# Topology/Bases

 Topology ← Topological Spaces Bases Points in Sets →

## Definition

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

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

## Conditions for Being a Base

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

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

## Constructing Topologies from Bases

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

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

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

## Semibases

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

## Exercises

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

 Topology ← Topological Spaces Bases Points in Sets →