Proving: Sets Proofs - Techniques & Examples

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Discussion Overview

The discussion revolves around proving set-theoretic statements, specifically focusing on two claims involving intersections and unions of sets. Participants seek techniques for formal proofs and clarification on logical reasoning within set theory.

Discussion Character

  • Homework-related
  • Exploratory
  • Technical explanation

Main Points Raised

  • One participant requests help with proving that if $A \cap B = \emptyset$, then $(A \times C) \cap (B \times C) = \emptyset$.
  • Another participant suggests starting the proof by assuming the contrary and explores the implications of $(x,y) \in (A \times C) \cap (B \times C)$.
  • For the second claim, $A \cup (B \cap C) = (A \cup B) \cap (A \cup C)$, a participant outlines a proof strategy involving cases based on membership in $A$ or $B \cap C$.
  • One participant expresses a lack of understanding of formal proof techniques and seeks guidance on how to present proofs more formally.
  • Another participant provides reasoning for why certain elements belong to unions of sets based on the definitions of those sets.

Areas of Agreement / Disagreement

Participants generally agree on the need for formal proof techniques, but there is no consensus on the specific methods or steps to take in completing the proofs. The discussion includes various approaches and suggestions without a clear resolution.

Contextual Notes

Some participants express uncertainty about filling in logical steps in their proofs, indicating potential gaps in understanding formal proof structures. There is also a reliance on definitions of set operations that may not be fully articulated.

Who May Find This Useful

This discussion may be useful for students learning set theory, particularly those struggling with formal proof techniques and logical reasoning in mathematics.

paulmdrdo1
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prove the followinga. prove that if $A\cap B=\emptyset$, then $(A\times C)\cap (B\times C)=\emptyset$
b. $A\cup(B\cap C)=(A\cup B)\cap (A\cup C)$

i don't have any idea how i would start proving this.
can you give me some techniques on proofs.
 
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paulmdrdo said:
prove the followinga. prove that if $A\cap B=\emptyset$, then $(A\times C)\cap (B\times C)=\emptyset$
b. $A\cup(B\cap C)=(A\cup B)\cap (A\cup C)$

i don't have any idea how i would start proving this.
can you give me some techniques on proofs.
a. Assume on the contrary that $(x,y)\in (A\times C)\cap (B\times C)$. Thus $x\in A\cap B$ and $y\in C$ (why?). Can you finish?

b. Let $x\in A\cup(B\cap C)$. Then $x\in A$ or $x\in B\cap C$.
Case 1. $x\in A$.
Here $x\in A\cup B$ and $x\in A\cup C$. (why?). Thus $x\in (A\cup B)\cap (A\cup C)$.

Case 2. $x\in B\cap C$.
Again, Here $x\in A\cup B$ and $x\in A\cup C$. (why?). Thus $x\in (A\cup B)\cap (A\cup C)$.

The above shows that any element of $A\cup (B\cap C)$ is in $(A\cup B)\cap(A\cup C)$. This means $A\cup (B\cap C)\subseteq (A\cup B)\cap(A\cup C)$. Can you show the reverse containment and finish?
 
caffeinemachine i know nothing about formal way of proving. i can say why that is true in words or verbally but not in a generalize way. i want a formal presentation of proofs like what you are doing. can you give some tips. because every time i encounter this kind of problems i always feel discouraged.

but this is what i tried

$x\in A\cap B$ and $y\in C$ because we assume that $(x,y)\in (A\times C)\cap (B\times C) $

i don't know howfill in the other (whys)
 
paulmdrdo said:
caffeinemachine i know nothing about formal way of proving. i can say why that is true in words or verbally but not in a generalize way. i want a formal presentation of proofs like what you are doing. can you give some tips. because every time i encounter this kind of problems i always feel discouraged.

but this is what i tried

$x\in A\cap B$ and $y\in C$ because we assume that $(x,y)\in (A\times C)\cap (B\times C) $

i don't know howfill in the other (whys)
The 'second why' is quite straightforward. If you know that $x\in A$, then $x$ also lies in any superset of $A$. Isn't it? Thus $x\in A$ gives $x\in A\cup B$ since $A\cup B$ is a superset of $A$. Similarly $x\in A$ gives $A\cup C$ too.

Now try to fill in for the third why.
 

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