Inverse Image Intersection Equality and Counterexample of a Function Proof

**Icehuu** · September 7th 2010, 05:43 PM

Suppose f is a function with sets A and B.
1. Show that:
$I_{f} \left(A \cap B\right) = I_{f} \left(A\right) \cap I_{f} \left(B\right)$

2. Show by giving a counter example that:
$f\left(A \cap B\right) \neq f\left(A\right) \cap f \left(B\right)$

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

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

2.
$f\left(A \cap B\right) \neq f\left(A\right) \cap f \left(B\right)$
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.

**Icehuu** · September 7th 2010, 07:18 PM

ii.)

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}.

Any suggestions on i.)?

**emakarov** · September 8th 2010, 12:27 AM

Sorry, what is $I_f$ ? This is not a standard notation. It sounds like something inverse to $f$ , but:

Let c be an element of $I_{f} \left(A \cap B\right)$ .
By the definition of $I_{f} \left(A \cap B\right)$ , there is a $d\in(A \cap B)$ so that $I_{f}(d)=c$ .

Is c an element of the domain, and is d an element of the codomain of f? But the inverse of a function is in general a relation. One can think of it as a function that maps elements of the codomain into subsets of the domain. Maybe I am thinking in the wrong direction...

Suppose f is a function with sets A and B.

What does this mean?

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