Continuity Test using Sub-Basis/Proof 1

Theorem
Let $\left({X_1, \tau_1}\right)$ and $\left({X_2, \tau_2}\right)$ be topological spaces.

Let $f: X_1 \to X_2$ be a mapping.

Let $\mathcal S$ be an analytic sub-basis for $\tau_2$.

Suppose that:
 * $\forall S \in \mathcal S: f^{-1} \left({S}\right) \in \tau_1$

where $f^{-1} \left({S}\right)$ denotes the preimage of $S$ under $f$.

Then $f$ is continuous.

Proof
Define:
 * $\displaystyle \mathcal B = \left\{{\bigcap \mathcal A: \mathcal A \subseteq \mathcal S, \, \mathcal A \text{ is finite}}\right\} \subseteq \mathcal P \left({X_2}\right)$

Let $B \in \mathcal B$.

Then there exists a finite subset $\mathcal A \subseteq \mathcal S$ such that:
 * $\displaystyle B = \bigcap \mathcal A$

Hence:

Define:
 * $\displaystyle \tau = \left\{{\bigcup \mathcal A: \mathcal A \subseteq \mathcal B}\right\} \subseteq \mathcal P \left({X_2}\right)$

By the definition of an analytic sub-basis, we have $\tau_2 \subseteq \tau$.

Let $U \in \tau_2$.

Then $U \in \tau$; therefore:
 * $\displaystyle \exists \mathcal A \subseteq \mathcal B: U = \bigcup \mathcal A$

Hence:

That is, $f$ is continuous.

Also see
As an analytic basis is also an analytic sub-basis, it is seen that Continuity Test using Basis is a special case of this theorem.