# Topology/Bases

Topology
 ← Topological Spaces Bases Points in Sets →

## Definition

Let ${\displaystyle (X,{\mathcal {T}})}$ be a topological space. A collection ${\displaystyle {\mathcal {B}}}$ of open sets is called a base for the topology ${\displaystyle {\mathcal {T}}}$ if every open set ${\displaystyle U}$ is the union of sets in ${\displaystyle {\mathcal {B}}}$.

Obviously ${\displaystyle {\mathcal {T}}}$ is a base for itself.

## Conditions for Being a Base

In a topological space ${\displaystyle (X,{\mathcal {T}})}$ a collection ${\displaystyle {\mathcal {B}}}$ is a base for ${\displaystyle {\mathcal {T}}}$ if and only if it consists of open sets and for each point ${\displaystyle x\in X}$ and open neighborhood ${\displaystyle U}$ of ${\displaystyle x}$ there is a set ${\displaystyle B\in {\mathcal {B}}}$ such that ${\displaystyle x\in B\subseteq U}$.

Proof:
We need to show that a subset ${\displaystyle U}$ of ${\displaystyle X}$ is open if and only if it is is a union of elements in ${\displaystyle B\in {\mathcal {B}}}$. However, the if part is obvious, from the facts that the elements in ${\displaystyle B\in {\mathcal {B}}}$ are open, and that so are arbitrary unions of open sets. Thus, we only have to prove, that any open set ${\displaystyle U}$ indeed is such a union.
Let ${\displaystyle U}$ be any open set. Consider any element ${\displaystyle x\in U}$. By assumption, there is at least one element in ${\displaystyle {\mathcal {B}}}$, which both contains ${\displaystyle x}$ and is a subset of ${\displaystyle U}$. By the axiom of choice, we may simultaneously for each ${\displaystyle x\in U}$ choose such an element ${\displaystyle B_{x}\in {\mathcal {B}}}$. The union of all of them indeed is ${\displaystyle U}$. Thus, any open set can be formed as a union of sets within ${\displaystyle {\mathcal {B}}}$.

## Constructing Topologies from Bases

Let ${\displaystyle X}$ be any set and ${\displaystyle {\mathcal {B}}}$ a collection of subsets of ${\displaystyle X}$. There exists a topology ${\displaystyle {\mathcal {T}}}$ on ${\displaystyle X}$ such that ${\displaystyle {\mathcal {B}}}$ is a base for ${\displaystyle {\mathcal {T}}}$ if and only if ${\displaystyle {\mathcal {B}}}$ satisfies the following:

1. If ${\displaystyle x\in X}$, then there exists a ${\displaystyle B\in {\mathcal {B}}}$ such that ${\displaystyle x\in B}$.
2. If ${\displaystyle B_{1},B_{2}\in {\mathcal {B}}}$ and ${\displaystyle x\in B_{1}\cap B_{2}}$, then there is a ${\displaystyle B\in {\mathcal {B}}}$ such that ${\displaystyle x\in B\subseteq B_{1}\cap B_{2}}$.

Remark : Note that the first condition is equivalent to saying that The union of all sets in ${\displaystyle {\mathcal {B}}}$ is ${\displaystyle X}$.

## Semibases

Let ${\displaystyle X}$ be any set and ${\displaystyle {\mathcal {S}}}$ a collection of subsets of ${\displaystyle X}$. Then ${\displaystyle {\mathcal {S}}}$ is a semibase if a base of X can be formed by a finite intersection of elements of ${\displaystyle {\mathcal {S}}}$.

## Exercises

1. Show that the collection ${\displaystyle {\mathcal {B}}=\{(a,b):a,b\in \mathbb {R} ,a of all open intervals in ${\displaystyle \mathbb {R} }$ is a base for a topology on ${\displaystyle \mathbb {R} }$.
2. Show that the collection ${\displaystyle {\mathcal {C}}=\{[a,b]:a,b\in \mathbb {R} ,a of all closed intervals in ${\displaystyle \mathbb {R} }$ is not a base for a topology on ${\displaystyle \mathbb {R} }$.
3. Show that the collection ${\displaystyle {\mathcal {L}}=\{(a,b]:a,b\in \mathbb {R} ,a of half open intervals is a base for a topology on ${\displaystyle \mathbb {R} }$.
4. Show that the collection ${\displaystyle {\mathcal {S}}=\{[a,b):a,b\in \mathbb {R} ,a of half open intervals is a base for a topology on ${\displaystyle \mathbb {R} }$.
5. Let ${\displaystyle a,b\in \mathbb {R} }$. A Partition ${\displaystyle {\mathcal {P}}}$ over the closed interval ${\displaystyle [a,b]\,\!}$ is defined as the ordered n-tuple ${\displaystyle a; the norm of a partition ${\displaystyle {\mathcal {P}}}$ is defined as ${\displaystyle \|{\mathcal {P}}\|=\sup\{x_{i}-x_{i-1}|2\leq i\leq n\}}$
For every ${\displaystyle \delta >0\,\!}$, define the set ${\displaystyle U_{\delta }=\{{\mathcal {P}}|\|{\mathcal {P}}\|\leq \delta \}}$.
If ${\displaystyle X\,\!}$ is the set of all partitions on ${\displaystyle [a,b]\,\!}$, prove that the collection of all ${\displaystyle U_{\delta }\,\!}$ is a Base over the Topology on ${\displaystyle X\,\!}$.

Topology
 ← Topological Spaces Bases Points in Sets →