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edited Dec 5, 2016 at 22:39

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I believe the answer is yes. Let $X\neq \emptyset$ and let $P,Q,R\in {\rm Part}(X)$.

First, note that if $|P|<|Q|$, then $P<_{\rm eff}Q$, since $|P-Q|\leq |P|<|Q|=|Q-P|$.

Now we are ready for the proof. Assume $P<_{\rm eff}Q<_{\rm eff}R$. Without loss of generality, we may remove $P\cap Q\cap R$ from all the sets, and reduce to the case that $P\cap Q\cap R=\emptyset$. By our previous paragraph, we also have $|P|\leq |Q|\leq |R|$. If inequality holds anywhere, then we have $P<_{\rm eff}R$ and we are done. Assume the contrary case holds, and so all three sets have equal cardinality. Note that $|P|$ must be infinite, else the computation from my comment above applies.

From the fact that $|P-Q|<|Q-P|$ and $|P|=|Q|$, we see that $|P\cap Q|=|P|=|Q|$. Now because $(P\cap Q)\cap R=\emptyset$, we have $|Q-R|\geq |P\cap Q|=|Q|=|R|\geq |R-Q|$, contradicting the fact that $Q<_{\rm eff}R$. Thus, this contrary case cannot actually happen after all.

I believe the answer is yes. Let $X\neq \emptyset$ and let $P,Q,R\in {\rm Part}(X)$.

First, note that if $|P|<|Q|$, then $P<_{\rm eff}Q$, since $|P-Q|\leq |P|<|Q|=|Q-P|$.

Now we are ready for the proof. Assume $P<_{\rm eff}Q<_{\rm eff}R$. Without loss of generality, we may remove $P\cap Q\cap R$ from all the sets, and reduce to the case that $P\cap Q\cap R=\emptyset$. By our previous paragraph, we also have $|P|\leq |Q|\leq |R|$. If inequality holds anywhere, then we have $P<_{\rm eff}R$ and we are done. Assume the contrary case holds, and so all three sets have equal cardinality.

From the fact that $|P-Q|<|Q-P|$ and $|P|=|Q|$, we see that $|P\cap Q|=|P|=|Q|$. Now because $(P\cap Q)\cap R=\emptyset$, we have $|Q-R|\geq |P\cap Q|=|Q|=|R|\geq |R-Q|$, contradicting the fact that $Q<_{\rm eff}R$. Thus, this contrary case cannot actually happen after all.

I believe the answer is yes. Let $X\neq \emptyset$ and let $P,Q,R\in {\rm Part}(X)$.

First, note that if $|P|<|Q|$, then $P<_{\rm eff}Q$, since $|P-Q|\leq |P|<|Q|=|Q-P|$.

Now we are ready for the proof. Assume $P<_{\rm eff}Q<_{\rm eff}R$. Without loss of generality, we may remove $P\cap Q\cap R$ from all the sets, and reduce to the case that $P\cap Q\cap R=\emptyset$. By our previous paragraph, we also have $|P|\leq |Q|\leq |R|$. If inequality holds anywhere, then we have $P<_{\rm eff}R$ and we are done. Assume the contrary case holds, and so all three sets have equal cardinality. Note that $|P|$ must be infinite, else the computation from my comment above applies.

From the fact that $|P-Q|<|Q-P|$ and $|P|=|Q|$, we see that $|P\cap Q|=|P|=|Q|$. Now because $(P\cap Q)\cap R=\emptyset$, we have $|Q-R|\geq |P\cap Q|=|Q|=|R|\geq |R-Q|$, contradicting the fact that $Q<_{\rm eff}R$. Thus, this contrary case cannot actually happen after all.

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created Dec 5, 2016 at 22:20

18.7k
4
75
137

I believe the answer is yes. Let $X\neq \emptyset$ and let $P,Q,R\in {\rm Part}(X)$.

First, note that if $|P|<|Q|$, then $P<_{\rm eff}Q$, since $|P-Q|\leq |P|<|Q|=|Q-P|$.

Now we are ready for the proof. Assume $P<_{\rm eff}Q<_{\rm eff}R$. Without loss of generality, we may remove $P\cap Q\cap R$ from all the sets, and reduce to the case that $P\cap Q\cap R=\emptyset$. By our previous paragraph, we also have $|P|\leq |Q|\leq |R|$. If inequality holds anywhere, then we have $P<_{\rm eff}R$ and we are done. Assume the contrary case holds, and so all three sets have equal cardinality.

From the fact that $|P-Q|<|Q-P|$ and $|P|=|Q|$, we see that $|P\cap Q|=|P|=|Q|$. Now because $(P\cap Q)\cap R=\emptyset$, we have $|Q-R|\geq |P\cap Q|=|Q|=|R|\geq |R-Q|$, contradicting the fact that $Q<_{\rm eff}R$. Thus, this contrary case cannot actually happen after all.