Sets and Relations - quick one

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The discussion centers on the conditions for the equality (A∪B)∩C = A∪(B∩C) to hold true. It is established that a necessary condition is A being a subset of C. The example provided, where A is the set of irrationals, B is the set of integers, and C is the set of reals, confirms that the equality holds since A is not a subset of B but is a subset of C. The conversation emphasizes the importance of understanding set relationships and suggests using Venn diagrams for clarity. Ultimately, the conditions for the equality involve the relationships between the sets A, B, and C in the context of real numbers.
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What are the conditons on A, B and C for (AuB)nC = Au(BnC) ?

Is it that AnBnC ?

Can someone explain if they are different and why? :confused:

Now If A = {irrationals}, B= {integers} and C={reals} does the equality from above hold in this case?

I answered yes the equality holds as AcBcC.

Any suggestions?
 
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Natasha1 said:
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Now If A = {irrationals}, B= {integers} and C={reals} does the equality from above hold in this case?
I answered yes the equality holds as AcBcC.
The irrationals are not a subset of the integers. :wink:
 
so for the first part of the question

What are the conditons on A, B and C for (AuB)nC = Au(BnC) ?

Is it that AnBnC ? This is correct

And the second part we get AcC only right?
 
Natasha1 said:
so for the first part of the question
What are the conditons on A, B and C for (AuB)nC = Au(BnC) ?
Is it that AnBnC ? This is correct
And the second part we get AcC only right?
A\cap B\cap C is not a complete sentence. Did you mean the condition for the equality to be true is that A\subseteq B \subseteq C ?
An easy way to check is to look at the condition of elements of each set. For example, if S = ((A\cup B)\cap C), then S = {x: (x\in A and x\in C) or (x\in B and x\in C)} which can be written more concisely (or draw a Venn diagram). Do the same for the other side.
For the second part, yes, A\subset C, but you didn't answer the question posed. :smile:
 
AnBnC isn't even a "condition"! It's a set and you have to say something about it to get a "condition". (AuB)nC consists of all things that are in either A or B and in C. Au(BnC) consists of all things that are in A or in both B and C. Drawing Venn diagrams might help. Suppose "z" is in A but not in C. then it would certainly be in Au(BnC) since it is in A. But since it is not in C it is not[\b] in (AuB)nC. So one possible condition is that A is a subset of C. Is that sufficient?
 
I did mean AcBcC actually not AnBnC sorry.

But actually the conditions for (AuB)nC = Au(BnC) in real numbers not our examble. Is that AcC and BcC right?

And in our example which is A= irrationals B= integers and C= reals is satisfies as AcC and BcC right?
 
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