Finite Weight Space has Basis equal to Image of Mapping of Intersections

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Theorem

Let $T = \left({X, \tau}\right)$ be a topological space with finite weight.

Then there exist a basis $\mathcal B$ of $T$ and a mapping $f:X \to \tau$ such that

$\mathcal B = \operatorname{Im} \left({f}\right)$ and
$\forall x \in X: \left({x \in f \left({x}\right) \land \forall U \in \tau: x \in U \implies f \left({x}\right) \subseteq U}\right)$.

where $\operatorname{Im} \left({f}\right)$ denotes the image of $f$.


Proof

By definition of weight there exists a basis $\mathcal B$ such that

$\left\vert {\mathcal B} \right\vert = w \left({T}\right)$

where

$w \left({T}\right)$ denotes the weight of $T$,
$\left\vert {\mathcal B} \right\vert$ denotes the cardinality of $\mathcal B$.

By assumption that weight is finite:

$\left\vert {\mathcal B} \right\vert$ is finite.

Then by Cardinality of Set is Finite iff Set is Finite:

$\mathcal B$ is finite.

Define a mapping $f: X \to \mathcal P \left({X}\right)$

$(1): \quad \forall x \in X: f \left({x}\right) = \bigcap \left\{{U \in \mathcal B: x \in U}\right\}$.

By definition of subset:

$\forall x \in X: \left\{{U \in \mathcal B: x \in U}\right\} \subseteq \mathcal B$.

By Subset of Finite Set is Finite

$\forall x \in X: \left\{{U \in \mathcal B: x \in U}\right\}$ is finite.

Then by General Intersection Property of Topological Space

$\forall x \in X: \bigcap \left\{{U \in \mathcal B: x \in U}\right\} \in \tau$.

So:

$f: X \to \tau$.


We will prove that

$(2): \quad \forall x \in X: \left({x \in f \left({x}\right) \land \forall U \in \tau: x \in U \implies f \left({x}\right) \subseteq U}\right)$.

Let $x \in X$.

By $(1)$:

$f \left({x}\right) = \bigcap \left\{{U \in \mathcal B: x \in U}\right\}$.

Thus by definition of intersection:

$x \in f \left({x}\right)$.

Let $U$ be an open set of $T$.

Let $x \in U$.

By definition of basis

$\exists V \in \mathcal B: x \in V \subseteq U$.

Then

$V \in \left\{{U \in \mathcal B: x \in U}\right\}$.

Hence by Intersection is Subset:

$f \left({x}\right) \subseteq V$.

Thus by Subset Relation is Transitive

$f \left({x}\right) \subseteq U$.

This ends the proof of $(2)$.


We will prove that $\operatorname{Im} \left({f}\right)$ is a basis of $T$.

By $f: X \to \tau$ and definition of image

$\operatorname{Im} \left({f}\right) \subseteq \tau$.

Let $U$ be an open set of $T$.

Let $x$ be a point $x \in X$ such that

$x \in U$.

By $(2)$:

$f \left({x}\right) \in \operatorname{Im} \left({f}\right) \land x \in f \left({x}\right) \subseteq U$.

By definition of basis this ends the proof of basis.

Thus the result.

$\blacksquare$


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