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August 14th, 2017, 12:22 AM   #1
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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.
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 August 14th, 2017, 01:25 AM #2 Math Team     Joined: Jul 2013 From: काठमाडौं, नेपाल Posts: 878 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: 232 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: 878 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
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Quote:
 Originally Posted by MATHEMATICIAN 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)
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: 878 Thanks: 60 Math Focus: सामान्य गणित By other sets I do not mean the universal set.
August 14th, 2017, 06:53 AM   #7
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Quote:
 Originally Posted by MATHEMATICIAN By other sets I do not mean the universal set.
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: 878 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
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Quote:
 Originally Posted by MATHEMATICIAN I agree with your explanation but I don't get the conclusion.
Well, the point is that $o(A)$ counts the number of elements in $A$ not in any other set. I've shown that every element of $A$ will always be contained in another (not universal, but in fact finite) set, so $o(A)$ must always be zero.

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: 878 Thanks: 60 Math Focus: सामान्य गणित Seems like I did not know about the correct notation

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