Show that if f: A->B, and A(1), A(2) are both subsets of A, then

  • Thread starter Thread starter Simkate
  • Start date Start date
  • Tags Tags
    Subsets
Click For Summary

Homework Help Overview

The discussion revolves around a proof involving functions and set theory, specifically focusing on the relationship between the images of the intersection of subsets and the intersection of their images. The original poster seeks to establish that if \( f: A \to B \) and \( A_1, A_2 \) are subsets of \( A \), then \( f(A_1 \cap A_2) \subseteq f(A_1) \cap f(A_2) \), and they request an example where this inclusion is strict.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning

Approaches and Questions Raised

  • Participants discuss the definitions and implications of the function and subsets involved. Some express confusion about the proof structure, while others attempt to clarify the reasoning behind the inclusion. There are also suggestions to provide examples of non-injective functions to illustrate the concept.

Discussion Status

There is an ongoing exploration of the proof, with some participants offering insights into the logical steps required. While some express uncertainty about the proof's clarity, others affirm the reasoning presented. The discussion reflects a mix of attempts to understand the problem and to clarify the proof's requirements.

Contextual Notes

Some participants note the need for a clearer understanding of the problem statement and the proof process, indicating that not all necessary information may have been provided initially. There is also mention of the requirement to show work before receiving further assistance.

Simkate
Messages
24
Reaction score
0
Show that if f: A-->B, and A(1), A(2) are both subsets of A, then

Show that if f: A-->B, and A(1), A(2) are both subsets of A, then
f(A1 ∩ A2) C(is the subset of) f(A1) ∩ f(A2).

Give an example of a situation where the inclusion is strict.
 
Physics news on Phys.org


For every element x in f(A1 ∩ A2), there must be an element y in A1 ∩ A2 mapping through f to x... and continue from there...

You need to show some semblance of work before we can really help you on a problem.
 


I really don't understand the whole proof
 


You haven't posted a proof, so we can't help you understand the "proof". If you mean that you don't understand the problem, what about it is tripping you up?
 


Sorry i meant to say i don't quite understand the problem. However i have gotten this much of understanding which i do know if it is right.

Let x be an element of f(A1 ∩ A2) and by definition of the f(A1 ∩ A2), there is a y element in ( A1 ∩ A2) so that f(y)=x.
Since y is an element in (A1 ∩ A2), y∈A1x∈A2. Since y,f(y)∈ f(A1).
This follows alongside y,f(y)f(A2)
and
Since f(y)=x∈f(A1) and f(y)=x∈f(A2),x= f(A1)(f(A2)

I don't know how to complete it
I would appreciate you help thank u
 


Your reasoning seems to be right. All you need to show is that any element x in f(A1 ∩ A2) is also in f(A1) ∩ f(A2). Do you understand why your argument does this?
 


How about f(x)=|x|, A11=R-,A2=R+. Any noninjective function really.
 


Though am i correct about the PROOF( for f(A1 ∩ A2) ⊆ f(A1) ∩ f(A2) )
that i have shown below?

Let x∈A1∩A2. Then x∈A1 or x∈A2, in which case f(x)∈f(A1) or f(A2) respectively; in any case, f(x)∈f(A1)∩f(A2), and so f(A1∩A2)⊆f(A1)∩f(A2).

On the other hand, let y∈f(A1)∩f(A2). Then y∈f(A1) or y∈f(A2), in which case there is some x∈A1 or A2 respectively with f(x)=y. Hence y∈f(A1∩A2), whence f(A1)∩f(A2)⊆f(A1∩

Therefore f(A1∩A2)⊆((A1)∩f(A2).
 
Last edited:

Similar threads

Replies
1
Views
2K
  • · Replies 2 ·
Replies
2
Views
2K
  • · Replies 9 ·
Replies
9
Views
2K
  • · Replies 5 ·
Replies
5
Views
2K
  • · Replies 4 ·
Replies
4
Views
1K
  • · Replies 1 ·
Replies
1
Views
2K
  • · Replies 4 ·
Replies
4
Views
3K
  • · Replies 2 ·
Replies
2
Views
2K
Replies
6
Views
2K
  • · Replies 2 ·
Replies
2
Views
1K