# Equivalence of Definitions of Analytic Basis

## Contents

## Theorem

The following definitions of the concept of **Analytic Basis** are equivalent:

### Definition 1

Let $\struct {S, \tau}$ be a topological space.

An **analytic basis for $\tau$** is a subset $\BB \subseteq \tau$ such that:

- $\displaystyle \forall U \in \tau: \exists \AA \subseteq \BB: U = \bigcup \AA$

That is, such that for all $U \in \tau$, $U$ is a union of sets from $\BB$.

### Definition 2

Let $\left({S, \tau}\right)$ be a topological space.

Let $\mathcal B \subseteq \tau$.

Then $\mathcal B$ is an analytic basis for $\tau$ if and only if:

- $\forall U \in \tau: \forall x \in U: \exists V \in \mathcal B: x \in V \subseteq U$

## Proof

### Definition 1 implies Definition 2

Let $\mathcal B$ be an analytic basis for $\tau$ by definition 1.

Let $U \in \tau$.

By definition 1 of an analytic basis, we can choose $\mathcal A \subseteq \mathcal B$ such that:

- $\displaystyle U = \bigcup \mathcal A$

By the definition of union:

- $\forall x \in U: \exists B \in \mathcal A: x \in B$

By Union is Smallest Superset:

- $\forall B \in \mathcal A: B \subseteq U$

Since $\mathcal A \subseteq \mathcal B$, the result follows.

Thus $\mathcal B$ is an analytic basis for $\tau$ by definition 2.

$\Box$

### Definition 2 implies Definition 1

Let $\mathcal B$ be an analytic basis for $\tau$ by definition 2.

Let $U \in \tau$.

Let $\mathcal A = \left\{{B \in \mathcal B: B \subseteq U}\right\}$.

Then $\mathcal A \subseteq \mathcal B$.

Let $x \in U$ be arbitrary.

Since $\mathcal B$ is an analytic basis for $\tau$ by definition 2, there is some $B_x \in \mathcal B$ such that:

- $x \in B_x \subseteq U$

Hence, by construction of $\mathcal A$, $B_x \in \mathcal A$.

Thus:

- $\displaystyle x \in \bigcup \mathcal A$

and it follows that:

- $\displaystyle U \subseteq \bigcup \mathcal A$

By Union is Smallest Superset applied to $\mathcal A$ and $U$:

- $\displaystyle \bigcup \mathcal A \subseteq U$

By definition of set equality and definition 1 of an analytic basis, the result follows.

Thus $\mathcal B$ is an analytic basis for $\tau$ by definition 1.

$\blacksquare$