Bases for Topologies

Section Bases for Topologies

It can be difficult to completely describe the open sets in a topology. Instead, we can describe the topology using a collection of sets that generate the topology. For example, if

(X, d)

is a metric space then the collection of open sets in

X

forms a topology on

X,

called the metric topology. We also saw that in a metric space, every open set in

X

is a union of open balls. For that reason we called the collection of open balls a basis for the open sets in

X .

We can do the same thing in any topological space. As a non-trivial example, an interesting topology defined on the positive integers is due to S.W. Golomb. One can use this topology to prove that there are infinitely many primes. This topology also makes the positive integers into a connected Hausdorff space (more on these concepts later). The Golomb topology is defined as follows. If

a

and

b

are coprime integers in

Z^{+}

(that is,

a

and

b

have no common positive factors other than 1 so that the greatest common divisor of

a

and

b

1

), let

B_{a, b} = {a + b n ∣ n \geq 0} .

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The collection of sets

B_{a, b}

is a basis for the Golomb topology, and the topological space

(Z^{+}, τ)

is called the Golomb space. It is an exercise in number theory to prove that the sets

B_{a, b}

form a basis for a topology, so we will not go into the details.

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Activity 12.3.

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Let

X = {a, b, c, d}

and let

τ = {\emptyset, {a}, {b}, {a, b}, {c, d}, {a, c, d}, {b, c, d}, X) .

You may assume that

τ

is a topology on

X .

Explain why any nonempty open set in the topological space

(X, τ)

can be written in terms of arbitrary unions and finite intersections of

{a},

{b},

and

{c, d} .

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Activity 12.3 shows that, just like the open balls in a metric space, a topology can have a collection of subsets whose unions make up all of the open sets in the topology. We do need to take a little care, though. A basis will generate the collection of open sets for a topology, so the basis sets we start with should themselves be open sets. In addition, every element in the topological space should be an element of one of the basis sets, and since the basis elements are to produce all of the open sets in the topology, every set in the topology (except the empty set) should be a union of sets in a basis. It also must be the case that we can ensure that any finite intersection of sets in the topology remains a set in the topology when we write the sets in the topology in terms of the sets in a basis. To make the last two conditions happen, we will see that it is enough to insist that for any point in the intersection of basis elements, there is another basis element in that intersection that contains the point. This is summarized in Theorem 12.2.

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Theorem 12.2.

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Let

X

be a set and let

B

be a collection of subsets of

X

such that

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For each $x \in X,$ there is a set in $B$ that contains $x .$
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If $x \in X$ is an element of $B_{1} \cap B_{2}$ for some $B_{1}, B_{2} \in B,$ then there is a set $B_{3} \in B$ such that $x \in B_{3} \subseteq B_{1} \cap B_{2} .$

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Then the set

τ

that consists of the empty set and unions of elements of

B

is a topology on

X .

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Before we prove Theorem 12.2, we will need to know one fact about the set

B .

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Activity 12.4.

🔗

Let

X

be a set and

B

a collection of subsets of

X

such that

🔗

(a)

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For each

x \in X,

there is a set in

B

that contains

x .

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(b)

🔗

x \in X

is an element of

B_{1} \cap B_{2}

for some

B_{1}, B_{2} \in B,

then there is a set

B_{3} \in B

such that

x \in B_{3} \subseteq B_{1} \cap B_{2} .

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Let

B_{1},

B_{2},

\dots,

B_{n}

be in

B .

Our goal in this activity is to extend property 2 and show that if

x \in ⋂_{1 \leq k \leq n} B_{k},

then there is a set

B \in B

such that

x \in B

and

B \subseteq ⋂_{1 \leq k \leq n} B_{k} .

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(i)

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Since the statement we want to prove depends on a positive integer

n,

we will use mathematical induction. Explain why the

n = 1

and

n = 2

cases are true.

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(ii)

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What is the inductive hypothesis and what do we want to prove in the inductive step?

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(iii)

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Use the inductive hypothesis and condition 2 to complete the proof of the following lemma.

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Now we can prove Theorem 12.2.

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Proof.

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Let

X

be a topological space, and let

B

and

τ

satisfy the given conditions. By definition,

\emptyset \in τ .

For each

x \in X

there is a set

B_{x} \in B

such that

x \in B_{x} .

Then

X = ⋃_{x \in X} B_{x},

and

X \in τ .

To complete our proof that

τ

is a topology on

X,

we need to demonstrate that

τ

is closed under arbitrary unions and finite intersections. We first consider unions. Let

{U_{α}}

be a collection of sets in

τ

for

α

in some indexing set

I .

By definition, each

U_{α}

is empty or is a union of elements of

B .

So either

U = ⋃_{α \in I} U_{α}

is empty, or is a union of sets in

B .

Thus,

U \in τ

and

τ

is closed under arbitrary unions.

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Now we show that

τ

is closed under finite intersections. Let

n

be a positive integer and let

{U_{k}}

a collection of sets in

τ

for

1 \leq k \leq n .

Let

U = U_{1} \cap U_{2} \cap \dots \cap U_{n} .

U_{k} = \emptyset

for any

k,

then

U = \emptyset

is in

τ .

So suppose that

U_{k} \neq \emptyset

for each

k

between

1

and

n .

Let

x \in U .

Then

x \in U_{k}

for each

k .

For every

m

between

1

and

n,

the fact that

U_{m}

is a union of elements in

B

implies that there exists

B_{m} \subseteq U_{m}

with

x \in B_{m} .

Thus,

x \in ⋂_{1 \leq m \leq n} B_{m} .

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Lemma 12.3 shows that there is a set

B_{x} \in B

such that

x \in B_{x}

and

B_{x} \subseteq ⋂_{1 \leq m \leq n} B_{m} \subseteq ⋂_{1 \leq m \leq n} U_{m} = U .

Since

x

is an arbitrary element of

U,

we must have

U \subseteq ⋃_{x \in U} B_{x} .

But each

B_{x}

is subset of

U,

⋃_{x \in U} B_{x} \subseteq U .

It follows that

U = ⋃_{x \in U} B_{x}

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and

U \in τ .

Therefore,

τ

is a topology on

X .

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Any collection

B

of sets as given in Theorem 12.2 is given a special name.

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Definition 12.4.

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Let

X

be a set. A set

B

is a basis for a topology (or just a basis) on

X

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For each $x \in X,$ there is a set in $B$ that contains $x .$
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If $x \in X$ is an element of $B_{1} \cap B_{2}$ for some $B_{1}, B_{2} \in B,$ then there is a set $B_{3} \in B$ such that $x \in B_{3} \subseteq B_{1} \cap B_{2} .$

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The elements of a basis

B

are called basis elements or the basic open sets. A basis for a topology on a set

X

defines a topology on

X

as shown in Theorem 12.2.

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Note that because of property (1) of Definition 12.4, the union of the sets in the basis must contain

X .

In other words, the sets in a basis cover the space. The second property ensures that if a point is in the intersection of two basic open sets, then there is a smaller basic open set that contains