Set Theory Problems: S1 U S2 = (S1' ∩ S2')' and S1 U S2 - (S1 ∩ S2') = S2

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Homework Help Overview

The discussion revolves around set theory problems, specifically demonstrating set equalities involving unions, intersections, and complements. The original poster attempts to show that \( S1 \cup S2 = (S1' \cap S2')' \) and to prove that \( S1 \cup S2 - (S1 \cap S2') = S2 \).

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning

Approaches and Questions Raised

  • Participants discuss the logic behind the first equality, with one participant expressing uncertainty about their proof and another pointing out a potential mistake in reasoning. There is also a request for hints on how to approach the second problem, indicating a need for conceptual clarification.

Discussion Status

The discussion is ongoing, with participants actively engaging in the reasoning process. Some guidance has been offered regarding the need to demonstrate both inclusions for the first equality, and there is an acknowledgment of the challenges faced in understanding set notation.

Contextual Notes

Participants note difficulties with set notation and the need for visual aids, suggesting that some foundational concepts may require further exploration. There is also an indication that the original poster has not yet attempted the second problem.

sbc824
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Homework Statement



show S1 U S2 = (S1' ∩ S2')'

The Attempt at a Solution



I'm pretty sure I have this right or I'm close

Let x ∈ S1 U S2
x ∈ S1 or x ∈ S2
Since x ∈ S1 or S2, then x ∉ S1' and S2'
If x ∉ S1' and S2', then x ∈ (S1' and S2')'
Therefore, S1 U S2 = (S1' ∩ S2')'

Homework Statement



show S1 U S2 - (S1 ∩ S2') = S2

The Attempt at a Solution



I have not attempted this as I'm not sure how to start this one...any help would be appreciated.
 
Last edited:
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sbc824 said:

Homework Statement



show S1 U S2 = (S1' ∩ S2')'

The Attempt at a Solution



I'm pretty sure I have this right or I'm close

Let x ∈ S1 U S2
x ∈ S1 or x ∈ S2
Since x ∈ S1 or x ∈ S2, then x ∉ S1' and x ∉ S2' This is not correct.

If x ∈ S1, then it does not need to be in S2. If it's not in S2, then x ∈ S2'.


If x ∉ S1' and x ∉ S2', then x ∈ (S1' and S2')'
Therefore, S1 U S2 = (S1' ∩ S2')'

Homework Statement



show S1 U S2 - (S1 ∩ S2') = S2

The Attempt at a Solution



I have not attempted this as I'm not sure how to start this one...any help would be appreciated.
You don't have the first part right.
 
SammyS said:
You don't have the first part right.

wow silly mistake thanks...any starting hints for 2? I can easily visualize it with a diagram...but I'm rusty with set notation.
 
sbc824 said:
wow silly mistake thanks...any starting hints for 2? I can easily visualize it with a diagram...but I'm rusty with set notation.
Another problem with your solution to part 1 is that you have only shown that S1 U S2 ⊆ (S1' ∩ S2')' (that is, if you have truly corrected your proof). To show equality, you also need to show that S1 U S2 ⊇ (S1' ∩ S2')' .
 

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