Equivalence of Definitions of Analytic Basis

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Theorem

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


Definition 1

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


An analytic basis for $\tau$ is a subset $\mathcal B \subseteq \tau$ such that:

$\displaystyle \forall U \in \tau: \exists \mathcal A \subseteq \mathcal B: U = \bigcup \mathcal A$


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

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.

$\blacksquare$