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Given two sets of points $\pmb{R_U}$, $\pmb{R_S}$ and a time interval $[a,b]$. Let $\pmb{R_U}(t)$, $\pmb{R_S}(t)$ denote the two sets of points occupied by $\pmb{R_U}$ and $\pmb{R_S}$, respectively, at some time $t \in [a,b]$. Then, $\pmb{R_U}$ overlaps $\pmb{R_S}$ in $[a,b]$ if and only if there exists a common point that lies within both $\pmb{R_U}(t)$ and $\pmb{R_S}(t)$ at a certain time $t \in [a,b]$, such that

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


How do you prove the above theorem?

Thanks

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@Asaf Is that sarcasm I detect? –  Quinn Culver Aug 17 '11 at 11:30
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@Asaf: why don't you retag? –  Willie Wong Aug 17 '11 at 13:17
    
Thanks for reminding. I am new to this site. I am sorry that I mistagged the questions. –  Elvis Aug 17 '11 at 13:31
    
@Willie: I wasn't even sure what the question was at the time of reading... (the blame lies in immense lack of sleep). –  Asaf Karagila Aug 17 '11 at 15:45
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1 Answer

up vote 0 down vote accepted

So in fact $R_U$ and $R_S$ are functions defined on $[a,b]$ such that for every $t$ in $[a,b]$, $R_U(t)$ and $R_S(t)$ are sets.

The LHS of the equivalence you try to prove is equivalent to: $\exists t\,(t\in[a,b]\land R_U(t)\cap R_S(t)\ne\varnothing)$.

The assertion $R_U(t)\cap R_S(t)\ne\varnothing$ is equivalent to: $\exists p\,(p\in R_U(t)\land p\in R_S(t))$.

Now, for any propositions $P$ and $Q$, the assertion: $\exists t\,(P(t)\land(\exists p\,Q(p,t))$ is equivalent to: $\exists (t,p)\,(P(t)\land Q(p,t))$.

Using $t\in[a,b]$ as $P(t)$ and $p\in R_U(t)\land p\in R_S(t)$ as $Q(p,t)$, one gets that the LHS is equivalent to the RHS.

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Thank you very much for your answer. Could you please help me with the following problem too? math.stackexchange.com/questions/58049/… –  Elvis Aug 17 '11 at 13:01
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