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PostPosted: Fri, 9 Mar 2012 19:46:35 UTC 
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Hello all

Union operation is a well-known operation used with sets.

Assume A, B, and C are three different sets.

1 - (|B U C|/ |A U B U C|)

How can we simplify the above formula to

|A|-|B| / |A|+|C|



Thanks for any help :)


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PostPosted: Fri, 9 Mar 2012 19:54:39 UTC 
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full_adder wrote:
Hello all

Union operation is a well-known operation used with sets.

Assume A, B, and C are three different sets.

1 - (|B U C|/ |A U B U C|)

How can we simplify the above formula to

|A|-|B| / |A|+|C|



Thanks for any help :)


What is your ||? Cardinality? Measure?

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PostPosted: Fri, 9 Mar 2012 20:20:40 UTC 
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Thanks Shadow
it is cardinality or set size


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PostPosted: Fri, 9 Mar 2012 20:24:37 UTC 
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full_adder wrote:
Thanks Shadow
it is cardinality or set size


Then what happens if some of them have infinite size? Are you assuming finite sets?

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PostPosted: Fri, 9 Mar 2012 20:53:13 UTC 
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Yes, I am assuming finite sets..


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PostPosted: Fri, 9 Mar 2012 21:57:35 UTC 
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full_adder wrote:
Hello all

Union operation is a well-known operation used with sets.

Assume A, B, and C are three different sets.

1 - (|B U C|/ |A U B U C|)

How can we simplify the above formula to

|A|-|B| / |A|+|C|



Thanks for any help :)


I don't buy it, let A be empty and B= {1}, C={2}, then |BUC|=2, and |AUBUC|=2, so the first equation is 0, but the second is -1/1=-1 which is false. Indeed, so long as |B|<|A| the second one is always negative, but the first one is clearly always positive.

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PostPosted: Fri, 9 Mar 2012 22:24:33 UTC 
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The formulas work with these conditions:
|A| > 0
|B| >= 0
|C| >= 0
and
|B| <= |A|


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PostPosted: Sat, 10 Mar 2012 00:53:51 UTC 
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full_adder wrote:
The formulas work with these conditions:
|A| > 0
|B| >= 0
|C| >= 0
and
|B| <= |A|


Still not true, let A=\{1,2,3\}, \; B=\{1,4\},\; C=\varnothing

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PostPosted: Sat, 10 Mar 2012 00:58:36 UTC 
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And if you think requiring |C|>0 will help, it doesn't. Use the same example and let C=\{5\} and it's still false.

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PostPosted: Sat, 10 Mar 2012 08:17:31 UTC 
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Shadow wrote:
And if you think requiring |C|>0 will help, it doesn't. Use the same example and let C=\{5\} and it's still false.


Or something like A=\{1,2,3,4\},B=\{5,6\},C=\{7\}.

In fact, to make the statement
\dfrac{\#A-\#B}{\#A+\#C}=1-\dfrac{\#(B\cup C)}{\#(A\cup B\cup C)}
true, you can choose the cardinalities of A\cap B\cap C^c,A\cap B^c\cap C,A\cap B^c\cap C^c,A^c\cap B\cap C,A^c\cap B\cap C^c (more-or-less) arbitrarily, only needing to adjust \#(A^c\cap B^c\cap C) to make sure \#(A\cap B\cap C) is a nonnegative integer.

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\begin{aligned}
Spin(1)&=O(1)=\mathbb{Z}/2&\quad&\text{and}\\
Spin(2)&=U(1)=SO(2)&&\text{are obvious}\\
Spin(3)&=Sp(1)=SU(2)&&\text{by }q\mapsto(\mathop{\mathrm{Im}}\mathbb{H}\ni p\mapsto qp\bar{q})\\
Spin(4)&=Sp(1)\times Sp(1)&&\text{by }(q_1,q_2)\mapsto(\mathbb{H}\ni p\mapsto q_1p\bar{q_2})\\
Spin(5)&=Sp(2)&&\text{by }\mathbb{HP}^1\cong S^4_{round}\hookrightarrow\mathbb{R}^5\\
Spin(6)&=SU(4)&&\text{by the irrep }\Lambda_+\mathbb{C}^4
\end{aligned}


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PostPosted: Sat, 10 Mar 2012 10:36:40 UTC 
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outermeasure wrote:
Shadow wrote:
And if you think requiring |C|>0 will help, it doesn't. Use the same example and let C=\{5\} and it's still false.


Or something like A=\{1,2,3,4\},B=\{5,6\},C=\{7\}.

In fact, to make the statement
\dfrac{\#A-\#B}{\#A+\#C}=1-\dfrac{\#(B\cup C)}{\#(A\cup B\cup C)}
true, you can choose the cardinalities of A\cap B\cap C^c,A\cap B^c\cap C,A\cap B^c\cap C^c,A^c\cap B\cap C,A^c\cap B\cap C^c (more-or-less) arbitrarily, only needing to adjust \#(A^c\cap B^c\cap C) to make sure \#(A\cap B\cap C) is a nonnegative integer.


Exactly, unless there are more data, though, this is quite false, and certainly for arbitrary sets this is very false.

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PostPosted: Sat, 10 Mar 2012 10:50:08 UTC 
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Shadow wrote:
outermeasure wrote:
Shadow wrote:
And if you think requiring |C|>0 will help, it doesn't. Use the same example and let C=\{5\} and it's still false.


Or something like A=\{1,2,3,4\},B=\{5,6\},C=\{7\}.

In fact, to make the statement
\dfrac{\#A-\#B}{\#A+\#C}=1-\dfrac{\#(B\cup C)}{\#(A\cup B\cup C)}
true, you can choose the cardinalities of A\cap B\cap C^c,A\cap B^c\cap C,A\cap B^c\cap C^c,A^c\cap B\cap C,A^c\cap B\cap C^c (more-or-less) arbitrarily, only needing to adjust \#(A^c\cap B^c\cap C) to make sure \#(A\cap B\cap C) is a nonnegative integer.


Exactly, unless there are more data, though, this is quite false, and certainly for arbitrary sets this is very false.


Yes.

An easy way to satisfy the conclusion is replacing the condition |B|<=|A| with B\subseteq A. But that is of no fun! :shock:

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\begin{aligned}
Spin(1)&=O(1)=\mathbb{Z}/2&\quad&\text{and}\\
Spin(2)&=U(1)=SO(2)&&\text{are obvious}\\
Spin(3)&=Sp(1)=SU(2)&&\text{by }q\mapsto(\mathop{\mathrm{Im}}\mathbb{H}\ni p\mapsto qp\bar{q})\\
Spin(4)&=Sp(1)\times Sp(1)&&\text{by }(q_1,q_2)\mapsto(\mathbb{H}\ni p\mapsto q_1p\bar{q_2})\\
Spin(5)&=Sp(2)&&\text{by }\mathbb{HP}^1\cong S^4_{round}\hookrightarrow\mathbb{R}^5\\
Spin(6)&=SU(4)&&\text{by the irrep }\Lambda_+\mathbb{C}^4
\end{aligned}


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PostPosted: Sat, 10 Mar 2012 10:53:39 UTC 
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outermeasure wrote:
Shadow wrote:
outermeasure wrote:
Shadow wrote:
And if you think requiring |C|>0 will help, it doesn't. Use the same example and let C=\{5\} and it's still false.


Or something like A=\{1,2,3,4\},B=\{5,6\},C=\{7\}.

In fact, to make the statement
\dfrac{\#A-\#B}{\#A+\#C}=1-\dfrac{\#(B\cup C)}{\#(A\cup B\cup C)}
true, you can choose the cardinalities of A\cap B\cap C^c,A\cap B^c\cap C,A\cap B^c\cap C^c,A^c\cap B\cap C,A^c\cap B\cap C^c (more-or-less) arbitrarily, only needing to adjust \#(A^c\cap B^c\cap C) to make sure \#(A\cap B\cap C) is a nonnegative integer.


Exactly, unless there are more data, though, this is quite false, and certainly for arbitrary sets this is very false.


Yes.

An easy way to satisfy the conclusion is replacing the condition |B|<=|A| with B\subseteq A. But that is of no fun! :shock:


YES! In fact that was the only way short of being very specific about the sizes of the piece of the Venn Diagram that actually made it true, and it was utterly trivial for that case.

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PostPosted: Sat, 10 Mar 2012 11:25:27 UTC 
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Thank you


Yes..replace |B| <= |A| with B\subseteq A


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PostPosted: Sat, 10 Mar 2012 15:01:57 UTC 
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Sorry brothers..

The last condition B\subseteq A is not enough because of

A=\{1,2,3\}, \; B=\{1,2\},\; C=\{1\}.. 1/3 vs. 1/4 !!

The last updated list of the conditions is:

|A| > 0

|B| >= 0

|C| >= 0

B\subseteq A

C\cap A =\varnothing

C\cap B =\varnothing


Hope no change later :)


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