# Basis

basis for a set

a basis for a set $$ X$$ is a collection $$ \mathcal{B}$$ 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 $$ {B}_{3}\subseteq {B}_{1}\cap {B}_{2}$$

collection generated by a basis

Given a basis $$ \mathcal{B}$$ we define the set $$ {G}_{\mathcal{B}}$$ by $$ U\in {G}_{\mathcal{B}}$$ iff for each $$ x\in U$$ there is a $$ B\in \mathcal{B}$$ such that $$ x\in B$$ and $$ B\subseteq U$$

$$ {G}_{\mathcal{B}}$$ is a topology

The collection $$ {G}_{\mathcal{B}}$$ is a topology

We will verify it's a topology from the definition

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

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

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

- base case
if $$ n=1$$ then the union of one element from $$ {G}_{\mathcal{B}}$$ (namely just itself) is also an element of $$ {G}_{\mathcal{B}}$$ trivially.

Before moving to the inductive step, notice that it also holds for $$ n=2$$, because given $$ {U}_{1},{U}_{2}\in {G}_{\mathcal{B}}$$, we can see that for any $$ x\in {U}_{1}\cap {U}_{2}$$, we know that we have a basis elements $$ {B}_{1},{B}_{2}\in \mathcal{B}$$ 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 $$ {U}_{1}\cap {U}_{2}\in {G}_{\mathcal{B}}$$

- inductive step
Let $$ k\in {\mathbb{N}}_{1}$$ and assume that the intersection of $$ k$$ elements from $$ {G}_{\mathcal{B}}$$ is also in $$ {G}_{\mathcal{B}}$$ 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 $$ ({U}_{1}\cap \dots \cap {U}_{k})\in {G}_{\mathcal{B}}$$ and therefore this $$ k+1$$ union is a union of two elements already in $$ {G}_{\mathcal{B}}$$ so by our $$ n=2$$ argument $$ {U}_{1}\cap \dots \cap {U}_{k+1}\in {G}_{\mathcal{B}}$$, as needed.

topology generated by a basis

Since a collection generated by a basis turns out to be a topology we define $$ {\mathcal{T}}_{\mathcal{B}}:={G}_{\mathcal{B}}$$ and call it the topology generated by $$ \mathcal{B}$$

basis for a topology

We say that $$ \mathcal{B}$$ is a basis for a topology $$ \mathcal{T}$$ when $$ {\mathcal{T}}_{\mathcal{B}}$$ $$ =\mathcal{T}$$. In this case we may also say that $$ \mathcal{B}$$ generates $$ \mathcal{T}$$

basis elements are open in a generated topology

Suppose we have the topology $$ {\mathcal{T}}_{\mathcal{B}}$$, then given any $$ B\in \mathcal{B}$$, $$ B\in {T}_{\mathcal{b}}$$, in other words $$ \mathcal{B}\subseteq {T}_{\mathcal{B}}$$

Considering a topology generated by a basis $$ \mathcal{B}$$ we can see that for any $$ B\in \mathcal{B}$$, we can use $$ B$$ itself such that for each $$ x\in B$$, we have $$ B$$ so that $$ B\subseteq B$$, to show that $$ B\in {\mathcal{T}}_{\mathcal{B}}$$

topology generated by a basis consists of unions

$$ {\mathcal{T}}_{\mathcal{B}}$$ equals the collection of all unions of elements of $$ \mathcal{B}$$

Symbolically we can write this collection of unions as $$ \mathcal{U}=\{\bigcup _{B\in \mathcal{S}}B:\mathcal{S}\subseteq \mathcal{C}\}$$

Suppose we have a subset of elements from $$ \mathcal{B}$$ called $$ \mathcal{S}$$, then for each $$ B\in \mathcal{S}$$ it is open in $$ X$$ because it is a basis element, and therefore by the definition of a topology, as $$ \bigcup _{B\subseteq \mathcal{S}}B$$ is an arbitrary union of open sets it is also open in $$ X$$, so we've shown that $$ \mathcal{U}\subseteq {\mathcal{T}}_{\mathcal{B}}$$

We'll show the converse now, so suppose that $$ U\in {\mathcal{T}}_{\mathcal{B}}$$, we want to prove that it's equal to an arbitrary union of elements of $$ \mathcal{B}$$.

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

basis criterion

Let $$ (X,\mathcal{T})$$ be a topological space. Suppose that $$ \mathcal{C}$$ is a collection of open sets of $$ X$$ such that for each open set $$ U$$ of $$ X$$ and each $$ x\in U$$, there is an element $$ C$$ of $$ \mathcal{C}$$ such that $$ x\in C\subseteq U$$. Then $$ \mathcal{C}$$ is a basis and $$ {\mathcal{T}}_{\mathcal{C}}=\mathcal{T}$$

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

Now let $$ {C}_{1},{C}_{2}$$ be elements of $$ \mathcal{C}$$ 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 $$ \mathcal{C}$$, there is some $$ C\in \mathcal{C}$$ 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 $$ \mathcal{C}$$ is a basis

Now we want to show that $$ {\mathcal{T}}_{\mathcal{C}}=\mathcal{T}$$. So suppose $$ U\in {\mathcal{T}}_{\mathcal{C}}$$, then it can be written as a union as elements from $$ \mathcal{C}$$, but since $$ \mathcal{C}$$ 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 $$ U\in \mathcal{T}$$ and consider some $$ x\in U$$ then by the definition of $$ \mathcal{C}$$, there is some $$ C\in \mathcal{C}$$ such that $$ x\in C\subseteq U$$, looking back at the definition for a topology generated by a basis, we can see this means $$ U\in {\mathcal{T}}_{\mathcal{C}}$$, as needed.

basis finer equivalence

Let

$$ \mathcal{B}$$ and

$$ {\mathcal{B}}^{\prime}$$ be bases for the topologies

$$ \mathcal{T}$$ and

$$ {\mathcal{T}}^{\prime}$$, respectively on

$$ X$$. Then the following are equivalent:

- $$ {\mathcal{T}}^{\prime}$$ is finer than $$ \mathcal{T}$$
- for each $$ x\in X$$ and each basis element $$ B\in \mathcal{B}$$ containing $$ x$$ there is a basis element $$ {B}^{\prime}\in {\mathcal{B}}^{\prime}$$ such that $$ x\in {B}^{\prime}\subseteq B$$

TODO

topology equality with basis

Given two bases

$$ \mathcal{B}$$ and

$$ {\mathcal{B}}^{{}^{\prime}}$$ for the toplogies

$$ \mathcal{T}$$ and

$$ {\mathcal{T}}^{{}^{\prime}}$$ on

$$ X$$, then

$$ \mathcal{T}={\mathcal{T}}^{{}^{\prime}}$$ if and only if all of the following hold:

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

TODO

standard topology on $$ {\mathbb{R}}^{n}$$

Let $$ \mathcal{B}$$ be the collection of all sets of the form $$ \{y\in {\mathbb{R}}^{n}:\Vert x-y\Vert <\u03f5\}$$ for any $$ x\in {\mathbb{R}}^{n}$$ and $$ \u03f5\in {\mathbb{R}}^{>0}$$, then the topology generated by $$ \mathcal{B}$$ is called the standard topology on the $$ {\mathbb{R}}^{n}$$. Whenever we consider $$ {\mathbb{R}}^{n}$$, 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 $$ \mathcal{T}$$ of all unions of finite intersections of elements of $$ S$$

topology generated by a subbasis

Given a subbasis $$ S$$, we know that this induces a basis $$ {\mathcal{B}}_{S}$$, and therefore a topology $$ {\mathcal{T}}_{{\mathcal{B}}_{S}}$$ which we call the topology generated by the subbasis $$ S$$.

subbasis for a topology

We say that $$ S$$ is a subbasis for the topology $$ \mathcal{T}$$ when $$ {\mathcal{T}}_{{\mathcal{B}}_{S}}$$$$ =\mathcal{T}$$