simplify an equation related to Sets

full_adder · **Joined:** Tue, 7 Aug 2007 11:37:58 UTC **Posts:** 120

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

Shadow · **Posted:** Fri, 9 Mar 2012 19:54:39 UTC

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?

full_adder · **Joined:** Tue, 7 Aug 2007 11:37:58 UTC **Posts:** 120

Thanks Shadow
it is cardinality or set size

Shadow · **Posted:** Fri, 9 Mar 2012 20:24:37 UTC

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?

full_adder · **Joined:** Tue, 7 Aug 2007 11:37:58 UTC **Posts:** 120

Yes, I am assuming finite sets..

Shadow · **Posted:** Fri, 9 Mar 2012 21:57:35 UTC

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.

full_adder · **Joined:** Tue, 7 Aug 2007 11:37:58 UTC **Posts:** 120

The formulas work with these conditions:
|A| > 0
|B| >= 0
|C| >= 0
and
|B| <= |A|

Shadow · **Posted:** Sat, 10 Mar 2012 00:53:51 UTC

full_adder wrote:

The formulas work with these conditions:
|A| > 0
|B| >= 0
|C| >= 0
and
|B| <= |A|

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

Shadow · **Posted:** Sat, 10 Mar 2012 00:58:36 UTC

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.

outermeasure · **Posted:** Sat, 10 Mar 2012 08:17:31 UTC

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.

Shadow · **Posted:** Sat, 10 Mar 2012 10:36:40 UTC

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.

outermeasure · **Posted:** Sat, 10 Mar 2012 10:50:08 UTC

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:

Shadow · **Posted:** Sat, 10 Mar 2012 10:53:39 UTC

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.

full_adder · **Joined:** Tue, 7 Aug 2007 11:37:58 UTC **Posts:** 120

Thank you

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

full_adder · **Joined:** Tue, 7 Aug 2007 11:37:58 UTC **Posts:** 120

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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simplify an equation related to Sets

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