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:
- $\displaystyle B_1 \cap B_2 = \bigcap \left({\mathcal F_1 \cup \mathcal F_2}\right)$
proof that $\displaystyle \bigcap_{i \mathop \in I} \bigcap \mathbb S_i$ is equal to $\displaystyle \bigcap \bigcup_{i \mathop \in I} \mathbb S_i$
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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): $\text{I}: \ \S 1$
