Changeset - 15598be578d9
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Merge
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Tom Bannink - 8 years ago 2017-06-01 18:00:22
tom.bannink@cwi.nl
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@@ -610,7 +610,7 @@ Note by Tom: So $A^{(\mathcal{P})}$ is the event that the set of all patches is
 
   	Otherwise if all elements of $S_{><}\setminus P$ are larger than $P_{\max}$ then we view the last summation as $\sum_{f'_{\overline{P}}\in\{0,1'\}^{|S\cap \overline{P}\setminus\{S_{\max}\}|}}\sum_{f''_{\overline{P}}\in\{0,1'\}^{1}}$ and use Lemma~\ref{lemma:probIndep} to conclude the cancellations pairwise regarding the filling of $S_{\max}$, i.e., the value of $f''_{\overline{P}}$. We proceed similarly when 
 
   	all elements of $S_{><}\setminus P$ are smaller than $P_{\min}$. In the last case we again proceed similarly, but now the cancellations will come from the interplay of $4$ fillings, corresponding to the possible filling of $S_{\min}$ and $S_{\max}$ simultaneously.
 
   	   
 
	I think the same arguments would directly translate to the torus and other translationally invariant objects, so we could go higher dimensional as Mario suggested. Then one would need to replace $|S_{><}|$ by the minimal number $k$ such that there is a $C$ set for which $S\cup C$ is connected.
 
	I think the same arguments would translate to the torus and other translationally invariant spaces, so we could go higher dimensional as Mario suggested. Then I think one would need to replace $|S_{><}|$ by the minimal number $k$ such that there is a $C$ set for which $S\cup C$ is connected. I am not entirely sure how to generalise Lemma~\ref{lemma:probIndep} though, which has key importance in the present proof.
 
    
 
    Questions:
 
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