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# basis (topology)

Let $(X,\mathcal{T})$ be a topological space. A subset $\mathcal{B}$ of
$\mathcal{T}$ is a *basis* for $\mathcal{T}$ if every member of $\mathcal{T}$ is a union of members of $\mathcal{B}$.

Equivalently, $\mathcal{B}$ is a basis if and only if whenever $U$ is open and $x\in U$ then there is an open set $V\in\mathcal{B}$ such that $x\in V\subseteq U$.

The topology generated by a basis $\mathcal{B}$ consists of exactly the unions of the elements of $\mathcal{B}$.

We also have the following easy characterization: (for a proof, see the attachment)

###### Proposition.

A collection of subsets $\mathcal{B}$ of $X$ is a basis for some topology on $X$ if and only if each $x\in X$ is in some element $B\in\mathcal{B}$ and whenever $B_{1},B_{2}\in\mathcal{B}$ and $x\in B_{1}\cap B_{2}$ then there is $B_{3}\in\mathcal{B}$ such that $x\in B_{3}\subseteq B_{1}\cap B_{2}$.

# 0.0.1 Examples

1. A basis for the usual topology of the real line is given by the set of open intervals since every open set can be expressed as a union of open intervals. One may choose a smaller set as a basis. For instance, the set of all open intervals with rational endpoints and the set of all intervals whose length is a power of $1/2$ are also bases. However, the set of all open intervals of length $1$ is not a basis although it is a subbasis (since any interval of length less than $1$ can be expressed as an intersection of two intervals of length $1$).

2. More generally, the set of open balls forms a basis for the topology on a metric space.

3. The set of all subsets with one element forms a basis for the discrete topology on any set.

## Mathematics Subject Classification

54A99*no label found*

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## Comments

## V ?

Line 2: 'Equivalently, B is a basis if and only if whenever U is open and x is an element of U then there is an open set V such that x is an element of V is a subset of U.'

What is the relationship between V and B? Should it be 'there is an open set V which is an element of B such that..' ?

## about proposition 1

there is at least one condition missing, since even the empty set fullfills the conditions of proposition 1. But a basis must have one element, (if X ist not empty).