Number of connected components of a disconnected space

Question

Let $H_1, H_2, H_3, H_4$ be four hyperplanes in $\mathbb{R}^3$. Then the maximum number of connected components of $\mathbb{R}^3-(H_1\cup H_2\cup H_3\cup H_4)$ 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.

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. — mathcounterexamples.net, Dec 05 '18 at 14:22
@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? — vidyarthi, Dec 05 '18 at 14:47

score 1 · Answer 1 · edited Dec 05 '18 at 15:27

1

Take following planes: $$\begin{cases} H_1 &\equiv z= 0 \\ H_2 &\equiv y= 0\\ H_3 &\equiv x=0 \\ H_4 &\equiv x+y = k\ , k\neq0 \end{cases}$$

You'll find that the number of connected components of $\mathbb{R}^3 \setminus (H_1\cup H_2\cup H_3\cup H_4)$ is equal to $14$ (look at the number of connected components of $H_1 \setminus (H_2\cup H_3\cup H_4)$ first which is equal to $7$).

edited Dec 05 '18 at 15:27

vidyarthi

7,315

answered Dec 05 '18 at 14:55

mathcounterexamples.net

71,758

but is the argument true or false? – vidyarthi Dec 05 '18 at 15:07
Circle division by lines is in 2D. The problem of your question is in 3D. At least I answered your first question. – mathcounterexamples.net Dec 05 '18 at 15:13
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? – vidyarthi Dec 05 '18 at 15:15
Have you as I suggested looked at the number of connected components of $H_1 \setminus (H_2\cup H_3\cup H_4)$? What is this number? – mathcounterexamples.net Dec 05 '18 at 15:16
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? – vidyarthi Dec 05 '18 at 15:22
Have a look to the picture... – mathcounterexamples.net Dec 05 '18 at 15:23
ok, yes, the number of components can be $7$, but not with $x+y=0$, but, $x+y=k, k\neq0$, isnt it? – vidyarthi Dec 05 '18 at 15:26
Indeed that is correct. – mathcounterexamples.net Dec 05 '18 at 15:40
see my answer below. Is it right? – vidyarthi Dec 05 '18 at 15:44

vidyarthi · Accepted Answer · 2018-12-05T18:42:20.377

1

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

edited Dec 05 '18 at 18:42

answered Dec 05 '18 at 15:42

vidyarthi

7,315

Number of connected components of a disconnected space

2 Answers2