Synthetic Sub-Basis and Analytic Sub-Basis are Compatible

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Theorem

Let $\left({X, \tau}\right)$ be a topological space.

Let $\mathcal S \subseteq \mathcal P \left({X}\right)$, where $\mathcal P \left({X}\right)$ denotes the power set of $X$.


Then $\mathcal S$ is an analytic sub-basis for $\tau$ if and only if $\tau$ is the topology on $X$ generated by the synthetic sub-basis $\mathcal S$.


Proof

Necessary Condition

Follows directly from the definitions of the generated topology and an analytic sub-basis.

$\Box$


Sufficient Condition

Suppose that $\mathcal S$ is an analytic sub-basis for $\tau$.

Let $\tau'$ be the topology on $X$ generated by the synthetic sub-basis $\mathcal S$.

By definition $1$ of the generated topology and the definition of an analytic sub-basis, we have $\tau \subseteq \tau'$.

By definition $2$ of the generated topology, it follows that $\tau' \subseteq \tau$.

By definition of set equality:

$\tau = \tau'$

$\blacksquare$