Answer
Solve:
a)
A = B = R, S = {x | x > 0}, T = {x | x < 0},
f (x) = x2
b)
It suffices to showthat f (S)∩f (T ) ⊆ f (S∩T ).
Let y ∈ B be an element of f (S) ∩ f (T ).
Then y ∈ f (S),
so y = f (x1) for some x1 ∈ S.
Similarly, y = f (x2) for some x2 ∈ T .
Because f is one-to-one, it follows that x1 = x2.
Therefore x1 ∈ S ∩ T , so y ∈ f (S ∩ T ).
Work Step by Step
Solve:
a)
A = B = R, S = {x | x > 0}, T = {x | x < 0},
f (x) = x2
b)
It suffices to showthat f (S)∩f (T ) ⊆ f (S∩T ).
Let y ∈ B be an element of f (S) ∩ f (T ).
Then y ∈ f (S),
so y = f (x1) for some x1 ∈ S.
Similarly, y = f (x2) for some x2 ∈ T .
Because f is one-to-one, it follows that x1 = x2.
Therefore x1 ∈ S ∩ T , so y ∈ f (S ∩ T ).
