Prove Intersection of A and B is Empty Set if B Subset of X/A

  • Thread starter Thread starter chocolatelover
  • Start date Start date
  • Tags Tags
    Intersection Proof
Click For Summary

Homework Help Overview

The discussion revolves around proving that the intersection of sets A and B is empty if and only if B is a subset of the complement of A in a universal set X. The subject area pertains to set theory and logical proofs.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning

Approaches and Questions Raised

  • Participants explore the idea of proving the statement by contrapositive and by contradiction. There are discussions about the implications of B being empty and the relationship between the sets involved.

Discussion Status

Some participants have provided guidance on how to approach the proof, suggesting different methods such as proving both directions of the "if and only if" statement. There is an ongoing exploration of logical reasoning and set relationships, with no explicit consensus reached.

Contextual Notes

Participants are navigating the requirements of the proof, including the need to demonstrate both directions of the statement and the implications of set membership. There is an acknowledgment of the complexity involved in the logical structure of the proof.

chocolatelover
Messages
238
Reaction score
0

Homework Statement


Prove A intersects B=empty set if and only if B is a subset of (X/A)


Homework Equations





The Attempt at a Solution



Would I prove the contrapositive in this case?

If B is not a subset of (X/A), then the intersection of A at B is not the empty set

Could someone please show me what to do?

Thank you very much
 
Physics news on Phys.org
"if and only if" means that you must prove the statement both ways. Proving by contradiction seems the easiest way for each. The approach you suggest will only get you half the proof.
 
Thank you very much

Could you show me where to go from there?

Thank you
 
Well since you wanted to use the contrapositive for A intersects B = empty -> B is a subset of X\A I'll explain that way. Let [itex]B=\{b_1,b_2,...\}[/itex]. If B is empty what we want is vacuously true since the empty set is a subset of every set (and thus B can never not be a subset of [itex]X \setminus A[/itex]). If [itex]B\not\subset X\setminus A[/itex] then at least one [itex]b_i \in B[/itex] is [itex]\in X\setminus (X \setminus A) = A[/itex] which implies A and B have these elements in common.

You can prove the other way in a similar fashion. My personal suggestion is contradiction for the other way.

On this note, you have another question that's pretty similar. Your questions really boil down to choosing elements of certain sets and then showing by logic that they must/must not exist in other sets. Try to proceed like this in your other question also.
 
Last edited:
Thank you very much

Regards
 

Similar threads

  • · Replies 1 ·
Replies
1
Views
2K
  • · Replies 4 ·
Replies
4
Views
1K
Replies
1
Views
2K
Replies
39
Views
5K
  • · Replies 8 ·
Replies
8
Views
2K
  • · Replies 4 ·
Replies
4
Views
2K
  • · Replies 8 ·
Replies
8
Views
7K
  • · Replies 1 ·
Replies
1
Views
1K
  • · Replies 7 ·
Replies
7
Views
2K
  • · Replies 4 ·
Replies
4
Views
2K