Continuity Test using Basis

Theorem
Let $\struct {X_1, \tau_1}$ and $\struct {X_2, \tau_2}$ be topological spaces.

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

Let $\BB$ be an analytic basis for $\tau_2$.

Suppose that:
 * $\forall B \in \BB: f^{-1} \sqbrk B \in \tau_1$

where $f^{-1} \sqbrk B$ 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:
 * $\ds \exists \AA \subseteq \BB: U = \bigcup \AA$

Hence:

The result follows from the definition of continuity.

Also see

 * Continuity Test using Sub-Basis for a general version.