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

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In summary, if f: A-->B, and A(1), A(2) are both subsets of A, then f(A1 ∩ A2) is a subset of f(A1) ∩ f(A2). This can be seen through the fact that for any element x in f(A1 ∩ A2), there must be an element y in A1 ∩ A2 mapping through f to x, and this element must also be in both A1 and A2, therefore belonging to both f(A1) and f(A2). This inclusion may be strict, as demonstrated by the example provided by f(x)=|x|, A1=R-,A2=R+.
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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.
 
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  • #2


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.
 
  • #3


I really don't understand the whole proof
 
  • #4


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?
 
  • #5


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
 
  • #6


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?
 
  • #7


How about f(x)=|x|, A11=R-,A2=R+. Any noninjective function really.
 
  • #8


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

What is the definition of a function?

A function is a relation between a set of inputs (called the domain) and a set of possible outputs (called the codomain), with the property that each input is related to exactly one output.

What does the notation f: A->B mean?

The notation f: A->B indicates that the function f maps elements from the set A to the set B. It means that for every element in A, there is a corresponding element in B.

What does it mean for A(1) and A(2) to be subsets of A?

If A(1) and A(2) are subsets of A, it means that all the elements in A(1) and A(2) are also elements of the set A. In other words, A(1) and A(2) are smaller sets contained within the larger set A.

What is the significance of f: A->B in the statement?

The notation f: A->B in the statement indicates that the function f maps elements from the set A to the set B. This is important because it defines the domain and codomain of the function and allows us to determine the possible inputs and outputs of the function.

What conclusion can be drawn from the statement "if f: A->B, and A(1), A(2) are both subsets of A"?

The statement indicates that the function f maps elements from the set A to the set B, and that A(1) and A(2) are both subsets of A. This means that the function f can only map elements from A(1) and A(2) to elements in B, and not to any other elements in A. In other words, the function is limited to only mapping elements from these subsets of A to the codomain B.

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