Inverse Images and Sets (union & intersection)

In summary, the conversation discusses the properties of inverse images and sets in relation to a function f. The first part shows that the inverse image of the intersection of two sets A and B is equal to the intersection of the inverse images of A and B. The second part provides a counterexample to show that the function f does not always follow this property, as there can be instances where the inverse image of the intersection of sets A and B is not equal to the intersection of the inverse images of A and B. This is demonstrated through a function that maps two non-intersecting sets to an intersecting set.
  • #1
NastyAccident
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Homework Statement


Suppose f is a function with sets A and B.
1. Show that:
[tex]I_{f} \left(A \cap B\right) = I_{f} \left(A\right) \cap I_{f} \left(B\right)[/tex]
Inverse Image of F (A intersects B) = Inverse Image of F (A) intersects Inverse Image of B.

2. Show by giving a counter example that:
[tex]f\left(A \cap B\right) \neq f\left(A\right) \cap f \left(B\right)[/tex]
F (A intersects B) does not equal F (A) intersects F(B)

Homework Equations



Knowledge of Sets and Inverse Images

The Attempt at a Solution


1.
Let c be an element of [tex]I_{f} \left(A \cap B\right)[/tex].
By the definition of [tex]I_{f} \left(A \cap B\right) [/tex], there is a [tex]d\in(A \cap B)[/tex] so that [tex]I_{f}(d)=c[/tex].
Since, [tex]d\in(A \cap B)[/tex], [tex] d \in A & d \in B[/tex]. Since [tex]d\inA, I_{f}(d)\in I_{f}(A)[/tex]. This follows alongside [tex]d\inB, I_{f}(d)\inI_{f}(B)[/tex].
Since [tex]I_{f}(d)=c \in I_{f}(A) [/tex] and [tex]I_{f}(d)=c \in I_{f}(B), c = I_{f}(A)\capI_{f}(B)[/tex].

Thoughts? Also would I need to show that the [tex]I_{f}(A)\capI_{f}(B) \in I_{f} \left(A \cap B\right)[/tex] to show true equality?

2.
[tex]f\left(A \cap B\right) \neq f\left(A\right) \cap f \left(B\right)[/tex]
I'm thinking either the absolute value function or a square function of some sort would show that it is not equal. Though, I'm not sure how to proceed with depicting the counter example.
NA
 
Last edited:
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  • #2
Okay, so my counterexample for No. 2 is:

f(x)=|x| {-3,...,-1} = A and {1,...,200} = B

There is no intersection between A ∩ B. However, there is an intersection with f(A) ∩ f(B) that gives the set {1,3}.

Thus, {null} != {1,3}.

Figured out the wording that I was missing =) Just need help with i now!
 

1. What is an inverse image?

An inverse image is a set of all elements in the domain that map to a specific element in the codomain under a given function. It is also known as the preimage of the element.

2. How is an inverse image represented mathematically?

The inverse image of an element y in the codomain is denoted by f-1(y) or {x ∈ domain | f(x) = y}. In set-builder notation, it can be written as {x : x ∈ domain and f(x) = y}.

3. What is the difference between the inverse image and the inverse of a function?

The inverse image is a set of elements in the domain that map to a specific element in the codomain, while the inverse of a function is a new function that reverses the mapping of the original function. The inverse of a function is denoted by f-1(x) and it maps elements from the codomain back to the domain.

4. What is the union of inverse images?

The union of inverse images is the set of all elements in the domain that map to any of the elements in the given set of the codomain under a given function. It is denoted by ∪f-1(A) or {x ∈ domain | f(x) ∈ A}.

5. How is the intersection of inverse images calculated?

The intersection of inverse images is the set of elements in the domain that map to all of the elements in the given set of the codomain under a given function. It is denoted by ∩f-1(A) or {x ∈ domain | f(x) ∈ A for all A in the given set}.

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