Maximum possible connected component by removing hyperplanes form $mathbb R^3$












0












$begingroup$


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







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$endgroup$












  • $begingroup$
    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
















0












$begingroup$


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







share|cite|improve this question









$endgroup$












  • $begingroup$
    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














0












0








0


0



$begingroup$


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







share|cite|improve this question









$endgroup$




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






share|cite|improve this question













share|cite|improve this question











share|cite|improve this question




share|cite|improve this question










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


















  • $begingroup$
    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










1 Answer
1






active

oldest

votes


















0












$begingroup$

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.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    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













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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









0












$begingroup$

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.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    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


















0












$begingroup$

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.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    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
















0












0








0





$begingroup$

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.






share|cite|improve this answer









$endgroup$



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.







share|cite|improve this answer












share|cite|improve this answer



share|cite|improve this answer










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




















  • $begingroup$
    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




















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