Power Set Unions and inclusion

Click For Summary
SUMMARY

The discussion centers on proving the statement that if \wp(A) ∪ \wp(B) = \wp(A ∪ B), then A ⊆ B or B ⊆ A, as presented in Velleman's "How to Prove It: A Structured Approach." The user is utilizing the Java applet "Proof Designer" to assist with the proof but is struggling with defining subsets and applying proof techniques such as contradiction and contrapositive. The conversation highlights the importance of visual aids like Venn diagrams in understanding set relationships and emphasizes the need for a structured approach to proofs.

PREREQUISITES
  • Understanding of set theory, specifically power sets and unions.
  • Familiarity with proof techniques including contradiction and contrapositive.
  • Experience using "Proof Designer" Java applet for formal proofs.
  • Basic knowledge of logical implications and disjunctions in mathematical reasoning.
NEXT STEPS
  • Study the properties of power sets and their relationships in set theory.
  • Learn advanced proof techniques, focusing on contrapositive proofs.
  • Explore the use of Venn diagrams for visualizing set operations and relationships.
  • Practice exercises from Velleman's "How to Prove It" to reinforce understanding of proof strategies.
USEFUL FOR

Students of mathematics, particularly those studying set theory and proof techniques, as well as educators looking for resources to teach structured proof methods.

Kolmin
Messages
66
Reaction score
0
I am studying Velleman's "How to prove it: a structured approach" and I have to say that it is one of the best decision I have taken. Right now I am working on the exercises that are on Velleman's page along with the java applet "Proof Designer" and I have the feeling I start to get a bit how proofs work. But still...

This proof is really making me think and I don't see how I can get out of it.

Assume \wp(A)\cup\wp(B)=\wp(A\cupB)
Prove that A\subseteqB or B\subseteqA.
[Suggested Exercise no.20]

I have to admit I tried basically everything (i.e. contradiction, cases) but I don't really go anywhere close to the solution.

For example, proof designer asks you to define a bounded variable if it is in the "given" section. I rephrase \wp(A)\cup\wp(B)=\wp(A\cupB) and I don't know how to define the subset X of A and B (out of desperation I put X=A but doesn't look a great idea).

Still I am not sure if it is a matter of not knowing how to properly use the software (which is actually quite easy) or I simply don't know how to work out the proof.
 
Physics news on Phys.org
How does your contradiction look?
 
Tried to use Venn Diagram?? It really helps in visualizing the sets, their union, intersections, etc.
 
Kolmin said:
I am studying Velleman's "How to prove it: a structured approach" and I have to say that it is one of the best decision I have taken. Right now I am working on the exercises that are on Velleman's page along with the java applet "Proof Designer" and I have the feeling I start to get a bit how proofs work. But still...

This proof is really making me think and I don't see how I can get out of it.

Assume \wp(A)\cup\wp(B)=\wp(A\cupB)
Prove that A\subseteqB or B\subseteqA.
[Suggested Exercise no.20]

I have to admit I tried basically everything (i.e. contradiction, cases) but I don't really go anywhere close to the solution.

For example, proof designer asks you to define a bounded variable if it is in the "given" section. I rephrase \wp(A)\cup\wp(B)=\wp(A\cupB) and I don't know how to define the subset X of A and B (out of desperation I put X=A but doesn't look a great idea).

Still I am not sure if it is a matter of not knowing how to properly use the software (which is actually quite easy) or I simply don't know how to work out the proof.
How about proving the contra positive?
Assume \displaystyle \text{ A} \not\subseteq \text{B and B} \not\subseteq \text{ A }.

Prove \displaystyle \ \ <br /> \wp\text{(A)}\cup\wp\text{(B)}\neq\wp\text{(A}\cup\text{B)}\,.<br />​
 
Last edited:
Well, my contradiction didn't look that great...

About the Venn diagrams, it's not a matter of understanding the result (at least this one...): the only problem is really to prove it.

Right now I am trying with the contrapositive (thanks for the suggestion...:smile: ) but it's a bit problematic. I don't know if you ever used "Proof Designer": I find it amazing, because it really forces you to not fly around.
The fact is that I guess I should build up a subset with certain properties but I am not really sure I can do it. [...or maybe I am completely wrong]
 
Btw, a question that should look silly but that I facing yesterday.

Let's imagine we have the following goal:

A not-\subseteqB \vee A\subseteqB

Instead of using the disjunction or the cases approach to prove the goal, can we consider it an implication?

I mean, basically it is the same of

\neg(A\subseteqB) \vee A\subseteqB

which is the same as

A\subseteqB\rightarrowA\subseteqB

I was wondering cause (if the trick is not wrong) you cannot use it in Proof Designer
 
Anyway, I finally got it.
Being a bit in a hurry for a travel, I will post the proof when I'll have time (hopefully before coming from the trip).
 

Similar threads

  • · Replies 15 ·
Replies
15
Views
3K
  • · Replies 5 ·
Replies
5
Views
3K
  • · Replies 3 ·
Replies
3
Views
1K
Replies
24
Views
3K
  • · Replies 2 ·
Replies
2
Views
2K
  • · Replies 2 ·
Replies
2
Views
1K
  • · Replies 6 ·
Replies
6
Views
2K
  • · Replies 10 ·
Replies
10
Views
3K
  • · Replies 10 ·
Replies
10
Views
2K
  • · Replies 10 ·
Replies
10
Views
4K