Maximum possible connected component by removing hyperplanes form $mathbb R^3$
$begingroup$
Let $A_1,A_2,A_3,A_4$ be 4 hyperplanes in $mathbb R^3$ then How many maximum possible connected component are present in $mathbb R^3$ after removing these 4 hyperplane
I can visuallise that if i remove 3 hyperplanes then maximum 8 connected component present.
I donot able to visualise for 4 hyperplanes .
Any help will be appreciated
real-analysis general-topology analysis connectedness
asked Nov 30 '18 at 4:53
MathLoverMathLover
51410
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add a comment |
$begingroup$
Let $A_1,A_2,A_3,A_4$ be 4 hyperplanes in $mathbb R^3$ then How many maximum possible connected component are present in $mathbb R^3$ after removing these 4 hyperplane
I can visuallise that if i remove 3 hyperplanes then maximum 8 connected component present.
I donot able to visualise for 4 hyperplanes .
Any help will be appreciated
real-analysis general-topology analysis connectedness
asked Nov 30 '18 at 4:53
MathLoverMathLover
51410
$endgroup$
$begingroup$
See cake number, the number for 4 planes is 15. There is an animation for this 4->15 configuration in above wiki entry.
$endgroup$
– achille hui
Nov 30 '18 at 6:59
$begingroup$
Why hyper?$;!$
$endgroup$
– Ivan Neretin
Nov 30 '18 at 7:27
add a comment |
$begingroup$
Let $A_1,A_2,A_3,A_4$ be 4 hyperplanes in $mathbb R^3$ then How many maximum possible connected component are present in $mathbb R^3$ after removing these 4 hyperplane
I can visuallise that if i remove 3 hyperplanes then maximum 8 connected component present.
I donot able to visualise for 4 hyperplanes .
Any help will be appreciated
real-analysis general-topology analysis connectedness
asked Nov 30 '18 at 4:53
MathLoverMathLover
51410
$endgroup$
Let $A_1,A_2,A_3,A_4$ be 4 hyperplanes in $mathbb R^3$ then How many maximum possible connected component are present in $mathbb R^3$ after removing these 4 hyperplane
I can visuallise that if i remove 3 hyperplanes then maximum 8 connected component present.
I donot able to visualise for 4 hyperplanes .
Any help will be appreciated
real-analysis general-topology analysis connectedness
real-analysis general-topology analysis connectedness
asked Nov 30 '18 at 4:53
MathLoverMathLover
51410
asked Nov 30 '18 at 4:53
MathLoverMathLover
51410
asked Nov 30 '18 at 4:53
MathLoverMathLover
51410
asked Nov 30 '18 at 4:53
MathLoverMathLover
51410
asked Nov 30 '18 at 4:53
MathLoverMathLover
51410
51410
$begingroup$
See cake number, the number for 4 planes is 15. There is an animation for this 4->15 configuration in above wiki entry.
$endgroup$
– achille hui
Nov 30 '18 at 6:59
$begingroup$
Why hyper?$;!$
$endgroup$
– Ivan Neretin
Nov 30 '18 at 7:27
add a comment |
$begingroup$
See cake number, the number for 4 planes is 15. There is an animation for this 4->15 configuration in above wiki entry.
$endgroup$
– achille hui
Nov 30 '18 at 6:59
$begingroup$
Why hyper?$;!$
$endgroup$
– Ivan Neretin
Nov 30 '18 at 7:27
$begingroup$
See cake number, the number for 4 planes is 15. There is an animation for this 4->15 configuration in above wiki entry.
$endgroup$
– achille hui
Nov 30 '18 at 6:59
$begingroup$
See cake number, the number for 4 planes is 15. There is an animation for this 4->15 configuration in above wiki entry.
$endgroup$
– achille hui
Nov 30 '18 at 6:59
$begingroup$
Why hyper?$;!$
$endgroup$
– Ivan Neretin
Nov 30 '18 at 7:27
$begingroup$
Why hyper?$;!$
$endgroup$
– Ivan Neretin
Nov 30 '18 at 7:27
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
In some sense, this is akin to removing three lines from $Bbb R^2$, where
you can get at most seven planar regions.
Consider a tetrahedron $T$ in $Bbb R^3$.
By removing the four planes forming the boundary of $T$ one gets fifteen regions
remaining.
Let's consider what happens when we remove the fourth plane $P$ in the general
case. Before then there are at most eight regions. We claim we can gain
at most seven regions at this stage. The earlier planes meet $P$ in
at most three lines. These lines divide the plane $P$ into at most seven
plane regions. But these plane regions correspond to regions in $Bbb R^3$
that are split in two by the fourth hyperplane: the number of regions gained
by the fourth plane is the number of plane regions the lines divide
$P$ into, and is at most seven.
answered Nov 30 '18 at 5:04
Lord Shark the UnknownLord Shark the Unknown
104k1160132
$endgroup$
$begingroup$
Nice Solution Sir. I counted for tetrahedron that there are 13 component outside and 4 component inside tetrahedorn .SO there are 17 total connected component? Is this true
$endgroup$
– MathLover
Nov 30 '18 at 5:20
add a comment |
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1 Answer
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active
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1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
In some sense, this is akin to removing three lines from $Bbb R^2$, where
you can get at most seven planar regions.
Consider a tetrahedron $T$ in $Bbb R^3$.
By removing the four planes forming the boundary of $T$ one gets fifteen regions
remaining.
Let's consider what happens when we remove the fourth plane $P$ in the general
case. Before then there are at most eight regions. We claim we can gain
at most seven regions at this stage. The earlier planes meet $P$ in
at most three lines. These lines divide the plane $P$ into at most seven
plane regions. But these plane regions correspond to regions in $Bbb R^3$
that are split in two by the fourth hyperplane: the number of regions gained
by the fourth plane is the number of plane regions the lines divide
$P$ into, and is at most seven.
answered Nov 30 '18 at 5:04
Lord Shark the UnknownLord Shark the Unknown
104k1160132
$endgroup$
$begingroup$
Nice Solution Sir. I counted for tetrahedron that there are 13 component outside and 4 component inside tetrahedorn .SO there are 17 total connected component? Is this true
$endgroup$
– MathLover
Nov 30 '18 at 5:20
add a comment |
$begingroup$
In some sense, this is akin to removing three lines from $Bbb R^2$, where
you can get at most seven planar regions.
Consider a tetrahedron $T$ in $Bbb R^3$.
By removing the four planes forming the boundary of $T$ one gets fifteen regions
remaining.
Let's consider what happens when we remove the fourth plane $P$ in the general
case. Before then there are at most eight regions. We claim we can gain
at most seven regions at this stage. The earlier planes meet $P$ in
at most three lines. These lines divide the plane $P$ into at most seven
plane regions. But these plane regions correspond to regions in $Bbb R^3$
that are split in two by the fourth hyperplane: the number of regions gained
by the fourth plane is the number of plane regions the lines divide
$P$ into, and is at most seven.
answered Nov 30 '18 at 5:04
Lord Shark the UnknownLord Shark the Unknown
104k1160132
$endgroup$
$begingroup$
Nice Solution Sir. I counted for tetrahedron that there are 13 component outside and 4 component inside tetrahedorn .SO there are 17 total connected component? Is this true
$endgroup$
– MathLover
Nov 30 '18 at 5:20
add a comment |
$begingroup$
In some sense, this is akin to removing three lines from $Bbb R^2$, where
you can get at most seven planar regions.
Consider a tetrahedron $T$ in $Bbb R^3$.
By removing the four planes forming the boundary of $T$ one gets fifteen regions
remaining.
Let's consider what happens when we remove the fourth plane $P$ in the general
case. Before then there are at most eight regions. We claim we can gain
at most seven regions at this stage. The earlier planes meet $P$ in
at most three lines. These lines divide the plane $P$ into at most seven
plane regions. But these plane regions correspond to regions in $Bbb R^3$
that are split in two by the fourth hyperplane: the number of regions gained
by the fourth plane is the number of plane regions the lines divide
$P$ into, and is at most seven.
answered Nov 30 '18 at 5:04
Lord Shark the UnknownLord Shark the Unknown
104k1160132
$endgroup$
In some sense, this is akin to removing three lines from $Bbb R^2$, where
you can get at most seven planar regions.
Consider a tetrahedron $T$ in $Bbb R^3$.
By removing the four planes forming the boundary of $T$ one gets fifteen regions
remaining.
Let's consider what happens when we remove the fourth plane $P$ in the general
case. Before then there are at most eight regions. We claim we can gain
at most seven regions at this stage. The earlier planes meet $P$ in
at most three lines. These lines divide the plane $P$ into at most seven
plane regions. But these plane regions correspond to regions in $Bbb R^3$
that are split in two by the fourth hyperplane: the number of regions gained
by the fourth plane is the number of plane regions the lines divide
$P$ into, and is at most seven.
answered Nov 30 '18 at 5:04
Lord Shark the UnknownLord Shark the Unknown
104k1160132
answered Nov 30 '18 at 5:04
Lord Shark the UnknownLord Shark the Unknown
104k1160132
answered Nov 30 '18 at 5:04
Lord Shark the UnknownLord Shark the Unknown
104k1160132
answered Nov 30 '18 at 5:04
Lord Shark the UnknownLord Shark the Unknown
104k1160132
104k1160132
$begingroup$
Nice Solution Sir. I counted for tetrahedron that there are 13 component outside and 4 component inside tetrahedorn .SO there are 17 total connected component? Is this true
$endgroup$
– MathLover
Nov 30 '18 at 5:20
add a comment |
$begingroup$
Nice Solution Sir. I counted for tetrahedron that there are 13 component outside and 4 component inside tetrahedorn .SO there are 17 total connected component? Is this true
$endgroup$
– MathLover
Nov 30 '18 at 5:20
$begingroup$
Nice Solution Sir. I counted for tetrahedron that there are 13 component outside and 4 component inside tetrahedorn .SO there are 17 total connected component? Is this true
$endgroup$
– MathLover
Nov 30 '18 at 5:20
$begingroup$
Nice Solution Sir. I counted for tetrahedron that there are 13 component outside and 4 component inside tetrahedorn .SO there are 17 total connected component? Is this true
$endgroup$
– MathLover
Nov 30 '18 at 5:20
add a comment |
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$begingroup$
See cake number, the number for 4 planes is 15. There is an animation for this 4->15 configuration in above wiki entry.
$endgroup$
– achille hui
Nov 30 '18 at 6:59
$begingroup$
Why hyper?$;!$
$endgroup$
– Ivan Neretin
Nov 30 '18 at 7:27