Continuity Test using Basis

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 $\mathcal B$ be an analytic basis for $\tau_2$.

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

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

Then $f$ is continuous.

Proof
Let $U \in \tau_2$.

By the definition of an analytic basis, it follows that:
 * $\displaystyle \exists \mathcal A \subseteq \mathcal B: U = \bigcup \mathcal A$

Hence:

The result follows from the definition of continuity.

Also see
As an analytic basis is also an analytic sub-basis, it is seen that Continuity Check using Sub-Basis is a generalization of this theorem.