alexander horned sphere (see en.wikipedia.org/wiki/Alexander_horned_sphere). It separates $S^3$ in two non-empty components $U, V$ as in your question, but one of these is not simply connected.
$endgroup$
– Paul Frost
Jan 8 at 17:22
$begingroup$
I wonder if it is possible to show that the common boundary of $U, V$ is connected?
$endgroup$
– Paul Frost
Jan 8 at 17:29
$begingroup$
@PaulFrost It's certainly true that the boundary is connected; the discrete method I know about for this problem also establishes that the boundary is connected, but it's not a very satisfying proof.
$endgroup$
– Milo Brandt
Jan 8 at 17:35
$begingroup$
I asked because one can show that the complement of compact connected $B subset S^2$ splits into the disjoint union of at most countably many open disks.
$endgroup$
– Paul Frost
Jan 8 at 17:42
$begingroup$
@PaulFrost Ah, that's interesting, I didn't know that. That seems like it would be most of the way to a proof - though I don't have a quick way to see that the boundary is indeed connected. (But that might be easier; it might even be doable via homology)
$endgroup$
– Milo Brandt
Jan 8 at 20:35