# 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

1. for each $x\in X$, there is at least one basis element $B$ containing $x$
2. 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 $\left({U}_{1}\cap \dots \cap {U}_{k}\right)\cap {U}_{k+1}$ so by our inductive hypothesis $\left({U}_{1}\cap \dots \cap {U}_{k}\right)\in {G}_{\mathcal{B}}$ 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 $\mathcal{U}=\left\{\bigcup _{B\in \mathcal{S}}B:\mathcal{S}\subseteq \mathcal{C}\right\}$

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 $\left(X,\mathcal{T}\right)$ 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:
1. ${\mathcal{T}}^{\prime }$ is finer than $\mathcal{T}$
2. 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 $\left\{y\in {\mathbb{R}}^{n}:‖x-y‖<ϵ\right\}$ for any $x\in {\mathbb{R}}^{n}$ and $ϵ\in {\mathbb{R}}^{>0}$, 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 $S$
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 ${\mathcal{B}}_{S}$
TODO
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}$