MHB Venn Diagram II: Check Answers & Understand Relationships

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SUMMARY

The discussion revolves around solving Venn diagram problems related to set theory, specifically addressing the relationships between sets A and B. Key points include the necessity of illustrating the universal set for complement operations and the correct representation of subset relationships, particularly that \(B \subset A\) when \(B\) is contained within \(A\). Participants emphasize the importance of demonstrating both the truth and falsehood of conditions in set relationships to ensure thorough understanding.

PREREQUISITES
  • Understanding of set theory concepts, including unions and intersections.
  • Familiarity with Venn diagrams for visualizing set relationships.
  • Knowledge of complement notation, specifically \(A'\) and \(B'\).
  • Ability to interpret and manipulate subset relationships.
NEXT STEPS
  • Study the properties of set operations, focusing on unions and intersections.
  • Learn how to effectively use Venn diagrams to represent complex set relationships.
  • Explore the concept of universal sets and their role in set theory.
  • Investigate the implications of subset relationships and their visual representations in diagrams.
USEFUL FOR

Students of mathematics, educators teaching set theory, and anyone interested in enhancing their understanding of Venn diagrams and set relationships.

bergausstein
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Another Venn diagram problem. just want to check if my answer is correct since i don't have a solutions manual of the book I'm using.

Use Venn diagrams to illustrate the following.

a. $\displaystyle A\cup B\,=\,A$ if and only if $\displaystyle B\subset A$
b. $\displaystyle A\cap B\,=\,B$ if and only if $\displaystyle B\subset A$
c. $\displaystyle B\subset A$ if and only if $\displaystyle A'\subset B'$
d. $\displaystyle \left(A'\right)'\,=\,A$

my answers
View attachment 1110
 

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bergausstein said:
Another Venn diagram problem. just want to check if my answer is correct since i don't have a solutions manual of the book I'm using.

Use Venn diagrams to illustrate the following.

a. $\displaystyle A\cup B\,=\,A$ if and only if $\displaystyle B\subset A$
b. $\displaystyle A\cap B\,=\,B$ if and only if $\displaystyle B\subset A$
c. $\displaystyle B\subset A$ if and only if $\displaystyle A'\subset B'$
d. $\displaystyle \left(A'\right)'\,=\,A$

my answers
View attachment 1110
I think you should have drawn the universal set too, especially while showing diagrams relating to complements.
 
caffeinemachine said:
I think you should have drawn the universal set too, especially while showing diagrams relating to complements.

i have drawn the universal set. w/c is set A. right?
 
bergausstein said:
i have drawn the universal set. w/c is set A. right?
What is 'w/c'??
 
caffeinemachine said:
What is 'w/c'??

i mean i have drawn the universal set. it's set A.
 
This is how I would draw them...I will leave the other cases for part c) for you to draw. :D

View attachment 1112

edit: caffeinemachine is correct, the universal set (the rectangles in my sketches) is needed for complementation.
 

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MarkFL said:
This is how I would draw them...I will leave the other cases for part c) for you to draw. :D

View attachment 1111

edit: caffeinemachine is correct, the universal set (the rectangles in my sketches) is needed for complementation.

why $\displaystyle A\subset B$ ? shouldn't it be $\displaystyle B\subset A$ since B is inside A?
 
bergausstein said:
why $\displaystyle A\subset B$ ?

Darn it...after all that work, I manage to foul it up...of course those should read $B\subset A$. I amaze myself sometimes...(Tongueout) Thank you for catching this error! (Yes)

edit: I edited the drawing, and reattached it. :D
 

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Last edited:
  • #10
The other two cases involve $B$ and $A$ having partial intersection and no intersection. You have drawn the same case I drew.
 
  • #11
here's my second attempt for C.
View attachment 1116

i just want to ask why do we have to show different cases for each problems?
 

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Last edited:
  • #12
Demonstrating the truth for the given condition is not enough. You must also show it is false for the other cases to be thorough. This is how I would demonstrate the second case is not true for c):

View attachment 1117

The area in blue is within $A'$ but not in $B'$, hence $A'\not\subset B'$
 

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