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August 14th, 2017, 12:22 AM  #1 
Senior Member Joined: Nov 2015 From: hyderabad Posts: 206 Thanks: 2  Cardinality of union of two sets proof
Hello all I am not able to write a proof of the property $\displaystyle n(A \cup B) = n(A) + n(B)  n(A \cap B) $. I was just able to draw sets diagram and one or two steps attached. Please help me with the proof. 
August 14th, 2017, 01:25 AM  #2 
Math Team Joined: Jul 2013 From: काठमाडौं, नेपाल Posts: 831 Thanks: 60 Math Focus: सामान्य गणित 
We know, $\displaystyle n (A) = o (A) +n (A \cap B)$ $\displaystyle n (B) = o (B) +n (A \cap B)$ $\displaystyle n (A \cup B)= o (A) + n (A\cap B)+ o (B) $ So we have, $\displaystyle n (A) + n (B) = o (A) + n (A\cap B)+ o (B) +n (A \cap B)$ $\displaystyle n (A) + n (B) = n (A\cup B)+ n (A \cap B)$ $\displaystyle n (A) + n (B)  n (A\cap B) = n (A \cup B)$ 
August 14th, 2017, 01:28 AM  #3 
Senior Member Joined: Nov 2015 From: hyderabad Posts: 206 Thanks: 2 
Thanks for the reply. May i know the difference between $o(A)$ and $n(A)$. Thank you 😊 
August 14th, 2017, 01:31 AM  #4 
Math Team Joined: Jul 2013 From: काठमाडौं, नेपाल Posts: 831 Thanks: 60 Math Focus: सामान्य गणित 
o (A) = number of elements that belong to only A (should not belong to other sets) n (A) = number of elements that belong to A(may belong to other sets) 
August 14th, 2017, 05:00 AM  #5 
Member Joined: Aug 2017 From: United Kingdom Posts: 90 Thanks: 26  The issue is that $o(A)$ is always zero according to this definition: given any element of $A$, you can always make a set not equal to $A$ that contains that element. That is, there are no elements only in A that are not in any other sets.

August 14th, 2017, 06:13 AM  #6 
Math Team Joined: Jul 2013 From: काठमाडौं, नेपाल Posts: 831 Thanks: 60 Math Focus: सामान्य गणित 
By other sets I do not mean the universal set.

August 14th, 2017, 06:53 AM  #7 
Member Joined: Aug 2017 From: United Kingdom Posts: 90 Thanks: 26  You do not need the universal set here. I'll explain: If $A$ has more than one element, then consider any element $x$ in $A$. $x$ belongs to the (not universal) set $\{x\} \neq A$, so $x$ is not only in $A$. Hence $o(A) = 0$. If $A$ just has one element, say $A = \{a\}$, consider the set $B = \{a,1,2\}$. $a$ belongs to the (not universal) set $B \neq A$, so $a$ is not only in $A$. Hence $o(A) = 0$ in this case, too. And clearly $o(A) = 0$ if $A$ is the empty set. Last edited by cjem; August 14th, 2017 at 07:05 AM. 
August 14th, 2017, 09:40 AM  #8 
Math Team Joined: Jul 2013 From: काठमाडौं, नेपाल Posts: 831 Thanks: 60 Math Focus: सामान्य गणित 
I agree with your explanation but I don't get the conclusion.

August 14th, 2017, 11:05 AM  #9  
Member Joined: Aug 2017 From: United Kingdom Posts: 90 Thanks: 26  Quote:
Instead of $o(A)$, you should have been using $n(A \setminus B)$, where $\setminus$ is set difference. Last edited by cjem; August 14th, 2017 at 11:13 AM.  
August 15th, 2017, 06:36 PM  #10 
Math Team Joined: Jul 2013 From: काठमाडौं, नेपाल Posts: 831 Thanks: 60 Math Focus: सामान्य गणित 
Seems like I did not know about the correct notation


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cardinality, proof, sets, union 
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