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Given a set of points $\pmb{R}$. Let $\pmb{R}(t)$ denote the set of points occupied by $\pmb{R}$ at some time $t \in [a,b]$. Let $\pmb{S}([a,b])$ denote the swept volume generated by the motion of $\pmb{R}$ over the time interval $[a,b]$. $\pmb{S}([a,b])$ can be defined as the infinite union of $\pmb{R}(t)$ in $[a,b]$, such that

$\displaystyle \pmb{S}([a,b]) = \bigcup _{t \in [a,b]} \pmb{R}(t) = \{\vec{p} \mid \vec{p} \in \pmb{R}(t), \exists t \in [a,b] \}.$

Let $I_i$ denote the subinterval of $[a,b]$, such that

$\displaystyle [a,b] = \bigcup^{n}_{i = 1} I_i$

where $I_i$ = $[t_{i-1},t_{i}]$ for $a = t_0 < t_i < \cdots < t_n = b$. Hence, $\pmb{S}([a,b])$ can also be expressed by the union of smaller swept volumes, such that

$\displaystyle \pmb{S}([a,b]) = \bigcup^{n}_{i = 1} \pmb{S} (I_i).$


Given two sets of points $\pmb{R_U}$ and $\pmb{R_S}$. Let $\pmb{R_U}(t)$ and $\pmb{R_S}_(t)$ denote the two sets of points occupied by $\pmb{R_U}$ and $\pmb{R_S}$, respectively, at some time $t \in [a,b]$, and $\pmb{S_U}([a,b])$ and $\pmb{S_S}([a,b])$ denote the swept volume generated by the motion of $\pmb{R_U}$ and $\pmb{R_S}$, respectively, over the time interval $[a,b]$.

It is proven that $\pmb{R_U}$ overlaps $\pmb{R_S}$ in $[a,b]$ if and only if there exists a common point $\vec{p}$ that lies within both $\pmb{R_U}(t)$ and $\pmb{R_S}(t)$ at a certain time $t \in [a,b]$, such that

$\displaystyle \bigcup_{t \in [a,b]} ( \pmb{R_U}(t) \cap \pmb{R_S}(t) ) \neq \emptyset \Leftrightarrow \exists(\vec{p},t) (\vec{p} \in \pmb{R_U}(t) \wedge \vec{p} \in \pmb{R_S}(t), t \in [a,b]).$

Now, prove the following:

  1. $\displaystyle \bigcup_{t \in [a,b]} ( \pmb{R_U}(t) \cap \pmb{R_S}(t) ) \subseteq \bigcup^{n}_{i = 1} ( \pmb{S_U} (I_i) \cap \pmb{S_S} (I_i) )$

  2. $\pmb{R_U}$ overlaps $\pmb{R_S}$ in $[a,b]$ only if $\displaystyle \bigcup^{n}_{i = 1} ( \pmb{S_U} (I_i) \cap \pmb{S_S} (I_i) ) \neq \emptyset$

Many thanks in advance for your help!

share|improve this question
    
As Asaf pointed out in your previous question, your tag choices are a bit suspect. Please read the tag wiki excerpt when you choose your tags (it is even displayed automatically now!); it is pretty clear that your question has absolutely nothing to do with (descriptive-set-theory), and on this site we reserve the (set-theory) tag for more abstract topics than what pertains to this question. –  Willie Wong Aug 17 '11 at 13:21
    
Thanks for reminding. I am sorry that I mistagged the questions. –  Elvis Aug 17 '11 at 13:32
    
No harm done. Since you are new here (Welcome, by the way!), I thought I'd point out to you one of the little quirks of ours on this forum. Enjoy! –  Willie Wong Aug 17 '11 at 13:43

1 Answer 1

Let us shorten condition 1. into: $A\subseteq B$, and condition 2. into: $R_U$ overlaps $R_S$ only if $B\ne\varnothing$.

Then, $p\in A$ is equivalent to: $\exists t\,(t\in[a,b]\land p\in R_U(t)\land p\in R_S(t))$, which is equivalent to $$ \exists (i,t)\,(t\in I_i\land p\in R_U(t)\land p\in R_S(t)). $$ Likewise, $p\in B$ is equivalent to: $\exists i\,(p\in S_U(I_i)\land p\in S_S(I_i))$. Since $p\in S_U(I_i)$ is equivalent to $\exists u\,(u\in I_i\land p\in R_U(u))$ and $p\in S_S(I_i)$ is equivalent to $\exists s\,(s\in I_i\land p\in R_S(s))$, $p\in B$ is equivalent to $$ \exists (i,u,s)\,(u\in I_i\land s\in I_i\land p\in R_U(u)\land p\in R_S(s)). $$ The first condition implies the second (take $u=s=t$) hence: $\forall p\,(p\in A\implies p\in B)$, which means $A\subseteq B$. This proves that 1. holds.

Furthermore, the property that $R_U$ overlaps $R_S$ is defined as the fact that $A\ne\varnothing$, hence 2. reads: $A\ne\varnothing\implies B\ne\varnothing$. Since $A\subseteq B$ and $A\ne\varnothing$ imply $B\ne\varnothing$, 1. implies 2., hence 2. holds.

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Hi Didier, many thanks for your help. I still have a question about (1). If $n \rightarrow \infty$, can I write $\displaystyle \bigcup_{t \in [a,b]} ( \pmb{R_U}(t) \cap \pmb{R_S}(t) ) = \bigcup^{n}_{i = 1} ( \pmb{S_U} (I_i) \cap \pmb{S_S} (I_i) )$ ? If so how do you prove it? –  Elvis Aug 18 '11 at 4:39
    
The answer is already written in my post, in a way: you must determine whether the second displayed assertion (which is a translation of $p\in B$) implies the first one (which is a translation of $p\in A$) or not. If it does, then $A=B$, otherwise $A\subseteq B$ and $A\ne B$ in general. So... what is your pick? –  Did Aug 18 '11 at 7:44
    
Reading your comment again, I fail to understand the meaning of "If $n\to\infty$" since the formula after "can I write" involves a fixed $n$. You could make more precise what is your question. –  Did Aug 22 '11 at 1:10

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