definition let f : a b and let x a and y b. the image (set) of x is f(x) = {y b : y = f(x) for...
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DEFINITION Let f : A B and let X A and Y B. The image (set) of X is
f(X) = {y B : y = f(x) for some x X}
and the inverse image of Y is
f –1(Y) = {x A : f(x) Y}.
Look at the examples and comments on pages 220, 221, and 222.
Theorem 4.5.1 Let f : A B, let C and D be subsets of A, and let E
and F be subsets of B. Then
(a) f(C D) f(C) f(D) ,
(b) f(C D) = f(C) f(D) ,
(c) f –1(E F) = f –1(E) f –1(F) ,
(d) f –1(E F) = f –1(E) f –1(F) .
Proof of (a)
Let b f(C D). Then b = f(a) for some a C D.
a C /\ a D
b f(C)
Theorem 4.5.1 Let f : A B, let C and D be subsets of A, and let E
and F be subsets of B. Then
(a) f(C D) f(C) f(D) ,
(b) f(C D) = f(C) f(D) ,
(c) f –1(E F) = f –1(E) f –1(F) ,
(d) f –1(E F) = f –1(E) f –1(F) .
___________________________
___________________________
b f(D) ___________________________
definition of C D
a C /\ b = f(a)
a D /\ b = f(a)
b f(C) f(D) ___________________________definition of f(C) f(D)
f(C D) f(C) f(D) b f(C D) b f(C) f(D)
Theorem 4.5.1 Let f : A B, let C and D be subsets of A, and let E
and F be subsets of B. Then
(a) f(C D) f(C) f(D) ,
(b) f(C D) = f(C) f(D) ,
(c) f –1(E F) = f –1(E) f –1(F) ,
(d) f –1(E F) = f –1(E) f –1(F) .
Proof of (b)
Let b f(C D). Then b = f(a) for some a C D.
[a C /\ b = f(a)] \/ [a D /\ b = f(a)]
[b f(C)] \/ [b f(D)]
___________________________
___________________________
b f(C) f(D) ___________________________
definition of C D
definition of image
definition of f(C) f(D)
f(C D) f(C) f(D) b f(C D) b f(C) f(D)
Proof of (b)
Let b f(C D). Then b = f(a) for some a C D.
[a C /\ b = f(a)] \/ [a D /\ b = f(a)]
[b f(C)] \/ [b f(D)]
___________________________
___________________________
b f(C) f(D) ___________________________
definition of C D
definition of image
definition of f(C) f(D)
f(C D) f(C) f(D) b f(C D) b f(C) f(D)
Let b f(C) f(D). Then b f(C) or b f(D).
f(C) f(D) f(C D), since b f(C) f(D) b f(C D)
If b f(C), then b = f(a) for some a C C D.
If b f(D), then b = f(a) for some a D C D.
In either case, we can say b = f(a) for some a _________ .C D
Since f(C D) f(C) f(D) and f(C) f(D) f(C D), we have f(C D) = f(C) f(D).
Theorem 4.5.1 Let f : A B, let C and D be subsets of A, and let E
and F be subsets of B. Then
(a) f(C D) f(C) f(D) ,
(b) f(C D) = f(C) f(D) ,
(c) f –1(E F) = f –1(E) f –1(F) ,
(d) f –1(E F) = f –1(E) f –1(F) .Proof of (c)a f –1(E F)f(a) E F
___________________________definition of inverse image
f(a) E /\ f(a) F
___________________________definition of intersection
a f –1(E) /\ a f –1(F)
___________________________definition of inverse image
a f –1(E) f –1(F)
___________________________definition of intersection
Theorem 4.5.1 Let f : A B, let C and D be subsets of A, and let E
and F be subsets of B. Then
(a) f(C D) f(C) f(D) ,
(b) f(C D) = f(C) f(D) ,
(c) f –1(E F) = f –1(E) f –1(F) ,
(d) f –1(E F) = f –1(E) f –1(F) .Proof of (d)a f –1(E F)f(a) E F
___________________________definition of inverse image
f(a) E \/ f(a) F
___________________________definition of union
a f –1(E) \/ a f –1(F)
___________________________definition of inverse image
a f –1(E) f –1(F)
___________________________definition of union
1 (b)
Exercises 4.5 (pages 223-225)
2 (b)
2 (d)
(e)
2 (f)
3 (a)
3 (b)
(c)
3 (d)
(e)
3 (f)
4 (a)
4 (d)
(e)
4 (f)
6