Property of set exclusion set.
Property of set exclusion set.
Let $T$ have the property that for all sets $A, B in T$ we have that $(Abackslash B) in T$.
How can I prove that $forall A,B in T, Acap B in T$?
I was thinking I should start with both expressions:
$(Abackslash B) in T$.
$(Bbackslash A) in T$.
and show that$ (A cup B)backslash((Abackslash B)cup (B backslash A)=Acap B in T$.
I'm not sure how to show that the final part is in that set. It doesn't say anything about unions.
2 Answers
2
Hint: $Acap B=Asetminus (Asetminus B)$
Beat me to it! +1
– Fimpellizieri
Sep 5 '18 at 18:48
I just had to show that these two were equivalent and that did the trick, thanks guys.
– Wesley Strik
Sep 5 '18 at 19:34
Write $Acap B = Asetminus(Asetminus B)$.
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Hold that thought.
– Wesley Strik
Sep 5 '18 at 18:48