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

[Simkate]

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.

 

[Tedjn]

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.

 

[Simkate]

I really don't understand the whole proof

 

[jgens]

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?

 

[Simkate]

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

 

[Tedjn]

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?

 

[losiu99]

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

 

[Simkate]

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).