Proving the Intersection of Functions: A Mathematical Study

[Simkate]

Are there any kinds of functions which satisfy f(A1 ∩ A2) =f(A1) ∩ f(A2)? Prove your claim.?

 

[l'Hôpital]

The identity function f(x) = x

 

[adriank]

The identity function?

 

[Petr Mugver]

This is true for every f:

$$f(A\cap B)\supset f(A)\cap f(B)$$

This is true for every injective f:

$$f(A\cap B)\subset f(A)\cap f(B)$$

So the answer to your question is: yes, every injective function.

 

[╔(σ_σ)╝]

Yes, namely: all one to one functions.

Now try proving this and show us your work.

 

[Mark44]

Are there any kinds of functions which satisfy f(A1 ∩ A2) =f(A1) ∩ f(A2)? Prove your claim.?

Please do not double-post your questions.

 

[Simkate]

In order to show that it is a one-to-one function( Injective) i have go the following steps but i don't know where to go with it after it is confusing...


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)

 

[Tedjn]

In this case, you want to show the other direction. That, if x is in f(A1) ∩ f(A2) and f is injective, then x is also in f(A1 ∩ A2).

 

[Simkate]

So can you tell me whether i am correct now??

So, let y∈f(A1)∩f(A2); then y∈f(A1) and y∈f(A2). Then there is an x1∈A1 and an x2∈A2 with f(x1)=f(x2)=y. But since f is one-to-one, x1=x2, and so y∈f(A1∩A2), completing the proof.

 

[adriank]

That's correct.

 

[Simkate]

Thank You:)