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I am trying to prove or disprove the following. Let $\mathcal{C}$ be a lower dimensional subset in $\mathbb{R}^n$. In addition, let $\mathcal{M}^{(i)}=\{\mathcal{M}^{(i)}_{1},\mathcal{M}^{(i)}_{2}\}$ be two subsets of $\mathcal{C}$ such that \begin{equation} \mathcal{M}^{(i)}_{1}\cup \mathcal{M}^{(i)}_{2}=\mathcal{C}, ~~\mathcal{M}^{(i)}_{1}\cap \mathcal{M}^{(i)}_{2}=\emptyset \end{equation} and \begin{equation} \mu(\mathcal{\mathcal{M}^{(i)}_{1}})=\mu(\mathcal{\mathcal{M}^{(i)}_{2}})=\frac{1}{2}\mu(\mathcal{C}) \end{equation} where $\mu(\mathcal{S})$ is the volume of the subset $\mathcal{S}\subset\mathcal{C}$. There are $m$ different couples $\{\mathcal{M_1}^{(1)},\mathcal{M_2}^{(1)}\},...,\{\mathcal{M}^{(m)}_{1},\mathcal{M}^{(m)}_{2}\}$, all satisfying the above equations.

Now, given a point $q\in\mathcal{M}^{(1)}_{k_1},\mathcal{M}^{(2)}_{k_2},...,\mathcal{M}^{(m)}_{k_m}$ where $k_i=1,2$. Can I always say that the volume of the union \begin{equation} \bigcup_{i=1}^m\mathcal{M}^{(i)}_{k_i} \end{equation} is always smaller (and not equal) than $\mu(\mathcal{C})$. In simple words, will a family of different half subsets of $\mathcal{C}$ that also contain a common element, cover the whole set $\mathcal{C}$. Is this true always or only in some special geometries? For example, if $\mathcal{M}^{(i)}_{k_i}$ is a hemisphere, than this is true.

Thanks.

Asaf Karagila
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Jeremy Span
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1 Answers1

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No, you cannot expect that. In fact, per your statement that $\mathcal C$ is of "lower dimension" than $\Bbb R^n$, that is never true. Why? Because by any definition of "lower dimension" that I am familiar with (I can think of four off the top of my head), it implies that $\mu(\mathcal C) = 0$. So the measure of your union cannot be any smaller.

But even if you drop that lower dimension restriction, this is still not true. Let $\mathcal C$ be some set with $\mu(\mathcal C) > 0$ and let $q \in \mathcal C$, and let $A,B$ be any two measurable sets with $A\cup B = \mathcal C \setminus \{q\}, A\cap B = \emptyset$.

Then you can set $\mathcal M^1_1 = A\cup\{q\}, \mathcal M^1_2 = B, \mathcal M^2_1 = B\cup\{q\}, \mathcal M^2_2 = A$.

So $ \mathcal M^1_1 \cup \mathcal M^2_1 = (A\cup\{q\}) \cup (B\cup\{q\}) = \mathcal C$

And futher, even though $\mathcal M^1_2 \cup \mathcal M^2_2 = \mathcal C \setminus \{q\}$, we still have $\mu(\mathcal M^1_2 \cup \mathcal M^2_2) = \mu(\mathcal C)$.

Added

Here is a more specific example, which also contradicts your statement about hemispheres. I'll use my suggested notation. Let $C$ be the unit circle in the plane. Define $$U_1 = \{(x,y) \in C \mid y \le 0\}, V_1 = \{(x,y) \in C \mid y > 0\}\\U_2 = \{(x,y) \in C \mid x \le 0\}, V_2 = \{(x,y) \in C \mid x > 0\}\\U_3 = \{(x,y) \in C \mid y \ge 0\}, V_3 = \{(x,y) \in C \mid y < 0\}\\U_4 = \{(x,y) \in C \mid x \ge 0\}, V_4 = \{(x,y) \in C \mid x < 0\}$$ Then for each $i, U_i \cap V_i = \emptyset$ and $U_i \cup V_i = C$. Further, for all $i, (0,0) \in U_i$. But $$\bigcup_{i=1}^4 U_i = C$$

Paul Sinclair
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  • Thank you for your answer. You are correct about the volume and it should be some other measure. However, I did not get the second part. If the union of sets A and B is C, then q can only be in one of them so that any M is half of C. Therefore, q cannot be both in M^1_1 and M^2_1. – Jeremy Span Mar 03 '17 at 05:35
  • My counter-example works for any measure that has a point $q$ for which ${q}$ is measurable. – Paul Sinclair Mar 03 '17 at 05:40
  • I get it. Still, if the union of sets A and B is C, then {q} can only be in one of them so that any M is half of C. Therefore, {q} cannot be both in M^1_1 and M^2_1. This does not add up for me... Thanks. – Jeremy Span Mar 03 '17 at 05:45
  • You are the one who required that the sets being unioned all contain $q$, and now you are claiming this is a problem? Remember that it is $M^1_1$ and $M^1_2$ that you required to be disjoint, not $M^1_1$ and $M^2_1$. – Paul Sinclair Mar 03 '17 at 05:49
  • My mistake. I should have written that $k_i$ equals 1 or 2, not both... – Jeremy Span Mar 03 '17 at 05:53
  • You would be better off abandoning the subscript/superscript orgy and simply calling your $i$th "halfspaces" $U_i$ and $V_i$, So for each $i$, $C$ is the disjoint union of $U_i$ and $V_i$, and $q \in \bigcap_i U_i$, and you are asking about $\bigcup_i U_i$. That is a much easier symbolism to follow. (The mathcals don't add anything to this problem either.) – Paul Sinclair Mar 03 '17 at 18:17
  • Ya, sorry about that, I copied it from my Latex paper and the indices are essential to my work. But not for the essence of the problem. Sorry again... – Jeremy Span Mar 04 '17 at 05:10
  • No reason to apologize. I just think you would find it easier to contemplate. Also, I've added a concrete example, to hopefully make the problem more clear to you. – Paul Sinclair Mar 04 '17 at 18:47