basis

Basis for a Set

a basis for a set $X$ is a collection $ℬ$ of subsets of $X$ called basis elements such that

for each $x \in X$ , there is at least one basis element $B$ containing $x$
if $x$ belongs in the intersection of two basis elements $B_{1}$ and $B_{2}$ , then there is a basis element $B_{3}$ containing $x$ such that $B_{3} \subseteq B_{1} \cap B_{2}$

Intervals With Rational Endpoints Are a Basis for the Real Numbers

The set

J : = {(a, b) : a < b \in Q}

is a basis for

R

Let $x \in R$ , and consider the points $x - 1, x, x + 1$ , so that we obain some $j, k \in Q$ such that $x - 1 < j < x < k < x + 1$ so $x \in (j, k) \in J$

Now suppose that $x \in J_{1} \cap J_{2}$ for some $J_{1}, J_{2} \in J$ since they are both intervals then so is their intersection, therefore take $J_{3} = J_{1} \cap J_{2}$ , to see why this works, if $J_{1} = (a, b), J_{2} = (c, d)$ then the possibilities for $J_{1} \cap J_{2}$ are $(a, b), (c, d), (c, b), (a, d)$ , in any such case we obtain an interval with rational end points, and thus this shows $x \in J_{3} \subseteq J_{1} \cap J_{2}$ and that $J_{3} \in J$

Half Open Intervals With Rational Endpoints Are a Basis for the Real Numbers

The set

{[a, b) : a < b \in Q}

is a basis for

R

For any

x \in R

then

x \in [x, x + 1)

, additionally given some

x \in [a, b) \cap [c, d)

then since their intersection is non-empty we deduce that

[a, b) \cap [c, d) = [i, j)

where

(i, j) \in {(a, b), (c, d), (a, d), (c, b)}

and therefore this will be our basis element contained in the intersection containing

x

collection generated by a basis

Given a basis

ℬ

we define the set

G_{ℬ}

U \in G_{ℬ}

iff for each

x \in U

there is a

B \in ℬ

such that

x \in B

and

B \subseteq U

G_{ℬ}

is a topology

The collection

G_{ℬ}

is a topology

We will verify it's a topology from the definition

$\emptyset$ is in $G_{ℬ}$ because it vacuously satisfies the condition. $X$ is in $G_{ℬ}$ since given any point $x \in X$ , we have a $B \in ℬ$ such that $x \in B \subseteq X$ by the definition of a basis

Now suppose we have an arbitrary union of elements from $G_{ℬ}$ , say $S \subseteq G_{ℬ}, M = ⋃_{U \in S} U$ . We need to verify that $M \in G_{ℬ}$ , so let $x \in M$ , therefore $x \in \overset{―}{U}$ for some $\overset{―}{U} \in S$ and since $\overset{―}{U}$ was assumed to be in $G_{ℬ}$ we know that there is some $B \in ℬ$ such that $x \in B \subseteq \overset{―}{U}$ but also $\overset{―}{U} \subseteq M$ since $M$ includes $\overset{―}{U}$ in the union so then $x \in B \subseteq M$ meaning by the definition that $M \in G_{ℬ}$

Now if we are given a finite intersection of elements from $G_{ℬ}$ we will prove by induction that it is also an element of $G_{ℬ}$

base case

if $n = 1$ then the union of one element from $G_{ℬ}$ (namely just itself) is also an element of $G_{ℬ}$ trivially.

Before moving to the inductive step, notice that it also holds for $n = 2$ , because given $U_{1}, U_{2} \in G_{ℬ}$ , we can see that for any $x \in U_{1} \cap U_{2}$ , we know that we have a basis elements $B_{1}, B_{2} \in ℬ$ such that $x \in B_{1} \subseteq U_{1}$ and $x \in B_{2} \subseteq U_{2}$ and thus $x$ also belongs to the intersection of $B_{1}$ and $B_{2}$ , so by the definition of a basis, we get a $B_{3} \subseteq B_{1} \cap B_{2}$ that contains $x$ since $B_{1} \cap B_{2} \subseteq U_{1} \cap U_{2}$ , we have $x \in B_{3} \subseteq U_{1} \cap U_{2}$ , which means that $U_{1} \cap U_{2} \in G_{ℬ}$

inductive step

Let $k \in ℕ_{1}$ and assume that the intersection of $k$ elements from $G_{ℬ}$ is also in $G_{ℬ}$ now we'll show it's true for $k + 1$ . So consider the intersection $U_{1} \cap \dots \cap U_{k + 1}$ , then we can re-write it as $(U_{1} \cap \dots \cap U_{k}) \cap U_{k + 1}$ so by our inductive hypothesis $(U_{1} \cap \dots \cap U_{k}) \in G_{ℬ}$ and therefore this $k + 1$ union is a union of two elements already in $G_{ℬ}$ so by our $n = 2$ argument $U_{1} \cap \dots \cap U_{k + 1} \in G_{ℬ}$ , as needed.

topology generated by a basis

Since a collection generated by a basis turns out to be a topology we define

𝒯_{ℬ} : = G_{ℬ}

and call it the topology generated by

ℬ

basis for a topology

We say that

ℬ

is a basis for a topology

𝒯

when

𝒯_{ℬ}

= 𝒯

. In this case we may also say that

ℬ

generates

𝒯

basis elements are open in a generated topology

Suppose we have the topology

𝒯_{ℬ}

, then given any

B \in ℬ

B \in T_{𝒷}

, in other words

ℬ \subseteq T_{ℬ}

Considering a topology generated by a basis

ℬ

we can see that for any

B \in ℬ

, we can use

B

itself such that for each

x \in B

, we have

B

so that

B \subseteq B

, to show that

B \in 𝒯_{ℬ}

Topology Generated by a Basis Consists of Unions

𝒯_{ℬ}

equals the collection of all unions of elements of

ℬ

, that is:

T_{B} = {⋃_{B \in 𝒮} B : 𝒮 \subseteq ℬ}

The the set on the right hand side be denoted as $𝒰$ , we will start by showing that $U \subseteq T_{B}$ .

Suppose we have a subset of elements from $ℬ$ called $𝒮$ , then for each $B \in 𝒮$ it is open in $X$ because it is a basis element, and therefore by the definition of a topology, as $⋃_{B \subseteq 𝒮} B$ is an arbitrary union of open sets it is also open in $X$ , so we've shown that $𝒰 \subseteq 𝒯_{ℬ}$

We'll show the converse now, so suppose that $U \in 𝒯_{ℬ}$ , we want to prove that it's equal to an arbitrary union of elements of $ℬ$ .

Recall the definition of a topology generated by a basis so since $U \in 𝒯_{ℬ}$ then for each $x \in U$ we have some $B_{x} \in ℬ$ with $x \in B_{x} \subseteq U$ , therefore $U = ⋃_{x \in U} B_{x}$ , and as a union of of elements of $B$ thus $U$ is an element of $𝒰$ so $𝒯_{ℬ} \subseteq 𝒰$ .

basis criterion

Let

(X, 𝒯)

be a topological space. Suppose that

𝒞

is a collection of open sets of

X

such that for each open set

U

X

and each

x \in U

, there is an element

C

𝒞

such that

x \in C \subseteq U

. Then

𝒞

is a basis and

𝒯_{𝒞} = 𝒯

We'll first show that $𝒞$ is a basis, so let $x \in X$ , but $X$ is open and so we can use $X$ as the element of $𝒞$ satisfying $x \in X \subseteq X$ .

Now let $C_{1}, C_{2}$ be elements of $𝒞$ and now consider $C_{1} \cap C_{2}$ , since it's a finite intersection, then we know it's also open in $X$ from the definition of a topology.

Since $C_{1} \cap C_{2}$ is open, let $x \in C_{1} \cap C_{2}$ , then by the definition of $𝒞$ , there is some $C \in 𝒞$ such that $x \in C \subseteq C_{1} \cap C_{2}$ , so we can take $B_{3} = C$ in the definition of basis, to see that the second condition is satisfied, and thus we know that $𝒞$ is a basis

Now we want to show that $𝒯_{𝒞} = 𝒯$ . So suppose $U \in 𝒯_{𝒞}$ , then it can be written as a union as elements from $𝒞$ , but since $𝒞$ was defined to consist of open sets of $X$ , then this union is also open with respect to $X$ by the definition of a topology.

Now suppose that $U \in 𝒯$ and consider some $x \in U$ then by the definition of $𝒞$ , there is some $C \in 𝒞$ such that $x \in C \subseteq U$ , looking back at the definition for a topology generated by a basis, we can see this means $U \in 𝒯_{𝒞}$ , as needed.

basis finer equivalence

Let

ℬ

and

ℬ^{'}

be bases for the topologies

𝒯

and

𝒯^{'}

, respectively on

X

. Then the following are equivalent:

$𝒯^{'}$ is finer than $𝒯$
for each $x \in X$ and each basis element $B \in ℬ$ containing $x$ there is a basis element $B^{'} \in ℬ^{'}$ such that $x \in B^{'} \subseteq B$

topology equality with basis

Given two bases

ℬ

and

ℬ^{^{'}}

for the toplogies

𝒯

and

𝒯^{^{'}}

X

, then

𝒯 = 𝒯^{^{'}}

if and only if all of the following hold:

for each $x \in X$ and each basis element $B \in ℬ$ containing $x$ there is a basis element $B^{'} \in ℬ^{'}$ such that $x \in B^{'} \subseteq B$
for each $x \in X$ and each basis element $B^{^{'}} \in ℬ^{^{'}}$ containing $x$ there is a basis element $B \in ℬ$ such that $x \in B \subseteq B^{^{'}}$

The Real Intervals With Rational Endpoints Generate Generate the Standard Topology

The countable basis for

R

{(a, b) : a < b \in Q}

generates the standard topology on

R

We will prove that they are equal by comparing the basis elements, so let

x \in R

and suppose that

x \in (a, b)

with

a < b \in Q

, then

(a, b)

itself is an element of the standard basis for

R

. Also note that given any

x \in R

and some open interval such that:

x \in (c, d)

where

c < d \in R

then by considering the pairs

c, x

and

x, d

we obtain some

j, k \in Q

between those pairs then

x \in (j, k) \subseteq (a, b)

therefore the topologies they generate are the same.

The Topology Generated by the Half Open Intervals With Rational Endpoints Is Strictly Corser Lower Limit Topology

Let

B = {[a, b) : a < b \in Q}

, then

T_{B} \subset T_{ℓ}

We will show that $T_{B} \subset T_{ℓ}$ by showing $T_{B} \subseteq T_{ℓ}$ and $T_{B} \neq T_{ℓ}$ .

Let $x \in R$ and $[a, b) \in B$ such that $x \in [a, b)$ then the interval $[a, b)$ considered with endpoints in $R$ is a basis element of the lower limit topology and is contained within it trivially and so we deduce that $T_{B} \subseteq T_{R}$ .

Now we'll prove that $T_{B} \neq T_{ℓ}$ , to do this we will use the contrapositive of this corollary. So consider $x = \sqrt{2}$ , then $[\sqrt{2}, 2)$ is a basis element from the lower limit topology that contains $\sqrt{2}$ but there is no element from the rational lower limit basis that contains this element and is also contained within $[\sqrt{2}, 2)$ to see why suppose for the sake of contradiction there was one, of the form $[q, p)$ if it contains $\sqrt{2}$ then we must have that $q \leq \sqrt{2} < p$ but since we require that $[q, p) \subseteq [\sqrt{2}, 2)$ it implies that $\sqrt{2} \leq q$ so we must have that $q = \sqrt{2}$ but that's a contradiction since $q \in Q$

standard topology on

ℝ^{n}

Let

ℬ

be the collection of all sets of the form

{y \in ℝ^{n} : ‖ x - y ‖ < ϵ}

for any

x \in ℝ^{n}

and

ϵ \in ℝ^{> 0}

, then the topology generated by

ℬ

is called the standard topology on the

ℝ^{n}

. Whenever we consider

ℝ^{n}

, we shall suppose it is given this topology unless we specifically state otherwise

lower limit topology on R

If B is the collection of all half-open intervals of the form [a, b) = {x | a ≤ x \lt b}, where a \lt b, the topology generated by B is called the lower limit topology on R. When R is given the lower limit topology, we denote it by Rl

K topology

Finally let K denote the set of all numbers of the form 1/n, for n ∈ Z+, and let B be the collection of all open intervals (a, b), along with all sets of the form (a, b) − K. The topology generated by B will be called the K-topology on R. When R is given this topology, we denote it by RK

subbasis

A subbasis

S

for a topology on

X

is a collection of subsets of

X

whose union equals

X

. The topology generated by the subbasis

S

is defined to be the collection

𝒯

of all unions of finite intersections of elements of

S

intersections of subbasis elements forms a basis

Given a subbasis

S

, then the collection of finite intersections of elements of

S

is a basis, which we denote by

ℬ_{S}

topology generated by a subbasis

Given a subbasis

S

, we know that this induces a basis

ℬ_{S}

, and therefore a topology

𝒯_{ℬ_{S}}

which we call the topology generated by the subbasis

S

subbasis for a topology

We say that

S

is a subbasis for the topology

𝒯

when

𝒯_{ℬ_{S}}

= 𝒯

The Topology Generated by a Basis Equals the Intersection of All Topologies That Contain That Basis

Let

A

be a basis, and let

A

be the collection of all topologies

T

such that

A \subseteq T

, then

T_{A} = ⋂ A

Recall that $A \subseteq T_{A}$ therefore $T_{A} \in A$ therefore $⋂ A \subseteq T_{A}$

Now let $U \in T_{A}$ , therefore it is a union of elements from $A$ , but since each $T \in A$ contains $A$ they also contain unions of elements of $A$ and therefore $U \in ⋂ T_{A}$

🏗️ ΘρϵηΠατπ🚧 (under construction)

🏗️ $Θ ρ ϵ η Π α τ π$ 🚧 (under construction)