Topology/Bases

Definition

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

Obviously $𝒯$ is a base for itself.

Conditions for Being a Base

In a topological space $(X,𝒯)$ a collection $ℬ$ is a base for $𝒯$ 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 $B\in ℬ$ such that $x\in B\subseteq U$.

Proof:
Let U be any open set. Consider any element x∈U. There is an open set within $ℬ$⊆U. The union of all of them is U. Thus, any open set can be formed as a union of sets within $ℬ$.

Constructing Topologies from Bases

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

1. If $x\in X$, then there exists a $B\in ℬ$ such that $x\in B$.
2. If $B_1,B_2\in ℬ$ and $x\in B_1\cap B_2$, then there is a $B\in ℬ$ 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 $ℬ$ is $X$.

Semibases

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

Exercises

1. Show that the collection $ℬ=\{(a,b):a,b\in ℝ,a<b\}$ of all open intervals in $ℝ$ is a base for a topology on $ℝ$.
2. Show that the collection $𝒞=\{[a,b]:a,b\in ℝ,a<b\}$ of all closed intervals in $ℝ$ is not a base for a topology on $ℝ$.
3. Show that the collection $ℒ=\{(a,b]:a,b\in ℝ,a<b\}$ of half open intervals is a base for a topology on $ℝ$.
4. Show that the collection $𝒮=\{[a,b):a,b\in ℝ,a<b\}$ of half open intervals is a base for a topology on $ℝ$.