# Synthetic Basis formed from Synthetic Sub-Basis

## Theorem

Let $X$ be a set.

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

Define:

- $\displaystyle \mathcal B = \left\{{\bigcap \mathcal F: \mathcal F \subseteq \mathcal S, \, \mathcal F \text{ is finite}}\right\}$

Then $\mathcal B$ is a synthetic basis on $X$.

## Proof

We consider $X$ as the universe.

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

- $\displaystyle \bigcap \varnothing = X \in \mathcal B$

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

- $\displaystyle X \subseteq \bigcup \mathcal B$

That is, axiom **B1** for a synthetic basis is satisfied.

We have that $\mathcal B \subseteq \mathcal P \left({X}\right)$.

Let $B_1, B_2 \in \mathcal B$. Then there exist finite $\mathcal F_1, \mathcal F_2 \subseteq \mathcal S$ such that:

- $\displaystyle B_1 = \bigcap \mathcal F_1$
- $\displaystyle B_2 = \bigcap \mathcal F_2$

It follows that:

- $\displaystyle B_1 \cap B_2 = \bigcap \left({\mathcal F_1 \cup \mathcal F_2}\right)$

By Union is Smallest Superset, $\mathcal F_1 \cup \mathcal F_2 \subseteq \mathcal S$.

We have that $\mathcal F_1 \cup \mathcal F_2$ is finite.

Hence $B_1 \cap B_2 \in \mathcal B$, so it follows by definition that axiom **B2** 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

- 1970: Lynn Arthur Steen and J. Arthur Seebach, Jr.:
*Counterexamples in Topology*... (previous) ... (next): $\text{I}: \ \S 1$