Sets and Relations - quick one

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Homework Help Overview

The discussion revolves around the conditions for the equality \((A \cup B) \cap C = A \cup (B \cap C)\) in the context of set theory, specifically examining sets A, B, and C. Participants explore whether specific sets, such as the irrationals, integers, and reals, satisfy this equality.

Discussion Character

  • Conceptual clarification, Assumption checking

Approaches and Questions Raised

  • Participants question the conditions under which the set equality holds and whether specific examples meet these conditions. There is discussion about the implications of subsets and the relationships between the sets involved.

Discussion Status

Several participants have offered insights into the conditions required for the equality to hold, with some suggesting that A must be a subset of C. There is ongoing exploration of the implications of the example sets provided, with no clear consensus reached on the sufficiency of the conditions discussed.

Contextual Notes

Some participants note that the original example of A, B, and C involves specific sets (irrationals, integers, reals) and question whether the conditions derived apply universally or only to this case. There is also mention of the need for clarity in defining conditions versus sets in the discussion.

Natasha1
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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:
...
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?
[itex]A\cap B\cap C[/itex] is not a complete sentence. Did you mean the condition for the equality to be true is that [itex]A\subseteq B \subseteq C[/itex] ?
An easy way to check is to look at the condition of elements of each set. For example, if [itex]S = ((A\cup B)\cap C)[/itex], then S = {x: ([itex]x\in A[/itex] and [itex]x\in C[/itex]) or ([itex]x\in B[/itex] and [itex]x\in C[/itex])} which can be written more concisely (or draw a Venn diagram). Do the same for the other side.
For the second part, yes, [itex]A\subset C[/itex], 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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