Proof of Injective Function Property

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Discussion Overview

The discussion revolves around proving a property of injective functions, specifically that if a function f:X-->Y is injective, then f(X\setminus A) is a subset of Y\setminus f(A) for all subsets A of X. The scope includes mathematical reasoning and proof techniques.

Discussion Character

  • Mathematical reasoning, Debate/contested

Main Points Raised

  • One participant, Kamataat, presents an argument attempting to prove the injective property by manipulating the sets involved and claims to show that elements in f(X\A) must be in Y\f(A).
  • Another participant challenges Kamataat's approach, stating that adjusting the set A is not permissible since the proof must hold for all subsets A of X.
  • This participant further suggests that to prove elements of f(X\A) are not in f(A), one could derive a contradiction based on the definition of injective functions.
  • Kamataat reiterates the challenge and seems to agree with the second participant's reasoning about the contradiction.

Areas of Agreement / Disagreement

Participants express disagreement regarding the validity of the initial proof attempt. There is no consensus on the correctness of the proof, as one participant believes it is incorrect while another supports the reasoning that leads to a contradiction.

Contextual Notes

The discussion highlights the importance of adhering to the conditions of the proof, specifically the requirement that the argument must apply to all subsets A of X without modification.

Kamataat
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I had this question on a test today.
Prove that if a function f:X-->Y is injective, then [itex]f(X\setminus A) \subset Y\setminus f(A), \forall A \subset X[/itex].

This is how I did it:
If x_1 is in A, then y_1=f(x_1) is in f(A). Because the function is injective, we can pick (cut Y into pieces) f(A) and f(X\A) so, that their intersection is empty (adjust A if needed). So, if x_2 is not in A = if x_2 is in X\A, then y_2=f(x_2) in f(X\A). From this we get that y_2 is in Y\f(A), since Y\f(A)=f(X\A). So, I've shown using the injective property, that y_2 in Y\f(A) follows from y_2 in f(X\A).

Right? Wrong?

- Kamataat
 
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At first glance, it appears to be wrong. You say that you might adjust A, but that's not "legal". You were asked to prove the given relation for /all/ subsets A of X.

It should be clear that any element of f(X\A) is in Y. To prove that any element of f(X\A) is not in f(A), one might try to derive a contradiction...
 
Muzza said:
At first glance, it appears to be wrong. You say that you might adjust A, but that's not "legal". You were asked to prove the given relation for /all/ subsets A of X.

It should be clear that any element of f(X\A) is in Y. To prove that any element of f(X\A) is not in f(A), one might try to derive a contradiction...
So if any element of f(X\A) were also in f(A), then that element must have two originals, one in A and one in X\A, and hence the function is not injective?

- Kamataat
 
Yep, that's it.
 

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