Synthetic Basis formed from Synthetic Sub-Basis
Theorem
Let $X$ be a set.
Let $\SS$ be a synthetic sub-basis on $X$.
Define:
- $\displaystyle \BB = \set {\bigcap \FF: \FF \subseteq \SS, \text{$\FF$ is finite} }$
Then $\BB$ is a synthetic basis on $X$.
Proof
We consider $X$ as the universe.
Thus, in accordance with Intersection of Empty Set, we take the convention that:
- $\displaystyle \bigcap \O = X \in \BB$
By Set is Subset of Union: General Result, it follows that:
- $\displaystyle X \subseteq \bigcup \BB$
That is, axiom B1 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:
- $\displaystyle B_1 = \bigcap \FF_1$
- $\displaystyle B_2 = \bigcap \FF_2$
It follows that:
- $\displaystyle B_1 \cap B_2 = \bigcap \paren {\FF_1 \cup \FF_2}$
proof that $\displaystyle \bigcap_{i \mathop \in I} \bigcap \mathbb S_i$ is equal to $\displaystyle \bigcap \bigcup_{i \mathop \in I} \mathbb S_i$
You can help $\mathsf{Pr} \infty \mathsf{fWiki}$ by explaining it. To discuss this point in more detail, feel free to use the talk page.
If you are able to explain it, then when you have done so you can remove this instance of {{Explain}} from the code.
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 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
- 1978: Lynn Arthur Steen and J. Arthur Seebach, Jr.: Counterexamples in Topology (2nd ed.) ... (previous) ... (next): Part $\text I$: Basic Definitions: Section $1$: General Introduction
