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# MATH 360 Review of Set Theory

Examples:

1. Let f : R → R be given by f(x) =x2 + 1 and let A = . Then fA : A → R
has range=(1,∞), whereas f : R → R has a range of [1,∞)

• Note that f is not one-to-one, because if ∈ R such that ,
the . But we needed .

• Note that fA is one-to-one, because if ∈ A = such that  , then , since both must be
positive reals .

2. Let g : R → R be given by g(x) = sin(π x) and let B = Z. Then : B → {0}.
The range of = {0}, whereas the range of g is [−1, 1]. Note that gB is onto
the set {0}. Note that is not one-to-one, and neither was g. (Why?)

Definition: Let f : X → Y be a one-to-one onto function. The inverse of f is the
function f -1 : Y → X given by f -1(y) = x iff f(x) = y. Note: the inverse of the
function is another function. Example: Let f : R → given by f(x) = ex. Then f -1(x) = ln(x).

Definition:
Let f : X → Y be a function and let V Y . The inverse image of V
is the set f -1(V ) = {x ∈ X : f(x) ∈ V }.

Very Important Note: The inverse image of a set (just like the image of a set) is
another set! The inverse of an element (just like the image of element) is an element.
Here’s the tricky part: the inverse of a function sometimes may not exist. (For exam-
ple, there is not inverse of the function h : R → R given by h(x) =x2, since h is not
1-to-1.) But the inverse image of a set ALWAYS EXISTS! (For example, the inverse
image of [−1, 3] is the set h-1([−1, 3]) = the set of all things in the domain that map
to the set .

Examples:

1. Let f : R → R by f(x) = x + 1. Then f-1([2, 3]) = [1, 2].

2. Let g : R → R by g(x) = 4 −x2. Then g-1([2, 3]) = .

Theorem: Suppose that f : X → Y is a function S X and T Y . Show that Proof:

1. Let y ∈ f(f -1(T)). Then there exists x ∈ f -1(T) such that y = f(x) (because
we know that y is in the image of the set f -1(T).) Now for x ∈ f -1(T) means
that f(x) ∈ T. But remember that f(x) = y! Thus we have y = f(x) ∈ T so
by definition of subsets, f(f -1(T)) T.

2. Let x ∈ S. Then f(x) ∈ f(S), by definition of the set f(S). Thus, by definition
of inverse image of a set, x ∈ f -1(f(S)). Thus S f -1(f(S)).

Note: In the first proof, my element is y and in the second it is x. This has to do
with where the elements are actually located: in the domain or in the codomain.

Theorem: Let f : X → Y be a function and let A be a subset of Y . Then f-1(A) =
X − f-1(Y − A)

Proof: Again, we need to show both and : : Let x ∈ f -1(A). Then f(x) ∈ A. Then f(x) Y −A, so x f -1(Y −A). But since
x ∈ f -1(A), by definition of inverse image of a set, x ∈ X. Thus x ∈ X−f -1(Y −A). : Let x ∈ X − f -1(Y − A). Then x ∈ X but x f -1(Y − A). Thus f(x) Y − A
by definition of inverse images. But if f(x) Y − A, and we know that f(x) ∈ Y ,
because that is the codomain, thus f(x) ∈ A. Thus, by definition of inverse images,
x ∈ f -1(A).

We have proven both and so the equality holds.

Theorem: Let f : X → Y be a function and let {Tα : α∈ } be an indexed
collection of subsets of Y . Then Proof: Suppose x ∈ X.
Then x ∈ f -1( {Tα }) f(x) ∈ {Tα } there exists β ∈ such that f(x) ∈ Tβ there exists β ∈ such that x ∈ f -1(Tβ ) x ∈ {f -1(Tα )}.

Note that the same holds for intersections: .

Example: Let f : R → R be defined by f(x) =x2 + 1 and let = (0, 1]. For each
α ∈ let Vα = [5 −α , 5 +α ]. (For example, V1 = [4, 6], V.5 = [4.5, 5.5].)

• We have {Vα } = {5}. Thus • We have {Vα } = [4, 6]. Thus Well, that is all of the set theory that I think we will need this semester. Cool stuff,
eh?

MATH 360 - Set Theory Homework Assignment

1. Let X = R, A = (−2, 4],B = [0, 3),C = (−∞, 0) and D = [0, 4]. Find each of
the following sets. In parts (j) and (k) sketch the region.

(a) A − B
(b) A ∪ C
(c) B ∩ C
(d) (A ∪ B) − D
(e) X − A
(f) X − C
(g) (A ∩ B) ∩ D
(h) (A ∪ B) ∩ D
(i) B − A
(j) X × A
(k) B × C

2. Let A and B be subsets of X. Prove the other one of DeMorgan' s Laws :

X − (A ∩ B) = (X − A) ∪ (X − B)

3. Prove the following: A ∩ (B ∪ C) = (A ∩ B) ∪ (A ∩ C)

4. Let A and B be subsets of X. Show that A ∩ B = iff A X − B.

5. Find the sets indicated.

(a) = and for each α ∈ let Aα = [−π ,α ). Find {Aα }
and {Aα }.
(b) = and for each α ∈ let Bα = ( α, α + 3). Find {Bα }
and {Bα }.

6. Let {Aα :α ∈ } be an indexed collection of subsets of X. Prove the other one
of DeMorgan’s Laws for indexed collections of sets: 7. Suppose {Aα :α ∈ } is an indexed collection of subsets of X and S X.
Then (Note: the following is also true, but is not a homework problem: 8. Let f : X → Y be a function and let {Aα :α ∈ } be an indexed collection of
subsets of X. Then .

9. Let f : X → Y be a function and let U X and V X. Prove that

f(U) − f(V ) f(U − V ).

10. Let f : X → Y, g : Y → Z be functions.

(a) Prove that if both f and g are onto, then g o f : X → Z is onto.
(b) Prove that if g o f is one-to-one, then f is one-to-one.

11. Find an example ( different than the one given in class) of a function f : X → Y

and A X such that

(a) f|A : A → Y is one-to-one, but f is not one-to-one.
(b) f : X → Y is onto and f|A : A → Y is not onto.

12. Let f : X → Y be a function and let {Tα : α ∈ } be an indexed collection of
subsets of Y . Show that .

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