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Lecture note /topology / lecturer :Zahir Dobeas AL -nafie
The Euclidean Topology
Introduction
In a movie or a novel there are usually a few central characters about whom the plot revolves.
In the story of topology, the euclidean topology on the set of real numbers is one of the central
characters. Indeed it is such a rich example that we shall frequently return to it for inspiration
and further examination.
Let R denote the set of all real numbers. In Chapter 1 we defined three topologies that can
be put on any set: the discrete topology, the indiscrete topology and the finite-closed topology.
So we know three topologies that can be put on the set R. Six other topologies on R
were
defined in Exercises 1.1 #5 and #9. In this chapter we describe a much more important and
interesting topology on R which is known as the euclidean topology.
An analysis of the euclidean topology leads us to the notion of "basis for a topologyÔ. In
the study of Linear Algebra we learn that every vector space has a basis and every vector is a
linear combination of members of the basis. Similarly, in a topological space every open set can
be expressed as a union of members of the basis. Indeed, a set is open if and only if it is a union
of members of the basis.
1
2
2.1
The Euclidean Topology on R
Definition.
A subset S of R is said to be open in the euclidean topology on R if
it has the following property:
()
Notation.
For each x  S, there exist a,b in R, with a < b, such that x  (a,b)  S.
Whenever we refer to the topological space R without specifying the topology, we
mean R with the euclidean topology.
Remarks.
(i)
The "euclidean topologyÔ

is a topology.
Proof.
We are required to show that

satisfies conditions (i), (ii), and (iii) of Definitions
.
We are given that a set is in
Firstly, we show that R 
.

if and only if it has property .
Let x  R. If we put a = x1 and b = x+1, then x  (a,b)  R;
that is, R has property  and so R 
.
Secondly, Ø 

as Ø has property  by default.
Now let {A j : j  J}, for some index set J, be a family of members of
S
show that
jJA j

;
that is, we have to show that
Then x  A k, for some k  J.
x  (a,b)  A k. As k  J, A k 
 and thus is in
,
jJ
.
Then we have to
A j has property . LetjJxA.j
as required.
Then y  A 1. As A 1 
,

. We have to prove that A 1  A 2 
.
So let y  A 1  A 2.
there exist a and b in R with a < b such that y  (a,b)  A 1. Also
 . So there exist c and d in R with c < d such that y  (c,d)  A 2. Let e be the greater
of a and c, and f the smaller of b and d. It is easily checked that e < y < f, and so y  (e,f). As
(e,f)  (a,b)  A 1 and (e,f)  (c,d)  A 2, we deduce that y  (e,f)  A 1  A 2. Hence A 1  A2
has property  and so is in
Thus

S
As A k   , there exist a and b in R with a < b such that
S jJ A j and so x  (a,b) 
S jJ A .jH ence S jJ A j has property
Finally, let A 1 and A 2 be in
y  A2 
S
.
is indeed a topology on R.
3
We now proceed to describe the open sets and the closed sets in the euclidean topology on
R . In particular, we shall see that all open intervals are indeed open sets in this topology and all
closed intervals are closed sets.
(ii)
Let r,s  R with r < s. In the euclidean topology  on R, the open interval (r,s) does
indeed belong to

and so is an open set.
Proof.
We are given the open interval (r,s).
We must show that (r,s) is open in the euclidean topology; that is, we have to show
that (r,s) has property () of Definition 2.1.1.
So we shall begin by letting x  (r,s). We want to find a and b in R with a < b such
that x  (a,b)  (r,s).
Let x  (r,s). Choose a = r and b = s. Then clearly
x  (a,b)  (r,s).
So (r,s) is an open set in the euclidean topology.
(iii)
The open intervals (r, ) and (,r) are open sets in R, for every real number r.
Proof.
Firstly, we shall show that (r, ) is an open set; that is, that it has property ().
To show this we let x  (r, ) and seek a,b  R such that
x  (a,b)  (r, ).
Let x  (r, ). Put a = r and b = x + 1. Then x  (a,b)  (r, ) and so (r, ) 
A similar argument shows that (,r) is an open set in R.
.
4
(iv)
It is important to note that while every open interval is an open set in R, the converse
is false. Not all open sets in R are intervals. For example, the set (1, 3)  (5, 6) is an open set in
S
R , but it is not an open interval. Even the set
(2n,
2n + 1) is an open set in R.
n=1
(v)
For each c and d in R with c < d, the closed interval [c,d] is not an open set in R.
Proof.
We have to show that [c,d] does not have property ().
To do this it sufces to find any one x such that there is no a,b having property ().
Obviously c and d are very special points in the interval [c,d]. So we shall choose
x = c and show that no a,b with the required property exist.
We use the method of proof called proof by contradiction. We suppose that a and
b exist with the required property and show that this leads to a contradiction, that is
something which is false. Consequently the supposition is false! Hence no such a and
b exist. Thus [c,d] does not have property () and so is not an open set.
Observe that c  [c,d]. Suppose there exist a and b in R with a < b such that c  (a,b)  [c,d].
Then c  (a,b) implies a < c < b and so a <
c +a
2
< c < b. Thus
c +a
2
 (a,b) and
c +a
2
/ [c,d].
Hencee舀
(a,b) 6 [c,d], which is a contradiction. So there do not exist a and b such that c  (a,b)  [c,d].
Hence [c,d] does not have property () and so [c,d] /

.