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.)?
Suppose f is a function with sets A and B.
1. Show that:
2. Show by giving a counter example that:
1.
Let c be an element of .
By the definition of , there is a so that .
Since, , . Since . This follows alongside .
Since and .
Thoughts? Also would I need to show that the to show true equality?
2.
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.
Sorry, what is ? This is not a standard notation. It sounds like something inverse to , but:
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...Let c be an element of .
By the definition of , there is a so that .
What does this mean?Suppose f is a function with sets A and B.