# Finding All Elements Not Shared by B & C

• scorpius1782
In summary, the problem states that if the intersection of sets B and C is a subset of set A, then the difference of the union of sets B and C and set A is a subset of the difference of the union of sets B and C and the intersection of sets B and C. The equations given are used to define the sets and the attempt at a solution involves manipulating these equations to show that the elements in the first statement must also be in the second statement.
scorpius1782

## Homework Statement

The whole problem:
If ##(B \cap C) \subseteq A## then ##[(B \cup C)-A] \subseteq [(B \cup C)-(B \cap C)]##

## Homework Equations

##A-B=A\cap \bar{B}##
##N \subseteq M=\bar{M} \subseteq \bar{N}##

## The Attempt at a Solution

Initially I'm having troubles with ##[(B \cup C)-(B \cap C)]##. I know it should be all the elements that are not shared between B and C but I don't know how to write this out in set notation. I'm hoping that once this is clear I can begin to work the problem backwards so to speak.

For reference the way I've thought of things so far:
I don't really know set problems at all yet so I've begun by thinking of elements inside the sets. So, to start here ##(B \cap C) \subseteq A## means there is an object 'x' in all sets A, B, C.
##[(B \cup C)-A]## Means there is an object 'y' that is in B or C but not in A. And finally: ##[(B \cup C)-(B \cap C)]## 'y' is in a set that is in B or C.

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scorpius1782 said:

## Homework Statement

The whole problem:
If ##(B \cap C) \subseteq A## then ##[(B \cup C)-A] \subseteq [(B \cup C)-(B \cap C)]##

## Homework Equations

##A-B=A\cap \bar{B}##
##N \subseteq M=\bar{M} \subseteq \bar{N}##

## The Attempt at a Solution

Initially I'm having troubles with ##[(B \cup C)-(B \cap C)]##. I know it should be all the elements that are not shared between B and C but I don't know how to write this out in set notation. I'm hoping that once this is clear I can begin to work the problem backwards so to speak.

For reference the way I've thought of things so far:
I don't really know set problems at all yet so I've begun by thinking of elements inside the sets. So, to start here ##(B \cap C) \subseteq A## means there is an object 'x' in all sets A, B, C.
##[(B \cup C)-A]## Means there is an object 'y' that is in B or C but not in A. And finally: ##[(B \cup C)-(B \cap C)]## 'y' is in a set that is in B or C.

No, ##(B \cap C) \subseteq A## means EVERY element x of A is also in B and C. If y is an element of ##[(B \cup C)-A]## then it's in B and C but not in A. And if y is an element of ##[(B \cup C)-(B \cap C)]## then it's in B and C but not in both.

Dick said:
##[(B \cup C)-(B \cap C)]## then it's in B and C but not in both.

I don't understand this. Did you mean to say B or C but not in both?

scorpius1782 said:
I don't understand this. Did you mean to say B or C but not in both?

Yes, that's what I meant. Sorry.

Thats okay, I'm still working on this. I'm just having a real hard time thinking about it.

I find everything after 'then' to be confusing. It seems like we're saying mutually exclusive things about 'y'. It can't be in B and C but not in A. And in B or C but not in both. Unless reusing 'y' for the final statement was incorrect.

Edit: Reading my son a bedtime story so won't be back for a bit.

scorpius1782 said:
Thats okay, I'm still working on this. I'm just having a real hard time thinking about it.

I find everything after 'then' to be confusing. It seems like we're saying mutually exclusive things about 'y'. It can't be in B and C but not in A. And in B or C but not in both. Unless reusing 'y' for the final statement was incorrect.

Edit: Reading my son a bedtime story so won't be back for a bit.

I might not be helping, I've found more typos in my previous quote. Let me try again.

No, ##(B \cap C) \subseteq A## means EVERY element x that's in both B and C is also in A. If y is an element of ##[(B \cup C)-A]## then it's in B or C but not in A. And if z is an element of ##[(B \cup C)-(B \cap C)]## then it's in B or C but not in both.

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Okay, then y may, or may not, equal z from those statements alone. However, y must equal z if ##y \subseteq z##. Then (x,y) must be in either B or C but not A. What does that have to do with (B∩C)⊆A? It seems to me that at a minimum A contains all the shared elements but it doesn't preclude the possibility that it is much larger and contains all the values they don't share as well. Unless that first statement is an attempt to define A somehow.

Trying to start the proof but I'm not sure I'm allowed to do things this way:
Using ##A-B=A\cap \bar{B}##
then ##(B \cap C) \subseteq A## is the same as ##(B \cap C) \cup \bar{A}##
and ##[(B \cup C)-(B \cap C)]## becomes ##[B \cup C \cap \bar{B \cap C}]##=##[B \cup C \cap \bar{B} \cup \bar{C}]##

So then: ##(B \cap C) \cup \bar{A} \subseteq [(B \cup C) \cap (\bar{B} \cup \bar{C})]##

Then from ##(B \cap C) \cup \bar{A}## y' is in either B or C but not A
and from ##[(B \cup C) \cap (\bar{B} \cup \bar{C})]## x' is in B or C but not Both.
From this I should be able to assert that y' must be in x'
But how does this relate?: ##(B \cap C) \subseteq A##turns into ##A \subseteq (\bar{B} \cup \bar{C})## from the other equation above.
Which says that w' is element in A that is not shared by B and C.

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scorpius1782 said:
Trying to start the proof but I'm not sure I'm allowed to do things this way:
Using ##A-B=A\cap \bar{B}##
then ##(B \cap C) \subseteq A## is the same as ##(B \cap C) \cup \bar{A}##

That's getting off to a bad start right there. ##A-B=A\cap \bar{B}## defines a set. ##(B \cap C) \subseteq A## isn't a set. It's a statement that one set is a subset of the other.

Try this. Can you show that if ##E \subseteq F## then for any set D, ##D-F \subseteq D-E##?

I think so:
There exists an element x in D but not F. And since E is in F that means x is also not in E. Therefore the statement is true.
I'm not sure how to write mathematically a proof for this though.

I made a mistake before. I should have written that [(B∪C)−A]=##(B∪C)\cap\bar{A}##

scorpius1782 said:
I think so:
There exists an element x in D but not F. And since E is in F that means x is also not in E. Therefore the statement is true.
I'm not sure how to write mathematically a proof for this though.

You understand it I think, but you aren't quite saying it right. If x is any element of D-F then x is in D but not in F. Since E is a subset of F then x is also not in E. So x is also an element of D-E. That proves D-F is a subset of D-E. Do you see how this applies to your problem?

I understand that because of the first condition we know that A contains all the shared elements from B and C. Then when we have [(B∪C)−A] we're only left with elements that are in either B or C but not both which is really the same set as [(B∪C)−(B∩C)] is all elements that they do not share. Because of [(B∪C)−A] this element may or may not be smaller depending on whether A contains any elements other than the shared elements between B and C. But clearly [(B∪C)−A]⊆[(B∪C)−(B∩C)] is true.

scorpius1782 said:
I understand that because of the first condition we know that A contains all the shared elements from B and C. Then when we have [(B∪C)−A] we're only left with elements that are in either B or C but not both which is really the same set as [(B∪C)−(B∩C)] is all elements that they do not share. Because of [(B∪C)−A] this element may or may not be smaller depending on whether A contains any elements other than the shared elements between B and C. But clearly [(B∪C)−A]⊆[(B∪C)−(B∩C)] is true.

Except for a few turns of phrase, like "this element may or may not be smaller" which are a little awkward, I think that's a perfectly fine proof.

1 person
Thank you, this helped me a lot. The class ramped up in difficulty for me rather quickly when they introduced sets. I'm a lot better at reading and understanding them with your help! Sorry for being a little dense, I'm an unbearably slow learner and this isn't the sort of thing I'm good at to begin with. Thanks, again.

## 1. What is the purpose of finding all elements not shared by B & C?

The purpose of finding all elements not shared by B & C is to identify the unique elements present in two sets, B and C. This can be useful in various data analysis and research applications, as well as in troubleshooting and debugging processes.

## 2. How is the process of finding all elements not shared by B & C carried out?

The process of finding all elements not shared by B & C involves comparing the elements of the two sets and identifying those that are present in one set but not the other. This can be done manually, by creating a Venn diagram or using mathematical operations such as set difference.

## 3. What are the potential applications of finding all elements not shared by B & C?

Some potential applications of finding all elements not shared by B & C include data analysis, research, troubleshooting, and debugging. This process can also be used in various mathematical and scientific fields, such as set theory, statistics, and genetics.

## 4. Can the process of finding all elements not shared by B & C be automated?

Yes, the process of finding all elements not shared by B & C can be automated using computer programming and algorithms. This can save time and effort compared to manual methods, especially when dealing with large datasets.

## 5. Are there any limitations to finding all elements not shared by B & C?

One limitation of finding all elements not shared by B & C is that it only considers two sets at a time. If there are more than two sets involved, additional steps or methods may be needed. Additionally, the accuracy of the results depends on the accuracy of the initial data and the method used for comparison.

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