Introduction to GENERAL TOPOLOGY (Finite and Closed Set)

 

1.3 The Finite-Closed Topology

It is usual to define a topology on a set by stating which sets are open. However,

sometimes it is more natural to describe the topology by saying which sets are closed.

The next definition provides one such example.

1.3.1 Definition.

Let X be any non-empty set. A topology τ on X is

called the finite-closed topology or the cofinite topology if the closed subsets

of X are X and all finite subsets of X; that is, the open sets are Ø and all subsets

of X which have finite complements.

Once again it is necessary to check that τ in Definition 1.3.1 is indeed a

topology; that is, that it satisfies each of the conditions of Definitions 1.1.1.

Note that Definition 1.3.1 does not say that every topology which has X and

the finite subsets of X closed is the finite-closed topology. These must be the only

closed sets. [Of course, in the discrete topology on any set X, the set X and all

finite subsets of X are indeed closed, but so too are all other subsets of X.]

In the finite-closed topology all finite sets are closed. However, the following

example shows that infinite subsets need not be open sets.

1.3.2 Example.

If N is the set of all positive integers, then sets such as {1},

{5, 6, 7}, {2, 4, 6, 8} are finite and hence closed in the finite-closed topology. Thus

their complements

{2, 3, 4, 5, . . .}, {1, 2, 3, 4, 8, 9, 10, . . .}, {1, 3, 5, 7, 9, 10, 11, . . .}

are open sets in the finite-closed topology. On the other hand, the set of even

positive integers is not a closed set since it is not finite and hence its complement,

the set of odd positive integers, is not an open set in the finite-closed topology.

So while all finite sets are closed, not all infinite sets are open.

1.3.3 Example.

Let τ be the finite-closed topology on a set X. If X has at

least 3 distinct clopen subsets, prove that X is a finite set.

Proof.

We are given that τ is the finite-closed topology, and that there are at least

3 distinct clopen subsets.

We are required to prove that X is a finite set.

Recall that τ is the finite-closed topology means that the family of all

closed sets consists of X and all finite subsets of X. Recall also that a set

is clopen if and only if it is both closed and open.

Remember that in every topological space there are at least 2 clopen

sets, namely X and Ø. (See the comment immediately following Definition

1.2.6.) But we are told that in the space (X, τ) there are at least 3 clopen

subsets. This implies that there is a clopen subset other than Ø and X. So

we shall have a careful look at this other clopen set!

As our space (X, τ) has 3 distinct clopen subsets, we know that there is a clopen

subset S of X such that S 6= X and S 6= Ø. As S is open in (X, τ), Definition 1.2.4

implies that its complement X \ S is a closed set.

Thus S and X \ S are closed in the finite-closed topology τ. Therefore S and

X \ S are both finite, since neither equals X. But X = S ∪ (X \ S) and so X is the

union of two finite sets. Thus X is a finite set, as required.

We now know three distinct topologies we can put on any infinite set – and

there are many more. The three we know are the discrete topology, the indiscrete

topology, and the finite-closed topology. So we must be careful always to specify

the topology on a set.

For example, the set {n : n > 10} is open in the finite-closed topology on the

set of natural numbers, but is not open in the indiscrete topology. The set of odd

natural numbers is open in the discrete topology on the set of natural numbers, but

is not open in the finite-closed topology

We shall now record some definitions which you have probably met before.

1.3.4 Definitions.

Let f be a function from a set X into a set Y .

(i) The function f is said to be one-to-one or injective if f(x1) = f(x2)

implies x1 = x2, for x1, x2 ∈ X;

(ii) The function f is said to be onto or surjective if for each y ∈ Y there

exists an x ∈ X such that f(x) = y;

(iii) The function f is said to be bijective if it is both one-to-one and onto.

1.3.5 Definitions.

Let f be a function from a set X into a set Y . The

function f is said to have an inverse if there exists a function g of Y into X

such that g(f(x)) = x, for all x ∈ X and f(g(y)) = y, for all y ∈ Y . The

function g is called an inverse function of f.

The proof of the following proposition is left as an exercise for you.

1.3.6 Proposition.

Let f be a function from a set X into a set Y .

(i) The function f has an inverse if and only if f is bijective.

(ii) Let g1 and g2 be functions from Y into X. If g1 and g2 are both inverse

functions of f, then g1 = g2; that is, g1(y) = g2(y), for all y ∈ Y .

(iii) Let g be a function from Y into X. Then g is an inverse function of f if

and only if f is an inverse function of g.

Warning.

It is a very common error for students to think that a function is oneto-one if “it maps one point to one point”.

All functions map one point to one point. Indeed this is part of the definition of a function.

A one-to-one function is a function that maps different points to different points.

We now turn to a very important notion that you may not have met before.

1.3.7 Definition.

Let f be a function from a set X into a set Y . If S is

any subset of Y , then the set f−1(S) is defined by

f−1(S) = {x : x ∈ X and f(x) ∈ S}.

The subset f−1(S) of X is said to be the inverse image of S.

Note that an inverse function of f : X → Y exists if and only if f is bijective.

But the inverse image of any subset of Y exists even if f is neither one-to-one nor

onto. The next example demonstrates this.

1.3.8 Example.

Let f be the function from the set of integers, Z, into itself

given by f(z) = |z|, for each z ∈ Z.

The function f is not one-to one, since f(1) = f(−1).

It is also not onto, since there is no z ∈ Z, such that f(z) = −1. So f is

certainly not bijective. Hence, by Proposition 1.3.6 (i), f does not have an inverse

function. However inverse images certainly exist. For example,

f−1({1, 2, 3}) = {−1, −2, −3, 1, 2, 3}

f−1({−5, 3, 5, 7, 9}) = {−3, −5, −7, −9, 3, 5, 7, 9}.

We conclude this section with an interesting example.

1.3.9 Example.

Let (Y, τ) be a topological space and X a non-empty set.

Further, let f be a function from X into Y . Put τ 1 = {f−1(S) : S ∈ τ }. Prove

that τ 1 is a topology on X.

Proof.

Our task is to show that the collection of sets, τ 1, is a topology on X;

that is, we have to show that τ 1 satisfies conditions (i), (ii) and (iii) of

Definitions 1.1.1

X ∈ τ 1 since X = f−1(Y ) and Y ∈ τ.

Ø ∈ τ 1 since Ø = f−1

(Ø) and Ø ∈ τ.

Therefore τ 1 has property (i) of Definitions 1.1.1.

To verify condition (ii) of Definitions 1.1.1, let {Aj : j ∈ J} be a collection

of members of τ 1 , for some index set J. We have to show that Sj∈J Aj ∈ τ 1.

As Aj ∈ τ 1, the definition of τ 1 implies that Aj = f−1(Bj), where Bj ∈ τ.

Also Sj∈J Aj = Sj∈J f−1(Bj

) = f−1 Sj∈J Bj . [See Exercises 1.3 # 1.]

Now Bj ∈ τ, for all j ∈ J, and so Sj∈J Bj ∈ τ, since τ is a topology on Y .

Therefore, by the definition of τ 1, f−1 Sj∈J Bj ∈ τ 1; that is, Sj∈J Aj ∈ τ 1.

So τ 1 has property (ii) of Definitions 1.1.1.

[Warning.

You are reminded that not all sets are countable. (See the Appendix

for comments on countable sets.) So it would not suffice, in the above argument,

to assume that sets A1, A2. . . . , An, . . . are in τ 1 and show that their union

A1 ∪ A2 ∪ . . . ∪ An ∪ . . . is in τ 1. This would prove only that the union of a

countable number of sets in τ 1 lies in τ 1, but would not show that τ 1 has property

(ii) of Definitions 1.1.1– this property requires all unions, whether countable or

uncountable, of sets in τ 1 to be in τ 1.]

Finally, let A1 and A2 be in τ 1. We have to show that A1 ∩ A2 ∈ τ 1.

As A1, A2 ∈ τ 1, A1 = f−1(B1) and A2 = f−1(B2), where B1, B2 ∈ τ.

A1 ∩ A2 = f−1(B1) ∩ f−1(B2) = f−1(B1 ∩ B2).

As B1 ∩ B2 ∈ τ, we have f−1(B1 ∩ B2) ∈ τ 1. Hence A1 ∩ A2 ∈ τ 1, and we

have shown that τ 1 also has property (iii) of Definitions 1.1.1.

So τ 1 is indeed a topology on X.

Exercises 1.3

1. Let f be a function from a set X into a set Y . Then we stated in Example 1.3.9 that

f−1 [j∈J

Bj
= [j∈J

f−1(Bj) (1)

and

f−1(B1 ∩ B2) = f−1(B1) ∩ f−1(B2) (2)

for any subsets Bj of Y , and any index set J.

(a) Prove that (1) is true.

[Hint. Start your proof by letting x be any element of the set on the lefthand side and show that it is in the set on the right-hand side. Then do the

reverse.]

(b) Prove that (2) is true.

(c) Find (concrete) sets A1, A2, X, and Y and a function f : X → Y such that

f(A1 ∩ A2) 6= f(A1) ∩ f(A2), where A1 ⊆ X and A2 ⊆ X.

2. Is the topology τ described in Exercises 1.1 #6 (ii) the finite-closed topology?

(Justify your answer.) T1-spaces

3. A topological space (X, τ) is said to be a T1-space if every singleton set {x}

is closed in (X, τ). Show that precisely two of the following nine topological

spaces are T1-spaces. (Justify your answer.)

(i) a discrete space;

(ii) an indiscrete space with at least two points;

(iii) an infinite set with the finite-closed topology;

(iv) Example 1.1.2;

(v) Exercises 1.1 #5 (i);

(vi) Exercises 1.1 #5 (ii);

(vii) Exercises 1.1 #5 (iii);

(viii) Exercises 1.1 #6 (i);

(ix) Exercises 1.1 #6 (ii).

4. Let τ be the finite-closed topology on a set X. If τ is also the discrete topology,

prove that the set X is finite.

T0-spaces and the Sierpinski Space

5. A topological space (X, τ) is said to be a T0-space if for each pair of distinct

points a, b in X, either2 there exists an open set containing a and not b, or there

exists an open set containing b and not a.

(i) Prove that every T1-space is a T0-space.

(ii) Which of (i)–(vi) in Exercise 3 above are T0-spaces? (Justify your answer.)

(iii) Put a topology τ on the set X = {0, 1} so that (X, τ) will be a T0-space but

not a T1-space. [The topological space you obtain is called the Sierpiński space.]

(iv) Prove that each of the topological spaces described in Exercises 1.1 #6 is

a T0-space. (Observe that in Exercise 3 above we saw that neither is a T1-space.)

Countable-Closed Topology

6. Let X be any infinite set. The countable-closed topology is defined to be the

topology having as its closed sets X and all countable subsets of X. Prove that

this is indeed a topology on X.

Intersection of Two Topologies

7. Let τ 1 and τ 2 be two topologies on a set X. Prove each of the following

statements.

(i) If τ 3 is defined by τ 3 = τ 1 ∪ τ 2, then τ 3 is not necessarily a topology on

X. (Justify your answer, by finding a concrete example.)

(ii) If τ 4 is defined by τ 4 = τ 1∩τ 2, then τ 4 is a topology on X. (The topology

τ 4 is said to be the intersection of the topologies τ 1 and τ 2.)

(iii) If (X, τ 1) and (X, τ 2) are T1-spaces, then (X, τ 4) is also a T1-space.

(iv) If (X, τ 1) and (X, τ 2) are T0-spaces, then (X, τ 4) is not necessarily a T0-

space. (Justify your answer by finding a concrete example.)

(v) If τ 1, τ 2, . . . , τ n are topologies on a set X, then τ =

n

T

i

=1

τi

is a topology

on X.

(vi) If for each i ∈ I, for some index set I, each τi

is a topology on the set X,

then τ = T

i

∈I

τi

is a topology on X.

Distinct T0-Topologies on a Finite Set

8. In Wikipedia in Exercises 1.2.7, it says that the number of topologies on a finite set with

n ∈ N points can be quite large, even for small n. This is also true even for T0-spaces as defined in Exercises 1.3 #5. Indeed that same Wikipedia source

says, that if n = 3, there are 19 distinct T0-spaces; for n = 4, there are 219 distinct T0-spaces; for n = 5, there are 4231 distinct T0-spaces. Prove, using

mathematical induction, that as n increases, the number of T0-spaces increases.

[Hint. It suffices to show that if there are M T0-spaces with n points, then there are at least M + 1 T0-spaces with n + 1 points.]

9. A topological space (X, τ) is said to be a door space if every subset of X is either an open set or a closed set (or both).

(i) Is a discrete space a door space?

(ii) Is an indiscrete space a door space?

(iii) If X is an infinite set and τ is the finite-closed topology, is (X, τ) a door space?

(iv) Let X be the set {a, b, c, d}. identify those topologies τ on X which make it into a door space?

Saturated Sets

10. A subset S of a topological space (X, τ) is said to be saturated if it is an intersection of open sets in (X, τ).

(i) Verify that every open set is a saturated set.

(ii) Verify that in a T1-space every set is a saturated set.

(iii) Give an example of a topological space which has at least one subset which is not saturated.

(iv) Is it true that if the topological space (X, τ) is such that every subset is saturated, then (X, τ) is a T1-space?

Compiled by Allen Oluwaseun (Ohm's-Jacobian)