Synthetic Basis formed from Synthetic Sub-Basis

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Theorem

Let $X$ be a set.

Let $\SS$ be a synthetic sub-basis on $X$.


Define:

$\ds \BB = \set {\bigcap \FF: \FF \subseteq \SS, \text{$\FF$ is finite} }$

Then $\BB$ is a synthetic basis on $X$.


Proof

We consider $X$ as the universal set.

Thus, in accordance with Intersection of Empty Set, we take the convention that:

$\ds \bigcap \O = X \in \BB$

By Set is Subset of Union: General Result, it follows that:

$\ds X \subseteq \bigcup \BB$

That is, axiom $(\text B 1)$ for a synthetic basis is satisfied.


We have that $\BB \subseteq \powerset X$.


Let $B_1, B_2 \in \BB$.

Then there exist finite $\FF_1, \FF_2 \subseteq \SS$ such that:

$\ds B_1 = \bigcap \FF_1$
$\ds B_2 = \bigcap \FF_2$

It follows that:

$\ds B_1 \cap B_2 = \bigcap \paren {\FF_1 \cup \FF_2}$


This article, or a section of it, needs explaining.
In particular: proof that $\ds \bigcap_{i \mathop \in I} \bigcap \mathbb S_i$ is equal to $\ds \bigcap \bigcup_{i \mathop \in I} \mathbb S_i$
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By Union is Smallest Superset, $\FF_1 \cup \FF_2 \subseteq \SS$.

We have that $\FF_1 \cup \FF_2$ is finite.

Hence $B_1 \cap B_2 \in \BB$, so it follows by definition that axiom $(\text B 2)$ for a synthetic basis is satisfied.

$\blacksquare$


Note

Note that by this construction, any collection of subsets of $X$ can form a synthetic basis and thus generate a topology on $X$.


Sources

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