# MATH 360 Review of Set Theory

**Examples:**

1. Let f : R → R be given by f(x) =x^{2} + 1 and let A =
. Then f_{A} :
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 f_{A} 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) = e^{x}.
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) =x

^{2}, 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 −x

^{2}. 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 {T_{α} : α∈
} be an indexed

collection of subsets of Y . Then

**Proof:** Suppose x ∈ X.

Then x ∈ f^{ -1}(
{T_{α} })

f(x) ∈ {T_{α}
}

there exists β ∈ such
that f(x) ∈ T_{β}

there exists β ∈
such that x ∈ f^{ -1}(T_{β} )

x ∈ {f^{ -1}(T_{α}
)}.

Note that the same holds for intersections: .

**Example:** Let f : R → R be defined by f(x) =x^{2} + 1 and let =
(0, 1]. For each

α
∈ let V_{α} = [5 −α , 5 +α ]. (For example, V_{1} = [4, 6], V_{.5}
= [4.5, 5.5].)

• We have {V_{α}
} = {5}. Thus

• We have
{V_{α}
} = [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 A_{α} = [−π ,α ). Find
{A_{α}
}

and {A_{α}
}.

(b) = and for each α ∈ let B_{α} = (
α, α + 3). Find
{B_{α}
}

and {B_{α}
}.

6. Let {A_{α}
:α ∈ } be an indexed collection of subsets of X. Prove the other one

of DeMorgan’s Laws for indexed collections of sets:

7. Suppose {A_{α}
: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 {A_{α}
: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 {T_{α}
: α ∈ } be an indexed collection of

subsets of Y . Show that .

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