1. science
2. mathematics
3. geometry

Math 285H Lecture Notes
April 29, 2015
2
Contents
1 Finite Topologies
1.1 Introduction to Sets and Notation . . . .
1.2 Topology and Topological Spaces . . . .
1.3 Bases and Subbases . . . . . . . . . . . .
1.4 Closed Sets . . . . . . . . . . . . . . . .
1.5 Interior, Exterior, Boundary, and Closure
Exercises . . . . . . . . . . . . . . . . . . . . .
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2 Infinite Sets, Sequences, and Limit Points
2.1 Topologies on Infinite Sets . . . . . . . . .
2.2 Neighborhoods . . . . . . . . . . . . . . .
2.3 Limit Points . . . . . . . . . . . . . . . . .
2.4 Functions . . . . . . . . . . . . . . . . . .
2.5 Sequences . . . . . . . . . . . . . . . . . .
2.6 Theorems on Limit Points and Closures . .
Exercises . . . . . . . . . . . . . . . . . . . . . .
3 The
3.1
3.2
3.3
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Real Numbers
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The Set N of Natural Numbers . . . . . . . . . . . . . . . . . 27
The Set Q of Rational Numbers . . . . . . . . . . . . . . . . . 32
The Set R of Real Numbers . . . . . . . . . . . . . . . . . . . 36
4 Proofs
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4.1 What Is a Proof? . . . . . . . . . . . . . . . . . . . . . . . . . 43
Index
49
3
4
CONTENTS
Chapter 1
Finite Topologies
1.1
Introduction to Sets and Notation
A set is a collection of elements. For example, if we consider the natural
numbers 1, 2, 3, 4, 5, . . . , then this constitutes a set. To specify such a set,
we can list its individual members and enclose them in curly brackets. That
is, N = {1, 2, 3, 4, 5, . . . }. Alternatively, we can write a description of our set
via a sentence with a variable. For example, a set A that contains only the
numbers 1 through 4 is written either as A = {1, 2, 3, 4} or as
A = {x : x is an integer and 1 ≤ x ≤ 4}
As in these two examples, it is standard to denote the names of sets with
capital letters. If p is an element of a set S, we write p ∈ S; and, if p is not
an element of S we write p ∈
/ S. For example, with A and N as above, 3 ∈ A,
5 ∈ N, but 5 ∈
/ A.
In any application of the theory of sets, all sets under investigation are
subsets of a fixed set. We call this set the universal set or the universe
of discourse and denote it by U . In topology, our universe of discourse is
more often denoted by X or Y and is called our space.
Definition 1.1. A set A is a subset of B, written A ⊆ B, if every element
in A is also in B. We can also say that A is contained in B or that B contains
A. If A ⊆ B we say that B is a superset of A, and write B ⊇ A.
Example 1. Let X = {2, 4} and Y = {1, 2, 3, 4}. Therefore, X is contained
in Y ; that is, X ⊆ Y . Furthermore, you can say Y contains X; that is,
Y ⊇ X.
5
6
CHAPTER 1. FINITE TOPOLOGIES
We will find it convenient to discuss subsets of our space with no elements
in them.
Definition 1.2. A set is empty or null if it contains no elements. We
denote an empty set by ∅.
Note. A = ∅ is an empty set while B = {∅} is a set containing the null set
as a member.
It is often convenient to collect together all subsets of a given set X.
Definition 1.3. The set of all subsets of a given set X is called its power
set, and is denoted P(X).
Example 2. Let X = {1, 2, 3}. Then
P(X) = {∅, {1}, {2}, {3}, {1, 2}, {1, 3}, {2, 3}, X}
Example 3. Let A = {1, {1, 2}}. Then P(A) = {∅, {1}, {{1, 2}}, A}.
In topology we will need the standard set operations.
Definition 1.4. The union of two sets A and B, denoted by A ∪ B, is the
set of all elements that belong either to A or to B or to both.
Example 4. Let A = {1, 2, 3} and B = {4, 5, 6}. Then A∪B = {1, 2, 3, 4, 5, 6}.
Example 5. Again let A = {1, 2, 3}, but now let B = {a, b, c, 1}. Then
A ∪ B = {1, 2, 3, a, b, c}. Notice we do not write 1 twice.
Definition 1.5. The intersection of A and B, denoted by A ∩ B, is the
set of all elements that belong to both A and B. So, if A = {1, 3, 5} and
B = {2, 4, 6}, then A ∩ B = ∅.
Example 6. Now let A = {a, b, c} and B = {c, d, e} then A ∩ B = {c}.
Definition 1.6. If our space is X, then the complement of A, denoted
either by Ac or by X\A, is the set of all elements in X that are not in A;
that is, Ac = X\A = {x ∈ X : x ∈
/ A}.
Note. In the above, X\A is called a set difference. It is analogous to
subtraction in algebra, but it operates on sets rather than numbers and
functions. It can be extended to any subsets of the universe of discourse. So
B\A = {x : x ∈ B, x ∈
/ A}. Because of its relationship to complement, set
difference is also called relative complement.
1.2. TOPOLOGY AND TOPOLOGICAL SPACES
7
Example 7. if X = {1, 2, 3, 4, 5}, A = {2, 4}, and B = {1, 2, 3}, then
Ac = {1, 3, 5}, B c = {1, 3, 5}, B\A = {1, 3}, and A\B = {2}.
Definition 1.7. A finite set is a set consisting of n elements where n is
some whole number. If a set S has n elements, we write #(S) = n and we
say S has cardinality n.
Example 8. S = {1, 2, 3, . . . , 100} is a finite set.
Example 9. Let A = ∅, B = {∅}, and C = {∅, {∅}} are finite sets with
cardinalities #(A) = 0, #(B) = 1, and #(C) = 2. Note further that {∅} is
both an element and subset of C.
A set with exactly one element is called a singleton. For example both
{1} and {∅} are singletons. A set with two elements is called a doubleton.
Hence both {1, 2} and {∅, {∅}} are doubletons. We note further that {1} is
read as “singleton one” and {1, 2} is read as “doubleton one, two.”
1.2
Topology and Topological Spaces
Finally we get to the star of our show: topology. Our initial definition of a
topology is very simple but not very intuitive. In this chapter we restrict our
space X to be a finite set. With this restriction we can define a topology to
be a collection of subsets of X closed under the two set operations of union
and intersection. Furthermore, for technical reasons that will become clear
in later chapters, we will require our topology to always include the space X
and the empty set. That is:
Definition 2.1. A collection T of subsets of a finite set X is a topology on
X if and only if each of the following axioms hold:
Axiom 1. T contains both the whole space X and empty set ∅.
Axiom 2. The union of any number of sets in T belongs to T . That is, T
is closed under unions.
Axiom 3. The intersection of any finite number of sets in T belongs to T .
That is, T is closed under finite intersections.
8
CHAPTER 1. FINITE TOPOLOGIES
Note. The pair X along with its topology T is called a topological space
and is denoted by (X, T ). A topology T is a collection of sets, and if a set G
is in a topology T , then we say that it is an open sets . We will denote open
sets with capital letters G, H, etc., usually starting at G (from the German
word for open “geoeffnet”.)
We now provide several examples of what are and what are not topologies.
Example 10. Let X = {1, 2, 3} and
T = {X, ∅, {1}, {3}, {1, 2}}
The set {1} ∪ {3} = {1, 3} is not in T . Since this violates Axiom 2, T is not
a topology.
Example 11. Let X = {1, 2, 3, 4} and
T = {X, ∅, {1}, {1, 2}, {1, 4}, {2, 4}, {1, 2, 4}}
The set {1, 2} ∩ {2, 4} = {2} is not in T . This violates Axiom 3, so T is not
a topology.
Example 12. Let X = {1, 2, 3, 4} and
T = {X, {1}, {2}, {4}, {1, 2}, {1, 4}, {2, 4}, {1, 2, 4}}
In this example, the empty set is missing. This violates Axiom 1, and so T
is not a topology.
Example 13. Let X = {1, 2, 3}.
1. One possible topology is T1 = {X, ∅, {1}, {2, 3}} since {1} ∪ {2, 3} = X
and {1} ∩ {2, 3} = ∅. Hence X, ∅, {1} and {2, 3} are the open sets in
T1 .
2. Another possible topology is D = {X, ∅, {1}, {2}, {3}, {1, 2}, {1, 3}, {2, 3}}.
That is, D = P(X), which clearly satisfies all three axioms. This topology is called the discrete topology on X.
3. A third topology is I = {X, ∅}. As X ∪ ∅ = X, and X ∩ ∅ = ∅. I also
clearly satisfies all three axioms. I is called the indiscrete topology
on X.
1.3. BASES AND SUBBASES
9
Note. As in this example, for any space X we call P(X) the discrete topology
on X and usually denote it by D. We call {X, ∅} the indiscrete topology and
denote it by I. Clearly, the discrete topology is the largest possible topology
on a space, and the indiscrete topology is the smallest.
Example 14. Let X = {1, 2, 3, 4}. We want to find a topology T on X
that is neither the discrete topology nor the indiscrete topology. By Axiom
1 above it is helpful to first write down X and ∅ as members of T (this
will insure that you never go through a lot of work only to forget that you
have not included one of these.) Next, keep in mind what the possible subsets of X are. You may choose any subsets you wish, but then need to
check that your list satisfies Axioms 2 and 3. If not, then add any necessary subsets to insure that Axioms 2 and 3 are satisfied. Say that you
choose X, ∅, {1}, {1, 2}, {3, 4} as your desired subsets to be in T . Checking Axiom 2 we see {1} ∪ {1, 2} = {1, 2}, which is in T . Next checking {1} ∪ {3, 4} = {1, 3, 4}, we see we need to add {1, 3, 4} to T . Since
{1, 2} ∪ {3, 4} = X, you now have X, ∅, {1}, {1, 2}, {3, 4}, {1, 3, 4} in T. Next
we check Axiom 3 (intersections). Clearly {1}∩{1, 2} = {1}, {1}∩{3, 4} = ∅,
{1} ∩ {1, 3, 4} = {1}, {1, 2} ∩ {3, 4} = ∅, and {1, 2} ∩ {1, 3, 4} = {1} and so
T = {X, ∅, {1}, {1, 2}, {3, 4}, {1, 3, 4}} is a solution.
1.3
Bases and Subbases
There are hundreds of other solutions to Example 14, such as T1 = {X, ∅, {1}},
T2 = {X, ∅, {3}}, T3 = {X, ∅, {1}, {1, 3}}, and T4 = {X, ∅, {1}, {3}, {1, 3}}.
Clearly, additional concepts are called for.
Definition 3.1. A collection of open subsets B of a space X is a base for a
topology T on X if and only if every open set G in T is the union of members
of B.
If we look again at our initial solution to Example 14, we chose the open
sets to be X, ∅, {1}, {1, 2}, {3, 4} to be in the topology. Collecting these as a
set forms a base for the topology T = {X, ∅, {1}, {1, 2}, {3, 4}, {1, 3, 4}}. In
this case, our topology only contains one more set than the base. Lets look
at another example with the same set X but a different topology.
Example 15. Let X = {1, 2, 3, 4}. The collection B = {X, ∅, {1, 2}, {3, 4}, {4}}
forms a base for the topology
T = {X, ∅, {4}, {1, 2}, {3, 4}, {1, 2, 4}}
10
CHAPTER 1. FINITE TOPOLOGIES
This can be seen by forming unions of the sets in B.
Example 16. Let X = {1, 2, 3, 4} and B = {X, ∅, {1, 2}, {2, 3}, {4}}. A
topology using these sets would have to include the set {2} in order to be
closed under intersections. But, {2} is not in B, ergo the topology cannot
be the union of members of B. Therefore, B is not a base. Remember that
every open set, i.e. every set in the topology, must be the union of members
of B. If there is even one case where this is false, then B is not a base.
Definition 3.2. A collection S of open subsets of X, i.e. S ⊆ T , is a subbase for a topology on X if and only if the collection of all finite intersections
of elements of S form a base for the topology.
Note. B in Example 13 not only constitutes a base for T , but it is also a
subbase for T . In general, a base is automatically a subbase; but, as the
next example shows, a subbase can quite usefully be smaller than a base.
Example 17. Let X = {1, 2, 3, 4} and S = {{1, 2}, {2, 3}, {3, 4}}. Taking
finite intersections of members of S yields the set
B = {∅, {2}, {3}, {1, 2}, {2, 3}, {3, 4}, X}
Now taking unions yields the topology
T = {∅, {2}, {3}, {1, 2}, {2, 3}, {3, 4}, {1, 2, 3}, {2, 3, 4}, X}
1.4
Closed Sets
Recall that elements of the topology are called open. We will also need the
complements of such sets.
Definition 4.1. A set is closed if and only if its complement is open.
Note. Since X c = ∅ and ∅c = X, both X and ∅ are both open and closed sets
in any topology. Such sets are called clopen.
Example 18. Let X = {1, 2, 3} and T = {X, ∅, {1}, {2}, {1, 2}}. Since {1}
is in the topology, it is open. Its complement is {2, 3} and is closed. Likewise,
{1, 3} and {3} are also closed.
1.5. INTERIOR, EXTERIOR, BOUNDARY, AND CLOSURE
11
Example 19. Let X = {1, 2.3, 4}. The set {1, 3} and its complement {2, 4}
both being subsets of X are open D, the descrete topology of X. Being
complements each other, {2, 4} and {1, 3} are both closed. Thus {1, 3} and
{2, 4} are clopen in is closed in D.
Note. In the discrete topology on any space X all subsets of X are both open
and closed (clopen.)
1.5
Interior, Exterior, Boundary, and Closure
Given a topology, there are special open and closed sets attached to any
subset of our space, called the interior, exterior, boundary, and closure of
the subset. This terminology, as we will see in Chapters Two and Three has
its roots in geometry, but today is used in many non-geometric contexts. In
this chapter these definitions are best taken as abstractions which are simply
memorized.
Definition 5.1. Let A be a subset of a topological space X. A point p ∈ A
is called an interior point of A if p belongs to an open set G contained in
A, that is,
p ∈ G ⊆ A where G is open.
The set of interior points of A, dentoted by
◦
int(A), A or A◦
is called the interior of A .
Proposition 5.2. The interior of a set A is the union of all open sets
contained in A. Moreover,
1. The interior of A is an open set.
2. The interior of A is the largest open set contained in A.
3. The set A is open if and only if A = A◦ .
Example 20. As in Example 19, let X = {1, 2, 3} and
T = {X, ∅, {1}, {2}, {1, 2}}.
Furthermore, let A = {1, 2}, B = {1, 3}, and C = {2, 3}. As A is already
open, A◦ = A. The set {1} is the largest open set in B, so B ◦ = {1}.
Similarly, C ◦ = {2}.
12
CHAPTER 1. FINITE TOPOLOGIES
Definition 5.3. The exterior of a set A, written ext(A), is the interior of
Ac . The boundary of A, written b(A) or ∂A, is the set of points which
belong to neither the interior nor the exterior of A.
Example 21. Consider the topology T = {X, ∅, {a}, {c, d}, {a, c, d}, {a, c, d, e}}
on X = {a, b, c, d, e} and the subset A = {b, c, d} of X.
1. Since c, d ∈ A and {c, d} is an open set, the points c and d are both
interior points of A. Since b ∈ A is in no open set belonging to A, it is
not an interior point of A. Accordingly, int(A) = {c, d}.
2. The complement of A is Ac = {a, e}. The only open set contained in
Ac is {a}. Therefore, the only interior point of Ac is a. This means
that a is the only exterior point of A.
3. If we remove both the interior of A and the exterior of A from X, we
have {b, e} left. Accordingly the boundary of A consists of the points
b and e; that is, ∂A = {b, e}.
Definition 5.4. The closure of A, written cl(A) or A, is the intersection of
all closed supersets of A. Alternatively, the closure of A can be thought of
as the smallest closed set containing A. Furthermore if p ∈ A, then we say p
is in the proximity of A or p is a proximate point of A.
Example 22. As in Example 21 let
T = {X, ∅, {a}, {c, d}, {a, c, d}, {a, c, d, e}}
be our topology on X = {a, b, c, d, e}. So the closed subsets on X are
∅, X, {b, c, d, e}, {a, b, e}, {b, e}, {a}. Accordingly, the closure of A = {b, c, d}
is {b, c, d, e}, the closure of B = {a, c} is all of X, and the closure of C = {b, d}
is {b, c, d, e}. Also b is a proximate point of A, whereas a is not a proximate
point of A.
Exercises
1. Find all four topologies on X = {1, 2}.
2. Which of the following are topologies on X = {1, 2, 3}?
(a) {X}
1.5. INTERIOR, EXTERIOR, BOUNDARY, AND CLOSURE
13
(b) {∅}
(c) {X, ∅}
(d) {X, ∅, {1}}
(e) {X, ∅, {1}, {2}}
(f) {X, ∅, 1, 2}
(g) {X, ∅, {1}, {1, 2}}.
3. Which of the following are bases for a topology on X = {1, 2, 3}?
(a) {X}
(b) {∅}
(c) {∅, {1}}
(d) {X, ∅, {1}}
(e) {X, ∅, {1}, {2}}
(f) {X, ∅, {1, 2}, {3}}
(g) {∅, {1, 2}, {3}}
(h) {{1}, {2}, {3}}.
4. Find the topologies generated by the following bases for X = {a, b, c}.
(a) {X}
(b) {∅}
(c) {X, ∅, {b}}
(d) {∅, {a, b}, {c}}.
5. Find the topologies generated by the following subbases for X = {1, 2, 3}.
(a) {X, ∅, {1}}
(b) {X, ∅, {1, 2}}
(c) {X, {1}, {2}}
(d) {∅, {1, 3}, {2}}.
14
CHAPTER 1. FINITE TOPOLOGIES
In problems 6 - 12 below let T = {X, ∅, {1}, {3}, {1, 3}, {2, 3}} be the
topology on the space X = {1, 2, 3}. Also let A = {2}, B = {1, 2}, and
C = {1, 3}.
6. Find the set of closed sets for the topology T .
7. Find the closures of A, B, and C.
8. Find the interiors of A, B, and C.
9. Find the exteriors of A, B, and C.
10. Find the boundaries of A, B, and C.
11. Classify each of 1, 2, and 3 as to whether or not they are proximate
points of A.
12. Prove that A ⊆ B if and only if B c ⊆ Ac .
Chapter 2
Infinite Sets, Sequences, and
Limit Points
2.1
Topologies on Infinite Sets
In the previous chapter we discussed topologies only on finite sets. In this
chapter we will introduce infinite sets. and modify our definition of topology
so that it that encompasses both finite sets and infinite sets.
We define the union of an arbitrary collection of sets. Let X be a set,
and suppose that I is another set such that for each i in I we have a subset
Gi ; the set I is often called an index set, and we refer to the collection of
subsets as {Gi : i ∈ I}. Then we define
[
Gi = {x ∈ X : x ∈ Gi for at leaast one value of i in I}
i∈I
We now give our definition of a topological space.
Definition 1.1. A topological space is a pair of objects (X, T ), where X
is a set T is a collection of subsets of X satisfying the following conditions:
1. ∅, X ∈ T ;
S
2. if {Gi : i ∈ I} ⊂ T , then i∈I Gi ∈ T ;
T
3. if G1 , . . . , Gn ∈ T , then nk=1 Gk ∈ T .
The collection T is called the topology on X and sets in T are called open
sets.
15
16 CHAPTER 2. INFINITE SETS, SEQUENCES, AND LIMIT POINTS
Example 1. We define the usual topology on R as the topological space
on R that has as a base the set of all finite open intervals (a, b) in R.
For any topological space (X, T ) with base B, a set G is open if and only
if for every element x ∈ G, there is an open set B in the base, such that
x ∈ B and B ⊆ G.
Therefore a set G is open in the usual topology on R if and only if for
any point x ∈ G, there exists some open interval (a, b) such that
x ∈ (a, b) ⊆ G.
Infinite open intervals of the form (a, ∞) or (−∞, b) are in the usual
topology on R. This is because the arbitrary union of open sets in a topology
is also in the topology. We see that
∞
[
(a, ∞) =
(a, a + n)
n=1
On the other hand the infinite intersection of open sets in a topology may
not necessarily be in the topology. For example, the set {0} is not an open
set in the usual topology on R (why?). However, we can also write {0} as
the infinite intersection of open intervals,
{0} =
∞
\
(−1/n, 1/n)
n=1
From this we see that infinite intersections of open sets need not be open.
Example 2. In the upper limit topology on the real numbers we define
the set of all intervals of the form (a, b] to be the base. So in the upper
limit topology on R, (1, 3] is open by definition of the base, whereas the sets
(1, 3] ∪ (4, 5] and (1, 3] ∪ (2, 4] ∪ (3, 5] ∪ · · · = (1, ∞) are open by forming
unions of base elements.
Definition 1.2. The set of all subsets of a space X whose complements in
X are finite sets along with the empty set form the cofinite topology on
X.
Note. Since X = X\∅ and the empty set has cardinality zero, X is automatically in the cofinite topology. [See also Exercise 2 below.]
2.2. NEIGHBORHOODS
17
Example 3. Consider N = {1, 2, 3, 4, 5 . . . } to be our space, and let T be
the cofinite topology on N. Then, the set {6, 7, 8, 9, 10 . . . } is open because
the set {1, 2, 3, 4, 5} is finite. Since the complement of the even numbers is
the odd numbers, and since both sets are infinite, neither set is open in the
cofinite topology on N.
2.2
Neighborhoods
Definition 2.1. Let p be a point in a topological space X and N be a subset
of X. The set N is a neighborhood of p if and only if N contains an open
set G containing p; that is, p ∈ G ⊆ N . The set of all neighborhoods of p is
called the neighborhood system of p, and is denoted by Np .
Note. The set N above does not have to be an open set in the topology; it
can be any subset of X. When a neighborhood of a point p is an open set in
the topology, we call it an open neighborhood of p.
Example 4. Let T = {X, ∅, {a}, {a, c}} be a topology on X = {a, b, c, d}.
Find N3 , the neighborhood system of the point c.
Solution. The only open sets containing c are {a, c} and X. We list the
subsets of X that contain these open sets. These sets are neigborhhods of c.
The supersets of {a, c} are
{a, c}, {a, b, c}, {a, c, d}, X
The only superset of X is X itself. So the neighborhoods of c are
{a, c}{a, b, c}, {a, c, d}, X
That is,
Nc = {X, {a, c}{a, b, c}, {a, c, d}
Example 5. Let T = {X, ∅, {5}, {1, 2}, {3, 4}, {1, 2, 5}, {3, 4, 5}, {1, 2, 3, 4}}
be a topology on X = {1, 2, 3, 4, 5}. Find Nc , the neighborhood system of
the point 3.
18 CHAPTER 2. INFINITE SETS, SEQUENCES, AND LIMIT POINTS
Solution. The open sets that contain 3 are
{3, 4}, {3, 4, 5}, {1, 2, 3, 4}, X
We now list all of the subsets of X that contain these open sets. These sets
are neigborhoods of 3. The supersets of {3, 4} are
{3, 4}, {2, 3, 4}, {1, 3, 4}, {1, 2, 3, 4}, {1, 3, 4, 5}, {3, 4, 5}, X
The supersets of {3, 4, 5} are
{3, 4, 5}, {2, 3, 4, 5}, {1, 3, 4, 5}, X
The supersets of {1, 2, 3, 4} are {1, 2, 3, 4} and X. So
N3 = {X, {3, 4}, {1, 3, 4}, {2, 3, 4}, {3, 4, 5}, {1, 2, 3, 4}, {1, 3, 4, 5}, {2, 3, 4, 5}}
Example 6. In the usual topology on R, a neighborhood of a point p is any
subset A of which contains an interval (a, b) such that p ∈ (a, b).
2.3
Limit Points
In the usual topology on the real numbers R, the sequence,
1
1 1
= 1, , , . . .
n
2 3
converges to zero. However zero is not actually in the set
1 1
1, , , . . .
2 3
This observation prompts the next four definitions.
Definition 3.1. Let A be a subset of a topological space (X, T ). A point
p in X is a limit point of A if and only if every open set containing p also
contains at least one point of A different from p itself.
That is, a point p is a limit point of a set A if and only if for every open
set G,
G\{p} ∩ A 6= ∅
.
2.3. LIMIT POINTS
19
Example 7. Define a topological space (X, T ) by X = {1, 2, 3, 4} and
T = {∅, X, {2}, {3}, {1, 2}, {2, 3}, {3, 4}, {1, 2, 3}, {2, 3, 4}}
. Let A = {2, 3, 4}. Find all of the limit points of A.
Solution. The number 1 is a limit point of A. The open sets containing 1 are
{1, 2} and {1, 2, 3}. We see that
{1, 2}\{1} ∩ A 6= ∅ and {1, 2, 3}\{1} ∩ A 6= ∅
The numbers 2 and 3 are not limit points of A. This is because both {2}
and {3} are open set and
{2}\{2} = ∅ and {3}\{3} = ∅
The number 4 is a limit point of A. The open sets containing 4 are {3, 4}
and {2, 3, 4}. We see that
{3, 4}\{4} =
6 ∅ and {2, 3, 4}\{4} =
6 ∅
Note. In general, if {p} is an open set, that is, {p} is in a topology, then p
cannot be a limit point of any subset of the space.
Example 8. In the usual topology on the real numbers, show that the number 3 is a limit point of A = [2, 3).
Solution. Any open set containing 3 in the usual topology will contain an
interval of the form (3 − r, 3 + r) for some r > 0. But
A ∩ (3 − r, 3 + r)\{3} =
6 ∅
It follows that for any open set G containing 3, the set G\{3} will contain
at least one point that is in A, which means that 3 is a limit point of A =
[2.3).
Definition 3.2. The derived set of a set A is the set of all limit points of
A and is denoted by A0 .
Example 9. Using (X, T ) and A from Example 7, find A0 , the derived set
of A.
20 CHAPTER 2. INFINITE SETS, SEQUENCES, AND LIMIT POINTS
Solution. The only limit point of A is 4, so the derived set of A is {4}. In
symbols, A0 = {4}.
Example 10. Let X = {1, 2, 3, 4}, T = {∅, X, {1, 2}, {3, 4}} and A = {1, 2, 3}.
Find A0 , the derived set of A.
Solution.
• The open sets containing 1 are {1, 2} and X. We see that
{1, 2}\{1} ∩ A = {2} =
6 ∅ and X\{1} ∩ A = {2, 3} =
6 ∅
So 1 is a limit point of A.
• The open sets containing 2 are {1, 2} and X. We see that
{1, 2}\{2} ∩ A = {1} and X\{2} ∩ A = {1, 3}
So 2 is a limit point of A.
• The open sets containing 3 are {3, 4} and X. We see that
{3, 4}\{3} ∩ A = ∅
So 3 is not a limit point of A.
The limit points of A are 1 and 2. That is, A0 = {1, 2}.
2.4
Functions
Very little interesting mathematics can be done without functions lurking
somewhere. In particular, this is true in topology.
We denote a function f from a set X into a set Y by
f :X→Y
This means that f assigns a unique element of Y to every element of X. In
particular, if f assigns a ∈ X to b ∈ Y , then we call b the image of a under
f , and we write either
f (a) = b or f : a 7→ b
2.4. FUNCTIONS
21
Example 11. Let X = {1, 2, 3, 4} and let Y = {1, 2, 3}. Define the function
f : X → Y by
1 7→ 1, 3 7→ 1, 2 7→ 2, 4 7→ 2
Then f (1) = f (3) = 1, and f (2) = f (4) = 2.
Suppose f : X → Y is a function, and A is a subset of X. Then the set
of all f (x) for x in A is called the image of A and is written f (A). The
image of X is called the range of f and is written Ran f . The set X is
called the domain of f , written Dom f . The set Y is called the codomain
of f , written Cod f . A function is called onto or surjective if its range and
codomain are equal. If f (a) = f (b) implies a = b for all a, b in X, then the
function f is called one-to-one or injective.
Example 12. Let X, Y , and f be as in Example 11, and let A = {1, 3}
and B = {1, 4}. Then f (A) = {1}, f (B) = {1, 2}, Ran f = {1, 2}, and Cod
f = {1, 2, 3}. Since Ran f 6= Cod f , f is not onto. Since f (1) = f (3), but
1 6= 3, f is not one-to-one.
Suppose that f : X → Y , and B is a subset of the range of f . Then the
set of all x ∈ X such that f (x) ∈ B is called the inverse image or pullback
of B under f , written A = f −1 (B).
Example 13. Let X, Y , and f be as in Example 11, and let A = {1, 3} ⊆ Y
and B = {2, 3} ⊆ Y . Find f −1 (A) and f −1 (B).
Solution.
f −1 (A) = {1, 3} and f −1 (B) = {2, 4}
Example 14. Let exp: R → R be the usual exponential function
exp : x 7→ ex , that is, exp(x) = ex
Is exp one-to-one? Is exp onto? Find the inverse image of the set {2, 3}.
Solution. The function exp is an increasing function. This means that if
a < b, then exp(a) < exp(b). It follows that if exp(x1 ) = exp(x2 ), then
x1 = x2 . Therefore exp is a one-to-one function. (How do we know that exp
is increasing?)
We have that Ran exp = (0, ∞), so exp is not an onto function. The
inverse image of {2, 3} is {ln 2, ln 3}.
22 CHAPTER 2. INFINITE SETS, SEQUENCES, AND LIMIT POINTS
2.5
Sequences
A special case of a function, familiar from calculus, is a sequence.
Definition 5.1. A sequence a in a space X is defined as a function from
the set of natural numbers into that space,
a:N→X
Note. As is usually the case, the elements of the range of the sequence a are
function values in X but are denoted by a1 , a2 . . . . That is,
a(1) = a1 , a(2) = a2 , a(3) = a3 , . . .
It is also common to denote the sequence a by ha1 , a2 , . . . i. So, for example,
the sequence of even numbers is denoted by h2, 4, 6 . . . i. This means
a1 = 2, a2 = 4, a3 = 6, . . .
Definition 5.2. We say a sequence a = han i converges to a point l in X,
denoted by
lim an = l or an → l
n→∞
if and only if for every open set G of the point l, there is some natural number
N0 such that if n > N0 , then an is in G.
In symbols,
an → l ⇐⇒ ∀G ∈ T , l ∈ G ⇒ ∃N0 ∈ N, s.t. n > N0 ⇒ an ∈ G
Example 15. In the usual topology for R, 1/2n → 0.
Example 16. If an = c for some constant c, then an → c in any topology
on a space X.
2.6
Theorems on Limit Points and Closures
Mathematics is not just definitions, examples and problems, but is also theorems about concepts and proofs of those theorems. The proof of the following
theorem is typical of a whole group of proofs in topology, geometry and analysis. The proof uses Exercise 12 of the last chapter, and warrants studying.
2.6. THEOREMS ON LIMIT POINTS AND CLOSURES
23
Theorem 6.1. A set is closed if and only if it contains all of its limit points.
Proof. (⇒) Suppoe that F is a closed set in T . Then, by the definition of a
closed set, F c is open. Let p be in F c . Then (F c \{p}) ∩ F ⊆ F c ∩ F = ∅.
So by definition, p is not a limit point. That is, p ∈
/ F 0 . As p is arbitrary,
F c ⊇ (F 0 )c . Equivalently, F 0 ⊆ F .
(⇐) Now suppose F contains all of its limit points. Let p be in F c . Then,
by assumption, p is not a limit point of F . Accordingly, there must be an
open set Gp of p for which (Gp \{p}) ∩ F = ∅. Since p is in F c , we have
Gp ∩ F = (Gp \{p}) ∩ F = ∅. That is, Gp ⊆ F c . It follows that
c
F c = ∪p∈F
/ {p} ⊆ ∪p∈F
/ Gp ⊆ F
But then F c is the union of open sets, and so is open. Hence F is closed.
In order to prove the next two theorems, we will need a pair of important
properties of sets, known as De Morgans Laws.
(A ∪ B)c = Ac ∩ B c and (A ∩ B)c = Ac ∪ B c
First, however, some examples.
Example 17. Let X = {1, 2, 3, 4, 5, 6}, A = {1, 3, 6} and B = {4, 5, 6}.
Then X\A = Ac = {2, 4, 5} and B c = {1, 2, 3}. Hence Ac ∩ B c = {2, 4, 5} ∩
{1, 2, 3} = {2} = (A ∪ B)c , and
Ac ∪ B c = {2, 4, 5} ∪ {1, 2, 3} = {1, 2, 3, 4, 5} = {6}c = (A ∪ B)c
Example 18. Let An = (−1/n, 1/n). Then ∩n∈N An = {0}, and so
(∩n∈N An )c = {0}c = (−∞, 0) ∪ (0, ∞)
Also, because An = (−1/n, 1/n), we have Acn = (−∞, −1/n] ∪ [1/n, ∞); accordingly, ∪n∈N Acn = (−∞, 0) ∪ (0, ∞). Consequently, (∩n∈N An )c = ∪n∈N Acn .
Our next theorem is an easy albeit important consequence of De Morgans
Laws. It generalizes Example 18.
Theorem 6.2. An arbitrary intersection of closed sets is closed.
24 CHAPTER 2. INFINITE SETS, SEQUENCES, AND LIMIT POINTS
Proof. Let F1 , F2 , . . . be closed sets. Then F1c , F2c , . . . are open sets, and so
∪n∈N Fnc is open. Then, by De Morgans Laws, ∩n∈N Fn = (∪n∈N Fnc )c, which is
closed.
Our last theorem is this chapter combines the idea of limit point with De
Morgans Laws.
Theorem 6.3. If A is an arbitrary subset of X and A0 is its derived set,
then A ∪ A0 is closed.
Proof. Let p ∈ (A ∪ A0 )c . By De Morgans Law p ∈
/ A and p ∈
/ A0 . As
0
p∈
/ A , there is an open neighborhood Gp of p such that (Gp \{p}) ∩ = ∅.
Since p ∈
/ A, we have Gp ∩ A = (Gp \{p}) ∩ A = ∅. Also, if q ∈ Gp , then
(Gp \{q}) ∩ A ⊆ Gp ∩ A = ∅. So q is not a limit point of A. As q is arbitrary,
Gp ∩A0 = ∅. This latter result, along with Gp ∩A = ∅, yields Gp ∩(A∪A0 ) = ∅.
Thus p is an exterior point of A ∪ A0 . Since p is arbitrary, it follows that
(A ∪ A0 )c is open, and so A ∪ A0 is closed.
Exercises
1. Define the base for the lower limit topology on R. [See Example 2]
2. Let X be an infinite set. Prove that the cofinite topology as defined
above truly is a topology. [Hint: De Morgans Laws]
3. What is meant by a closed neighborhood of a point?
In problems 4 through 6 below let T = {X, ∅, {a}, {a, c}} be our topology on X = {a, b, c, d}.
4. Find all closed neighborhoods of the point c.
5. Show that c is a limit point of A = {a, b}.
6. Find the derived set of A = {a, b}.
In problems 7 and 8 below let R have its usual topology.
7. Show that 2 and 3 are limit points of (1, 3).
8. Find the derived set of (1, 3).
2.6. THEOREMS ON LIMIT POINTS AND CLOSURES
25
9. What is the derived set of the empty set?
10. Let X have the discrete topology and let an → b. Prove that the
sequence must be eventually constant; that is, the sequence must be of
the form ha1 , a2 , . . . , ak , b, b, b, . . . i.
11. Let X have the indiscrete topology. Prove that any sequence in X has
any point of X as a limit. That is, for any sequence han i and for any
b ∈ X, an → b.
12. Prove for any f : X → Y and A ⊆ X that A ⊆ f −1 (f (A)).
13. Prove A0 ⊆ A.
14. Prove that A = A ∪ A0 [Hints: Theorem 6.3 and Exercise 13.]
15. Prove, if A ⊆ B, then A0 ⊆ B 0 .
16. Prove, if A ⊆ B, then cl(A) ⊆ cl(B).
26 CHAPTER 2. INFINITE SETS, SEQUENCES, AND LIMIT POINTS
Chapter 3
The Real Numbers
3.1
The Set N of Natural Numbers
We denote {1, 2, 3, . . . } of all natural numbers by N. Elements of N will
also be called the positive integers. Each number n has a successor, namely
n + 1. Thus the successor of 2 is 3, and 37 is the successor of 36.
The following properties of the natural numbers are called the Peano
Axioms. All the properties of N can be proved based on these five axioms.
N1. 1 belongs to N
N2. If n belongs to N, then its successor n + 1 belongs to N.
N3. 1 is not the successor of any element in N.
N4. If n and m in N have the same successor, then n = m.
N5. A subset N contains 1, and which contains n + 1 whenever it contains
n, must equal N.
Axiom N5 is the basis of mathematical induction. Let P1 , P2 , P3 , . . .
be a list of statements or propositions that may or may not be true. The principle of mathematical induction asserts that all the statements P1 , P2 , P3 , . . .
are true provided
(I1 ) P1 is true.
(I2 ) Pn+1 is true whenever Pn is true.
27
28
CHAPTER 3. THE REAL NUMBERS
We will refer to (I1 ), i.e., the fact that P1 is true, as the basis for induction and we will refer to (I2 ) as the induction step.
Example 1. Prove that for all natural numbers n,
1
1 + 2 + · · · + n = n(n + 1)
2
Solution.
3.1. THE SET N OF NATURAL NUMBERS
29
Example 2. Prove the assertion that all numbers of the form 7n − 2n are
divisible by 5.
Solution.
Example 3. Show that | sin nx| ≤ n| sin x| for all natural numbers n and all
real numbers x.
Solution.
30
CHAPTER 3. THE REAL NUMBERS
Exercises
1. Prove 12 + 22 + · · · + n2 = n(n + 1)(2n + 1)/6 for all natural numbers
n.
2. Prove 3 + 11 + · · · + (8n − 5) = 4n2 − n for all natural numbers n.
3. Prove 13 + 23 + · · · + n3 = (1 + 2 + · · · + n)2 for all natural numbers n.
4. (a) Guess the formula for 1 + 3 + · · · + (2n − 1) by evaluating the sum
for n = 1, 2, 3 and 4. [For n = 1, the sum is simply 1.]
(b) Prove your formula using mathematical induction.
5. Prove 1 + 1/2 + 1/4 + · · · + 1/2n = 2 − 1/2n for all natural numbers n.
6. Prove that (11)n − 4n is divisible by 7 when n is a natural number.
7. Prove that 7n − 6n − 1 is divisible by 36 for all positive integers n.
8. The principle of mathematical induction can be extended as follows.
A list Pm , Pm+1 , . . . of propositions is true provided (i) Pm is true, (ii)
Pn+1 is true whenever Pn is true and n ≥ m.
(a) Prove that n2 ≥ n + 1 for all integers n ≥ 2.
(b) Prove that n! ≥ n2 for all integers n ≥ 4.
9. (a) Decide for which integers the inequality 2n ≥ n2 is true.
(b) Prove your claim in (a) by mathematical induction.
10. Prove
(2n + 1) + (2n + 3) + (2n + 5) + · · · + (4n − 1) = 3n2
for all positive integers n.
11. For each n ∈ N, let Pn denote the assertion:
n2 + 5n + 1 is an even integer.
(a) Prove that Pn+1 is true whenever Pn is true.
(b) For which n is Pn actually true? What is the moral of the exercise?
3.1. THE SET N OF NATURAL NUMBERS
31
12. For n ∈ N denote 1 · 2 · 3 · · · n. Also let 0! = 1 and define
n
n!
=
k
k!(n − k)!
for k = 0, 1, . . . , n.
The binomial theorem asserts that
n n
n n−1
n n−2 2
n
n n
n
n−1
(a+b) =
a +
a b+
a b +· · ·+
ab +
b
0
1
2
n−1
n
1
= an + nan−1 b + an−2 b2 + · · · + nabn−1 + bn .
2
(a) Verify the binomial theorem for n = 1, 2 and 3.
(b) Show that
n
n
n+1
+
=
for k = 1, 2, . . . , n
k
k−1
k
(c) Prove the binomial theorem using mathematical induction and
part (b).
(d) Pascal’s Triangle is another method for finding the coefficients
of a binomial expansion.
1
1
1
1
2
3
4
1
3
6
1
4
1
..
.
Show using mathematical induction that Pascal’s Triangle gives
the same values for the coefficients of a binomial expansion as the
formula of the binomial theorem.
32
CHAPTER 3. THE REAL NUMBERS
3.2
The Set Q of Rational Numbers
We expand the set of natural numbers to include 0 and negatives of natural
numbers. This is the set of integers, which consists of numbers
Z = {0, 1, −1, 2, −2, 3, −3, . . . }
We then study the space Q of rational numbers which consist of all numbers of the form m/n where m, n ∈ Z and n 6= 0. The rational numbers
includes all terminating decimals such as
3.59 =
359
.
100
Consider a square with sides of length 1 and diagonal d.
d
1
1
Then by the Pythagorean Theorem,
12 + 12 = d2
so that d2 = 2. There must be a real number d that satisfies d2 = 2 because
we can look at the unit square and see that there is a length d for the diagonal.
We can also consider the graph of the function y = x2 − 2. We see that
this graph passes through the x-axis, and this must be at the points (±d, 0)
where d satisfies the equation x2 = 2.
3.2. THE SET Q OF RATIONAL NUMBERS
33
From these two examples we know that there are at least two numbers
that satisfy the equation
x2 = 2
Are these numbers rational numbers? The answer is no. Let’s try to prove
this.
Definition 2.1. A number is called an algebraic number if it satisfies a
polynomial equation
an xn + an−1 xn−1 + · · · + a1 x + a0 = 0
where the coefficients a0 , a1 , . . . , an are integers, an 6= 0 and n ≥ 1.
Rational numbers are always algebraic numbers. If r = m/n is a rational
number, m, n ∈ Z and n 6= 0, then it satisfies the equation nx − m =
√ √
0. Numbers defined in terms of , 3 , etc. or fractional exponents, and
ordinary algebraic operations on the rational numbers are invariably algebraic
numbers.
Example 4. Find a polynomial with integer coefficients such that
(2 + 51/3 )1/2
is a zero of the polynomial.
Solution.
Theorem 2.2 (Rational Zeros Theorem). Suppose that a0 , a1 , . . . , an are
integers and that r is a rational number satisfying the polynomial equation
an xn + an−1 xn−1 + · · · + a1 x + a + a0 = 0
where n ≥ 1 and a0 6= 0. Write p/q where p, q are integers having no common
factors and q 6= 0. Then q divides an and p divides a0 .
34
CHAPTER 3. THE REAL NUMBERS
Proof.
Example 5. Show that
Solution.
√
2 cannot represent a rational number.
3.2. THE SET Q OF RATIONAL NUMBERS
Example 6. Show that
√
35
17 cannot represent a rational number.
Solution.
Example 7. Show that a = (2+51/3 )1/2 cannot represent a rational number.
Solution.
36
CHAPTER 3. THE REAL NUMBERS
√
Example 8. Show that b = ((4 − 2 3)/7)1/2 cannot represent a rational
number.
Solution.
Exercises
1. Show that
√
3,
√
5,
√ √
√
7, 24, and 31 are not rational numbers.
2. Show that 21/3 , 51/7 , and (13)1/4 do not represent rational numbers.
√
3. Show that (2 + 2)1/2 does not represent a rational number.
√
4. Show that (5 − 3)1/3 does not represent a rational number.
√
5. Show that (3 + 2)2/3 does not represent a rational number.
3.3
The Set R of Real Numbers
The set of rational numbers, Q satisfies the following properties.
A1. a + (b + c) = (a + b) + c for all a, b, c. (Associate Law)
A2. a + b = b + a for all a, b. (Commutative Law)
3.3. THE SET R OF REAL NUMBERS
37
A3. a + 0 = a for all a.
A4. For each a, there is an element −a such that a + (−a) = 0.
M1. a(bc) = (ab)c for all a, b, c. (Associate Law)
M2. ab = ba for all a, b. (Communtative Law)
M3. a · 1 = a for all a.
M4. For each a 6= 0, there is an element a−1 such that aa−1 = 1.
DL. a(b + c) = ab + ac for all a, b, c. (Distributive Law)
A system that has more than one element and satisfies these nine properties is called a field.
The set Q also has an order structure ≤ satisfying
O1. Given a and b, either a ≤ b or b ≤ a.
O2. If a ≤ b and b ≤ a, then a = b.
O3. If a ≤ b and b ≤ c, then a ≤ c.
O4. If a ≤ b, then a + c ≤ b + c.
O5. If a ≤ b and 0 ≤ c, then ac ≤ bc.
The set of real numbers, R, includes all rational numbers, all algebraic
numbers, π, e, and more. Every real number will correspond to a point on
the number line, and every point on the number line corresponds to a real
number. Unlike Q, the set R will have no “gaps.” The set of real numbers
satisfies properties A1 through A4, M1 throught M4, and DL. The set R has
an order structure ≤ that satisfies properties O1 throught O5. Thus, like the
set of rational numbers, Q, the set of real numbers, R, is an ordered field.
The previous statements about R do not provided a precise definition of
R as a mathematical object. The set R can defined entirely in terms of the
set Q of rational numbes. However, this is a long and tedious task and will
not be shown here.
In the remainder of this section we will obtain some results for R that are
valid for any orederd field.
38
CHAPTER 3. THE REAL NUMBERS
Theorem 3.1. The following are consequences of the field properties:
(i) a + c = b + c implies a = b
(ii) a · 0 = 0 for all a
(iii) (−a)b = −ab for all a, b
(iv) (−a)(−b) = ab for all a, b
(v) ac = bc and c 6= 0 imply a = b
(vi) ab = 0 implies either a = 0 or b = 0 for all a, b, c ∈ R
Proof.
(i) a + c = b + c implies a + c + (−c) = b + c + (−c), which implies
a + [c + (−c)] = b + [c + (−c)]. This reduces to a + 0 = b + 0, which
implies that a = b.
(ii) a · 0 = a · (0 + 0) = a · 0 + a · 0. So 0 + a · 0 = a · 0 + a · 0. By (i) this
implies a · 0 = 0.
(iii) Since a + (−a) = 0, whe have ab + (−a)b = [a + (−a)] · b = 0 · b = 0 =
ab + (−(ab)). From (i) we obtain (−a)b = −ab.
(iv) is left as an exercise.
(v) is left as an exercies.
(vi) If ab = 0 and b 6= 0, then 0 = b−1 · 0 = 0 · b−1 = (ab) · b−1 = a(bb−1 ) =
a · 1 = a.
Theorem 3.2. The following are consequences fo the properties of an ordered
field.
(i) if a ≤ b, then −b ≤ −a
(ii) if a ≤ b, and c ≤ 0, then bc ≤ ac
(iii) if 0 ≤ a and 0 ≤ b, then 0 ≤ ab
3.3. THE SET R OF REAL NUMBERS
39
(iv) 0 ≤ a2 for all a
(v) 0 < 1
(vi) if 0 < a, then 0 < a−1
(vii) if 0 < a < b, then 0 < b−1 < a−1 for a, b, c ∈ R
Proof.
(i) Suppose a ≤ b, then a+[(−a)+(−b)] ≤ b+[(−a)+(−b)], which implies
[a + (−a)] + −b ≤ [b + (−b)] + (−a). Then 0 + (−b) ≤ 0 + (−a), which
implies −b ≤ −a.
(ii) If a ≤ b, and c ≤ 0, then 0 ≤ −c by (i). By O5, a(−c) ≤ b(−c), so
that −ac ≤ −bc. It follows from (i) that −(−bc) ≤ −(−ac), so that
bc ≤ ac.
(iii) If 0 ≤ a and 0 ≤ b, then by O5, 0 · b ≤ ab, which implies 0 ≤ ab.
(iv) For any a, either a ≥ 0 or a ≤ 0 by O1. If a ≥ 0, then a2 ≥ 0 by (iii).
If a ≤ 0, then −a ≥ 0 by (i), so that a(−a) ≤ 0(−a), which implies
−a2 ≤ 0. Then 0 ≤ a2 by (i).
(v) is left as an exercise.
(vi) Suppose that 0 < a, but a−1 < 0. Then −a−1 > 0 by (i) and 0 ≤
a(−a−1 ) by (iii). But this implies that 0 ≤ −1, which implies 1 ≤ 0 by
(i). This contradicts (v).
(vii) is left as an exercise.
40
CHAPTER 3. THE REAL NUMBERS
We define the notion of absolute value.
Definition 3.3. We define
|a| = a if a ≥ 0 and |a| = −a if a ≤ 0
|a| is called the absolute value of a.
We often think of the absolute value of a as the distance between 0 and
a, but in fact we will define the idea of “distance” in terms of the “absolute
value.”
Definition 3.4. For numbers a and b we define dist(a, b) = |a − b|. The
notation dist(a, b) represents the distance between a and b.
The basic properties of absolute value are given by the next theorem.
Theorem 3.5.
(i) |a| ≥ 0 for all a ∈ R.
(ii) |ab| = |a| · |b| for all a, b ∈ R.
(iii) |a + b| ≤ |a| + |b| for all a, b ∈ R.
Proof. (i) This follows from the definition. If a ≥ 0, then |a| = a ≥ 0. If
a ≤ 0, then |a| = −a ≥ 0.
(ii) There are four cases. If a ≥ 0 and b ≥ 0, then ab ≥ 0, and |ab| = ab =
|a| · |b|. If a ≤ 0 and b ≤ 0, then −a ≥ 0, −b ≥ 0 and (−a)(−b) ≥ 0 so
that |a| · |b| = (−a)(−b) = ab = |ab|. If a ≥ 0 and b ≤ 0, then −b ≥ 0
and a(−b) ≥ 0 so that |a| · |b| = a(−b) = −(ab) = |ab|. If a ≤ 0 and
b ≥ 0, then −a ≥ 0 and (−a)b ≥ 0 so that |a|·|b| = (−a)b = −ab = |ab|.
(iii) The inequalities −|a| ≤ a ≤ |a| are immediate, since either a = |a| or
a = −|a|. Similarly −|b| ≤ b ≤ |b|. Four applications of O5 yield
−|a| + (−|b|) ≤ −|a| + b ≤ a + b ≤ |a| + b ≤ |a| + |b|
so that
−(|a| + |b|) ≤ a + b ≤ |a| + |b|
3.3. THE SET R OF REAL NUMBERS
41
This tells us that a + b ≤ |a| + |b| and also that −(a + b) ≤ |a| + |b|.
Since |a + b| is either equal to either a + b or −(a + b), we conclude that
|a + b| ≤ |a| + |b|.
Corollary 3.6. dist(a, c) ≤ dist(a, b) + dist(b.c) for all a, b, c ∈ R.
Proof. We apply (iii) of Theorem 3.5 to a − b and b − c to obtain |(a − b) +
(b − c)| ≤ |a − b| + |b − c| or
dist(a, c) = |a − c| ≤ |a − b| + |b − c| = dist(a, b) + dist(b, c)
The inequality in Corollary 3.6 and (iii) of Theorem 3.5 are called the
Triangle Inequality.
|a + b| ≤ |a| + |b| for all a, b
Exercises
1. (a) Which of the properties A-A4, M1-M4, DL, O1-O5 fail for N?
(b) which of these properties fail for Z?
2. (a) Show that |b| ≤ a if and only if −a ≤ b ≤ a.
(b) Prove that ||a| − |b|| ≤ |a − b| for all a, b ∈ R.
3. (a) Prove that |a + b + c| ≤ |a| + |b| + |c| for all a, b, c ∈ R. Hint:
Apply the triangle inequality twice. Do not consider eight cases.
(b) Use induction to prove
|a1 + a2 + · · · + an | ≤ |a1 | + |a2 | + · · · + |an |
for n numbers a1 , a2 , . . . , an
4. (a) Show that |b| < a if and only if −a < b < a.
(b) Show that |a − b| < c if and only if b − c < a < b + c.
(c) Show that |a − b| ≤ c if and only if b − c ≤ a ≤ b + c.
5. Let a, b ∈ R. Show that if a ≤ b1 for every b1 > b, then a ≤ b.
42
CHAPTER 3. THE REAL NUMBERS
Chapter 4
Proofs
4.1
What Is a Proof ?
A proof is a convincing argument expressed in the language of mathematics
that a statement is true. In mathematics, a statment is a sentence that is
either true of false. Some examples follow:
1. Two parallel lines in a plane have the same slope.
2. 1 = 0.
3. The real number x > 0.
4. There is an anlge t such that cos(t) = t.
A proof should contain enough details to be convincing to the person(s) to
whome the proof is addressed. Your proofs should contain enough details to
be convincing to someone else at your own mathematical level (for example,
a classmate).
Given two statements A and B, each of which may be either true or
false, a fundamental problem of interest in mathematics is to show that the
following conditional statement – also called an implication– is true:
If A is true, then B is true.
Mathematicians have developed a symbolic shorthand notation and would
write “A ⇒ B” instead of “A imples B.” When working with the implication “A implies B, ” it is important to realize that there are three separate
43
44
CHAPTER 4. PROOFS
statements: the statement A which is called the hypothesis, the statement
B which is called the conclusion, and the statement “A imples B.”
The conditions under which “A imples B” are true depend on whether A
and B themselves are true. There are four possble cases to consider:
1. A is true and B is true.
2. A is true and B is false.
3. A is false and B is true.
4. A is false and B is false.
Suppose, for example, that your friend made the statement,
“If you study hard, then you will get a good grade.”
To determine when a statement “A imples B” is false, ask yourself in which
conditions of the four foregoing cases you would be willing to call your friend
a liar. Below is Table 1.1. It is an example of a truth table.
Table 1.1 The Truth Table for “A implies B.”
A
True
True
False
False
B
A imples B
True True
False False
True True
False True
The first step in doing a proof is to identify the hypothesis A and the
conclusion B.
Example 1. The sum of the first n positive integers is n(n + 1)/2.
Hypothesis: n is a positive integer.
Conclusion: The sum of the first n positive integers is n(n + 1)/2.
Example 2. The quadratic equation ax3 + bx + c = 0 has two real roots
provided b2 − 4ac > 0, where a 6= 0, b, and c are given real numbers.
Hypothesis: a 6= 0, b, and c are real numbers and b2 − 4ac > 0.
4.1. WHAT IS A PROOF?
45
Conclusion: The quadratic equation ax3 + bx + c = 0 has two real roots.
Example 3. Two lines tangent to to the endpoints of the diameter of a circle
are parallel.
Hypothesis: L1 and L2 are two lines that are tangent to the endpoints
of the diameter of a circle.
Conclusion: L1 and L2 are parallel.
Example 4. There is a real number x such that x = 2−x .
Hypothesis: None, othere than your previous knowledge of mathematics.
Conclusion: There is a real number x such that x = 2−x .
Excercises
1.1 Which of the following are mathematical statements?
(a)
(b)
ax2 + bx + c = 0
√
(−b + b2 − 4ac)/(2a)
(c) Triangle XY Z is similar to triangle RST
(d)
3 + n + n2
(e) For every angle t, sin2 t + cos2 t = 1
1.3 For each of the following problems, identify the hypothesis (what you
can assume is true) and the conlusion (what you are trying to show is
true).
(a) If a right triangle XY Z with sides of lengths x and y and hypotenuse of length z has an area of z 2 /4, then the triangle XY Z
is isosceles.
(b) n2 is an even integer provided that n is an even integer.
(c) Let a, b, c, d, e and f be real numbers. You can solve the two
linear equations ax + by = e and cx + dy = f for x and y when
ad − bc 6= 0.
46
CHAPTER 4. PROOFS
1.5 For each of the following problems, identify the hypothesis (what you
can assume true) and the conclusion (what you are trying to show is
true).
(a) Suppose that A and B are sets of real numbers with A ⊆ B. For
any set C of real numbers, it follows that A ∩ C ⊆ B ∩ C.
(b) For a positive integer n, define the following function:
n/2
if n is even.
f (n) =
3n + 1 if n is odd
Then for any positive integer n, there is an integer k > 0 such
that f k (n) = 1, where f k−1 (f (n)), and f 1 (n) = f (n).
(c) When x is a real numbrer, the minimum value of x(x−1) ≥ −1/4.
1.6 “If I do not get my car fixed, I will miss my job interview,” says Jack.
Later, yo ucome to know that Jack’s car was repaired but that he
missed his job interview. Was Jack’s statement true or false? Explain.,
1.9 Determine whether the conditions on the hypothesis A and conclusion
B under which the following statements are true and false and give
your response.
(a) If 2 > 7, then 1 > 3.
(b) If x = 3, then 1 < 2.
1.11 If you are trying to prove that “A implies B” is true and you know
that B is false, do you want to show that A is true or false? Explain.
1.13 Using Table 1.1, prepare a true table for “A imples (B implies C).”
1.14 Using Table 1.1, prepare a true table for “(A imples B) implies C).”
1.15 Using Table 1.1, prepare a true table for “B ⇒ A.” Is this statement
true under the same conditions for which “A ⇒ B” is true?
1.16 Suppose that you want to show that A ⇒ B is false. According to Table
1.1, how shoudl you do this? What should you try to show about the
truth of A and B? (Doing this is referred to as a counterexample to
A ⇒ B.
4.1. WHAT IS A PROOF?
47
1.17 Apply your answer to Exercise 1.16 to show that each of the following
statements is false by constructing a counterexample.
(a) If x > 0, then log10 (x) > 0.
(b) If n is a positive integer, then n2 ≥ n! (where n! = n(n − 1) · · · 1).
1.18 Apply your answer to Exercise 1.16 to show that each of the following
statements is false by constructing a counterexample.
(a) If n is a positive integer, then 3n ≥ n!.
(b) If x is a positive real number between 0 and 1, then the first three
decimal digits of x are not equal to the first three decimal digits
of 2−x .
48
CHAPTER 4. PROOFS
Index
absolute value, 40
algebraic number, 33
field, 37
finite set, 7
base, 9
basis for induction, 28
binomial theorem, 31
boundary, 12
hypothesis, 44
cardinality, 7
clopen, 10
closed, 10
closure, 12
codomain, 21
cofinite topology, 16
complement, 6
conclusion, 44
conditional statement, 43
converge, 22
counterexample, 46
De Morgan’s Laws, 23
derived set, 19
descrete topology, 9
discrete topology, 8
distance, 40
domain, 21
doubleton, 7
elements, 5
empty set, 6
exterior, 12
image, 20, 21
implication, 43
indescrete topology, 8, 9
index set, 15
induction step, 28
injective, 21
integers, 32
interior, 11
interior point, 11
intersection, 6
inverse image, 21
limit point, 18
mathematical induction, 27
natural numbers, 27
neighborhood, 17
neighborhood system, 17
null set, 6
one-to-one, 21
onto, 21
open neighborhood, 17
open sets, 8, 15
Pascal’s Triangle, 31
Peano Axioms, 27
49
50
positive integers, 27
proof, 43
proximate point, 12
proximity, 12
pullback, 21
range, 21
rational numbers, 32
real numbers, 37
relative complement, 6
sequence, 22
set, 5
set difference, 6
singleton, 7
space, 5
statement, 43
subbase, 10
subset, 5
superset, 5
surjective, 21
topological space, 8, 15
topology, 7, 15
triangle inequality, 41
truth table, 44
union, 6
universal set, 5
universe of discourse, 5
upper limit topology, 16
usual topology, 16
INDEX