Number of connected components of a disconnected space
$begingroup$
Let $H_1, H_2, H_3, H_4$ be four hyperplanes in $mathbb{R}^3$. Then the maximum number of connected components of $mathbb{R}^3-(H_1cup H_2cup H_3cup H_4)$ is $14$.
The question is whether the above assertion is true. I think it is false. I think it is equivalent to the circle cutting problem where the lines that cut can be regarded as the hyperplanes, which, in this case correspond to planes. Can this be generalized to higher dimensions?Thanks beforehand.
combinatorics geometry
asked Dec 5 '18 at 14:00
vidyarthividyarthi
2,9291832
$endgroup$
add a comment |
$begingroup$
Let $H_1, H_2, H_3, H_4$ be four hyperplanes in $mathbb{R}^3$. Then the maximum number of connected components of $mathbb{R}^3-(H_1cup H_2cup H_3cup H_4)$ is $14$.
The question is whether the above assertion is true. I think it is false. I think it is equivalent to the circle cutting problem where the lines that cut can be regarded as the hyperplanes, which, in this case correspond to planes. Can this be generalized to higher dimensions?Thanks beforehand.
combinatorics geometry
asked Dec 5 '18 at 14:00
vidyarthividyarthi
2,9291832
$endgroup$
$begingroup$
When you say it is false, do you think that the maximum number of connected components is higher? Lower? The number of connected components can indeed be equal to 14.
$endgroup$
– mathcounterexamples.net
Dec 5 '18 at 14:22
$begingroup$
@mathcounterexamples.net but, according to circle cutting problem, the number of connected components should be about 11. Could you show how it is $14$. So circle cutting is not valid in higher dimensions, I suppose?
$endgroup$
– vidyarthi
Dec 5 '18 at 14:47
add a comment |
$begingroup$
Let $H_1, H_2, H_3, H_4$ be four hyperplanes in $mathbb{R}^3$. Then the maximum number of connected components of $mathbb{R}^3-(H_1cup H_2cup H_3cup H_4)$ is $14$.
The question is whether the above assertion is true. I think it is false. I think it is equivalent to the circle cutting problem where the lines that cut can be regarded as the hyperplanes, which, in this case correspond to planes. Can this be generalized to higher dimensions?Thanks beforehand.
combinatorics geometry
asked Dec 5 '18 at 14:00
vidyarthividyarthi
2,9291832
$endgroup$
Let $H_1, H_2, H_3, H_4$ be four hyperplanes in $mathbb{R}^3$. Then the maximum number of connected components of $mathbb{R}^3-(H_1cup H_2cup H_3cup H_4)$ is $14$.
The question is whether the above assertion is true. I think it is false. I think it is equivalent to the circle cutting problem where the lines that cut can be regarded as the hyperplanes, which, in this case correspond to planes. Can this be generalized to higher dimensions?Thanks beforehand.
combinatorics geometry
combinatorics geometry
asked Dec 5 '18 at 14:00
vidyarthividyarthi
2,9291832
asked Dec 5 '18 at 14:00
vidyarthividyarthi
2,9291832
asked Dec 5 '18 at 14:00
vidyarthividyarthi
2,9291832
asked Dec 5 '18 at 14:00
vidyarthividyarthi
2,9291832
asked Dec 5 '18 at 14:00
vidyarthividyarthi
2,9291832
2,9291832
$begingroup$
When you say it is false, do you think that the maximum number of connected components is higher? Lower? The number of connected components can indeed be equal to 14.
$endgroup$
– mathcounterexamples.net
Dec 5 '18 at 14:22
$begingroup$
@mathcounterexamples.net but, according to circle cutting problem, the number of connected components should be about 11. Could you show how it is $14$. So circle cutting is not valid in higher dimensions, I suppose?
$endgroup$
– vidyarthi
Dec 5 '18 at 14:47
add a comment |
$begingroup$
When you say it is false, do you think that the maximum number of connected components is higher? Lower? The number of connected components can indeed be equal to 14.
$endgroup$
– mathcounterexamples.net
Dec 5 '18 at 14:22
$begingroup$
@mathcounterexamples.net but, according to circle cutting problem, the number of connected components should be about 11. Could you show how it is $14$. So circle cutting is not valid in higher dimensions, I suppose?
$endgroup$
– vidyarthi
Dec 5 '18 at 14:47
$begingroup$
When you say it is false, do you think that the maximum number of connected components is higher? Lower? The number of connected components can indeed be equal to 14.
$endgroup$
– mathcounterexamples.net
Dec 5 '18 at 14:22
$begingroup$
When you say it is false, do you think that the maximum number of connected components is higher? Lower? The number of connected components can indeed be equal to 14.
$endgroup$
– mathcounterexamples.net
Dec 5 '18 at 14:22
$begingroup$
@mathcounterexamples.net but, according to circle cutting problem, the number of connected components should be about 11. Could you show how it is $14$. So circle cutting is not valid in higher dimensions, I suppose?
$endgroup$
– vidyarthi
Dec 5 '18 at 14:47
$begingroup$
@mathcounterexamples.net but, according to circle cutting problem, the number of connected components should be about 11. Could you show how it is $14$. So circle cutting is not valid in higher dimensions, I suppose?
$endgroup$
– vidyarthi
Dec 5 '18 at 14:47
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
Take following planes:
$$begin{cases}
H_1 &equiv z= 0 \
H_2 &equiv y= 0\
H_3 &equiv x=0 \
H_4 &equiv x+y = k , kneq0
end{cases}$$
You'll find that the number of connected components of $mathbb{R}^3 setminus (H_1cup H_2cup H_3cup H_4)$ is equal to $14$ (look at the number of connected components of $H_1 setminus (H_2cup H_3cup H_4)$ first which is equal to $7$).
edited Dec 5 '18 at 15:27
vidyarthi
2,9291832
answered Dec 5 '18 at 14:55
mathcounterexamples.netmathcounterexamples.net
26k21955
$endgroup$
$begingroup$
but is the argument true or false?
$endgroup$
– vidyarthi
Dec 5 '18 at 15:07
$begingroup$
Circle division by lines is in 2D. The problem of your question is in 3D. At least I answered your first question.
$endgroup$
– mathcounterexamples.net
Dec 5 '18 at 15:13
$begingroup$
but, i think the number of componnenrs in your example equals $12$ right, because it bisects $4$ octants and leaves the remaing four octants free. so total is $12$, right?
$endgroup$
– vidyarthi
Dec 5 '18 at 15:15
$begingroup$
Have you as I suggested looked at the number of connected components of $H_1 setminus (H_2cup H_3cup H_4)$? What is this number?
$endgroup$
– mathcounterexamples.net
Dec 5 '18 at 15:16
$begingroup$
the number of components, i suppose, $6$, because, the plane $x+y=0$ bisects the 2nd and 4th quadrants and leaves the other quadrants in the plane $z=0$ free. Thus, the number of components equals $4+2=6$, right?
$endgroup$
– vidyarthi
Dec 5 '18 at 15:22
|
show 4 more comments
$begingroup$
The claim is false. The required number of components is the same as the maximum number of parts $4$ planes can cut the space to. It is given by the formula $frac{n^3+5n+6}{6}$ for $n$ planes. Substituting, we obtain the components to be $15$. Proof here. Example here
answered Dec 5 '18 at 15:42
vidyarthividyarthi
2,9291832
$endgroup$
add a comment |
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2 Answers
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active
oldest
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2 Answers
2
active
oldest
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active
oldest
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active
oldest
votes
$begingroup$
Take following planes:
$$begin{cases}
H_1 &equiv z= 0 \
H_2 &equiv y= 0\
H_3 &equiv x=0 \
H_4 &equiv x+y = k , kneq0
end{cases}$$
You'll find that the number of connected components of $mathbb{R}^3 setminus (H_1cup H_2cup H_3cup H_4)$ is equal to $14$ (look at the number of connected components of $H_1 setminus (H_2cup H_3cup H_4)$ first which is equal to $7$).
edited Dec 5 '18 at 15:27
vidyarthi
2,9291832
answered Dec 5 '18 at 14:55
mathcounterexamples.netmathcounterexamples.net
26k21955
$endgroup$
$begingroup$
but is the argument true or false?
$endgroup$
– vidyarthi
Dec 5 '18 at 15:07
$begingroup$
Circle division by lines is in 2D. The problem of your question is in 3D. At least I answered your first question.
$endgroup$
– mathcounterexamples.net
Dec 5 '18 at 15:13
$begingroup$
but, i think the number of componnenrs in your example equals $12$ right, because it bisects $4$ octants and leaves the remaing four octants free. so total is $12$, right?
$endgroup$
– vidyarthi
Dec 5 '18 at 15:15
$begingroup$
Have you as I suggested looked at the number of connected components of $H_1 setminus (H_2cup H_3cup H_4)$? What is this number?
$endgroup$
– mathcounterexamples.net
Dec 5 '18 at 15:16
$begingroup$
the number of components, i suppose, $6$, because, the plane $x+y=0$ bisects the 2nd and 4th quadrants and leaves the other quadrants in the plane $z=0$ free. Thus, the number of components equals $4+2=6$, right?
$endgroup$
– vidyarthi
Dec 5 '18 at 15:22
|
show 4 more comments
$begingroup$
Take following planes:
$$begin{cases}
H_1 &equiv z= 0 \
H_2 &equiv y= 0\
H_3 &equiv x=0 \
H_4 &equiv x+y = k , kneq0
end{cases}$$
You'll find that the number of connected components of $mathbb{R}^3 setminus (H_1cup H_2cup H_3cup H_4)$ is equal to $14$ (look at the number of connected components of $H_1 setminus (H_2cup H_3cup H_4)$ first which is equal to $7$).
edited Dec 5 '18 at 15:27
vidyarthi
2,9291832
answered Dec 5 '18 at 14:55
mathcounterexamples.netmathcounterexamples.net
26k21955
$endgroup$
$begingroup$
but is the argument true or false?
$endgroup$
– vidyarthi
Dec 5 '18 at 15:07
$begingroup$
Circle division by lines is in 2D. The problem of your question is in 3D. At least I answered your first question.
$endgroup$
– mathcounterexamples.net
Dec 5 '18 at 15:13
$begingroup$
but, i think the number of componnenrs in your example equals $12$ right, because it bisects $4$ octants and leaves the remaing four octants free. so total is $12$, right?
$endgroup$
– vidyarthi
Dec 5 '18 at 15:15
$begingroup$
Have you as I suggested looked at the number of connected components of $H_1 setminus (H_2cup H_3cup H_4)$? What is this number?
$endgroup$
– mathcounterexamples.net
Dec 5 '18 at 15:16
$begingroup$
the number of components, i suppose, $6$, because, the plane $x+y=0$ bisects the 2nd and 4th quadrants and leaves the other quadrants in the plane $z=0$ free. Thus, the number of components equals $4+2=6$, right?
$endgroup$
– vidyarthi
Dec 5 '18 at 15:22
|
show 4 more comments
$begingroup$
Take following planes:
$$begin{cases}
H_1 &equiv z= 0 \
H_2 &equiv y= 0\
H_3 &equiv x=0 \
H_4 &equiv x+y = k , kneq0
end{cases}$$
You'll find that the number of connected components of $mathbb{R}^3 setminus (H_1cup H_2cup H_3cup H_4)$ is equal to $14$ (look at the number of connected components of $H_1 setminus (H_2cup H_3cup H_4)$ first which is equal to $7$).
edited Dec 5 '18 at 15:27
vidyarthi
2,9291832
answered Dec 5 '18 at 14:55
mathcounterexamples.netmathcounterexamples.net
26k21955
$endgroup$
Take following planes:
$$begin{cases}
H_1 &equiv z= 0 \
H_2 &equiv y= 0\
H_3 &equiv x=0 \
H_4 &equiv x+y = k , kneq0
end{cases}$$
You'll find that the number of connected components of $mathbb{R}^3 setminus (H_1cup H_2cup H_3cup H_4)$ is equal to $14$ (look at the number of connected components of $H_1 setminus (H_2cup H_3cup H_4)$ first which is equal to $7$).
edited Dec 5 '18 at 15:27
vidyarthi
2,9291832
answered Dec 5 '18 at 14:55
mathcounterexamples.netmathcounterexamples.net
26k21955
edited Dec 5 '18 at 15:27
vidyarthi
2,9291832
edited Dec 5 '18 at 15:27
vidyarthi
2,9291832
edited Dec 5 '18 at 15:27
vidyarthi
2,9291832
2,9291832
answered Dec 5 '18 at 14:55
mathcounterexamples.netmathcounterexamples.net
26k21955
answered Dec 5 '18 at 14:55
mathcounterexamples.netmathcounterexamples.net
26k21955
answered Dec 5 '18 at 14:55
mathcounterexamples.netmathcounterexamples.net
26k21955
26k21955
$begingroup$
but is the argument true or false?
$endgroup$
– vidyarthi
Dec 5 '18 at 15:07
$begingroup$
Circle division by lines is in 2D. The problem of your question is in 3D. At least I answered your first question.
$endgroup$
– mathcounterexamples.net
Dec 5 '18 at 15:13
$begingroup$
but, i think the number of componnenrs in your example equals $12$ right, because it bisects $4$ octants and leaves the remaing four octants free. so total is $12$, right?
$endgroup$
– vidyarthi
Dec 5 '18 at 15:15
$begingroup$
Have you as I suggested looked at the number of connected components of $H_1 setminus (H_2cup H_3cup H_4)$? What is this number?
$endgroup$
– mathcounterexamples.net
Dec 5 '18 at 15:16
$begingroup$
the number of components, i suppose, $6$, because, the plane $x+y=0$ bisects the 2nd and 4th quadrants and leaves the other quadrants in the plane $z=0$ free. Thus, the number of components equals $4+2=6$, right?
$endgroup$
– vidyarthi
Dec 5 '18 at 15:22
|
show 4 more comments
$begingroup$
but is the argument true or false?
$endgroup$
– vidyarthi
Dec 5 '18 at 15:07
$begingroup$
Circle division by lines is in 2D. The problem of your question is in 3D. At least I answered your first question.
$endgroup$
– mathcounterexamples.net
Dec 5 '18 at 15:13
$begingroup$
but, i think the number of componnenrs in your example equals $12$ right, because it bisects $4$ octants and leaves the remaing four octants free. so total is $12$, right?
$endgroup$
– vidyarthi
Dec 5 '18 at 15:15
$begingroup$
Have you as I suggested looked at the number of connected components of $H_1 setminus (H_2cup H_3cup H_4)$? What is this number?
$endgroup$
– mathcounterexamples.net
Dec 5 '18 at 15:16
$begingroup$
the number of components, i suppose, $6$, because, the plane $x+y=0$ bisects the 2nd and 4th quadrants and leaves the other quadrants in the plane $z=0$ free. Thus, the number of components equals $4+2=6$, right?
$endgroup$
– vidyarthi
Dec 5 '18 at 15:22
$begingroup$
but is the argument true or false?
$endgroup$
– vidyarthi
Dec 5 '18 at 15:07
$begingroup$
but is the argument true or false?
$endgroup$
– vidyarthi
Dec 5 '18 at 15:07
$begingroup$
Circle division by lines is in 2D. The problem of your question is in 3D. At least I answered your first question.
$endgroup$
– mathcounterexamples.net
Dec 5 '18 at 15:13
$begingroup$
Circle division by lines is in 2D. The problem of your question is in 3D. At least I answered your first question.
$endgroup$
– mathcounterexamples.net
Dec 5 '18 at 15:13
$begingroup$
but, i think the number of componnenrs in your example equals $12$ right, because it bisects $4$ octants and leaves the remaing four octants free. so total is $12$, right?
$endgroup$
– vidyarthi
Dec 5 '18 at 15:15
$begingroup$
but, i think the number of componnenrs in your example equals $12$ right, because it bisects $4$ octants and leaves the remaing four octants free. so total is $12$, right?
$endgroup$
– vidyarthi
Dec 5 '18 at 15:15
$begingroup$
Have you as I suggested looked at the number of connected components of $H_1 setminus (H_2cup H_3cup H_4)$? What is this number?
$endgroup$
– mathcounterexamples.net
Dec 5 '18 at 15:16
$begingroup$
Have you as I suggested looked at the number of connected components of $H_1 setminus (H_2cup H_3cup H_4)$? What is this number?
$endgroup$
– mathcounterexamples.net
Dec 5 '18 at 15:16
$begingroup$
the number of components, i suppose, $6$, because, the plane $x+y=0$ bisects the 2nd and 4th quadrants and leaves the other quadrants in the plane $z=0$ free. Thus, the number of components equals $4+2=6$, right?
$endgroup$
– vidyarthi
Dec 5 '18 at 15:22
$begingroup$
the number of components, i suppose, $6$, because, the plane $x+y=0$ bisects the 2nd and 4th quadrants and leaves the other quadrants in the plane $z=0$ free. Thus, the number of components equals $4+2=6$, right?
$endgroup$
– vidyarthi
Dec 5 '18 at 15:22
|
show 4 more comments
$begingroup$
The claim is false. The required number of components is the same as the maximum number of parts $4$ planes can cut the space to. It is given by the formula $frac{n^3+5n+6}{6}$ for $n$ planes. Substituting, we obtain the components to be $15$. Proof here. Example here
answered Dec 5 '18 at 15:42
vidyarthividyarthi
2,9291832
$endgroup$
add a comment |
$begingroup$
The claim is false. The required number of components is the same as the maximum number of parts $4$ planes can cut the space to. It is given by the formula $frac{n^3+5n+6}{6}$ for $n$ planes. Substituting, we obtain the components to be $15$. Proof here. Example here
answered Dec 5 '18 at 15:42
vidyarthividyarthi
2,9291832
$endgroup$
add a comment |
$begingroup$
The claim is false. The required number of components is the same as the maximum number of parts $4$ planes can cut the space to. It is given by the formula $frac{n^3+5n+6}{6}$ for $n$ planes. Substituting, we obtain the components to be $15$. Proof here. Example here
answered Dec 5 '18 at 15:42
vidyarthividyarthi
2,9291832
$endgroup$
The claim is false. The required number of components is the same as the maximum number of parts $4$ planes can cut the space to. It is given by the formula $frac{n^3+5n+6}{6}$ for $n$ planes. Substituting, we obtain the components to be $15$. Proof here. Example here
answered Dec 5 '18 at 15:42
vidyarthividyarthi
2,9291832
edited Dec 5 '18 at 18:42
answered Dec 5 '18 at 15:42
vidyarthividyarthi
2,9291832
answered Dec 5 '18 at 15:42
vidyarthividyarthi
2,9291832
answered Dec 5 '18 at 15:42
vidyarthividyarthi
2,9291832
2,9291832
add a comment |
add a comment |
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$begingroup$
When you say it is false, do you think that the maximum number of connected components is higher? Lower? The number of connected components can indeed be equal to 14.
$endgroup$
– mathcounterexamples.net
Dec 5 '18 at 14:22
$begingroup$
@mathcounterexamples.net but, according to circle cutting problem, the number of connected components should be about 11. Could you show how it is $14$. So circle cutting is not valid in higher dimensions, I suppose?
$endgroup$
– vidyarthi
Dec 5 '18 at 14:47