Cardinality proof by indicating a bijection

Click For Summary

Homework Help Overview

The discussion revolves around proving the cardinality relationship |AB∪C| = |AB × AC| by demonstrating a bijection between the two sets. Participants explore the implications of the sets A, B, and C being finite or infinite and the conditions under which the bijection can be established.

Discussion Character

  • Exploratory, Assumption checking, Conceptual clarification

Approaches and Questions Raised

  • Participants discuss the existence of functions between the sets and the conditions under which these functions are onto or one-to-one. They question the necessity of the disjoint condition for B and C and explore the implications of the sets being finite or infinite.

Discussion Status

There is an ongoing exploration of the conditions required for establishing a bijection. Some participants suggest that if A, B, and C are infinite, the statement may hold, while others emphasize the need for clarity on the disjoint nature of B and C. Guidance is provided regarding the relationship between the cardinalities of unions and disjoint unions.

Contextual Notes

Participants note that the homework question does not specify whether the sets are finite or infinite, leading to uncertainty in proving the statement. The discussion highlights the importance of understanding the nature of the sets involved.

embemilyy
Messages
8
Reaction score
0

Homework Statement


Prove that |AB\cupC|=|ABx AC| by demonstrating a bijection between the two sets.



Homework Equations



Two sets have equivalent cardinality if there is a bijection between them/

The Attempt at a Solution



Essentially I can prove that there is a function from AB\cupC to ABx AC, defined by <f restricted to B, f restricted to C> but it's only onto if B\capC=0.

Or I can prove a function the other way but it's only one-to-one if the same condition holds.
Other solutions I've seen online also say that this condition is necessary, but it's not included in the homework question. Any ideas?
 
Physics news on Phys.org
embemilyy said:

Homework Statement


Prove that |AB\cupC|=|ABx AC| by demonstrating a bijection between the two sets.

Homework Equations



Two sets have equivalent cardinality if there is a bijection between them/

The Attempt at a Solution



Essentially I can prove that there is a function from AB\cupC to ABx AC, defined by <f restricted to B, f restricted to C> but it's only onto if B\capC=0.

Or I can prove a function the other way but it's only one-to-one if the same condition holds.
Other solutions I've seen online also say that this condition is necessary, but it's not included in the homework question. Any ideas?

Homework Statement


Homework Equations


The Attempt at a Solution


If B and C are not disjoint and all of the sets A, B and C are finite, then the statement is definitely false. If they didn't give you any information about the sets, then you can't prove it. If some of the sets are infinite, then it might be true. It depends on the details.
 
thanks for responding!
If A, B, and C are infinite, I think the statement holds (just by results of cardinal arithmetic that I've seen in the textbook), but I'm still not able to find a bijection.
 
embemilyy said:
thanks for responding!
If A, B, and C are infinite, I think the statement holds (just by results of cardinal arithmetic that I've seen in the textbook), but I'm still not able to find a bijection.

You need a bijection between the union of B and C and the 'disjoint union' of B and C. And, yes, if B U C is infinite then its cardinality is the max(card(B),card(C)). Same for the disjoint union, so there's definitely a bijection. But that's just an indirect conclusion. I think to actually write down the bijection easily you need to assume B and C are disjoint. If not, the point is not to write it down but to prove it exists.
 

Similar threads

Replies
1
Views
2K
  • · Replies 1 ·
Replies
1
Views
2K
  • · Replies 2 ·
Replies
2
Views
4K
Replies
2
Views
3K
Replies
4
Views
2K
  • · Replies 9 ·
Replies
9
Views
3K
  • · Replies 3 ·
Replies
3
Views
2K
  • · Replies 2 ·
Replies
2
Views
1K
  • · Replies 5 ·
Replies
5
Views
2K
  • · Replies 7 ·
Replies
7
Views
3K