Equivalence of Definitions of Weight of Topological Space
Theorem
Let $T$ be a topological space.
Let $\mathbb B$ be the set of all bases of $T$.
The following definitions of the weight of $T$ are equivalent:
Definition 1
The weight of $T$ is defined as:
- $\ds \map w T := \bigcap_{\BB \mathop \in \mathbb B} \card \BB$
where $\card \BB$ denotes the cardinality of $\BB$.
Definition 2
The weight of $T$ is the smallest cardinality of the elements of $\mathbb B$:
- $\map w T := \min \set {\card \BB: \BB \in \mathbb B}$
Proof
By Class of All Cardinals is Subclass of Class of All Ordinals, the set:
- $M = \set {\card \BB: \BB \in \mathbb B}$
is a subclass of the class of all ordinals.
By Class of All Ordinals is Well-Ordered by Subset Relation:
- $M$ is well ordered by the $\subseteq$ relation.
By Class of All Ordinals is Well-Ordered by Subset Relation there exists a smallest element $m_0 \in M$:
- $\forall m \in M: m_0 \subseteq m$
Hence by Smallest Element is Minimal there exists a basis $\BB_0$ of $T$ which has minimal cardinality:
- $m_0 = \map {w_2} T$.
Let:
- $\ds \map {w_1} T = \bigcap_{\BB \mathop \in \mathbb B} \card \BB$
- $\ds \map {w_1} T = \bigcap M \subseteq m_0$
But by Intersection is Largest Subset:
- $\ds \mathfrak m_0 \subseteq \bigcap M$
By definition of set equality:
- $\map {w_1} T = \map {w_2} T$
and hence the result.
$\blacksquare$