Continuity Test using Sub-Basis/Proof 2

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 $\SS$ be an analytic sub-basis for $\tau_2$.

Suppose that:

$\forall S \in \SS: f^{-1} \sqbrk S \in \tau_1$

where $f^{-1} \sqbrk S$ denotes the preimage of $S$ under $f$.

Then $f$ is continuous.

Proof

Let $\tau$ be the final topology on $X_2$ with respect to $f$.

By hypothesis, $\SS \subseteq \tau$.

By Synthetic Sub-Basis and Analytic Sub-Basis are Compatible, we have that $\tau_2$ is the topology generated by the synthetic sub-basis $\SS$.

By the definition of the generated topology, we have $\tau_2 \subseteq \tau$.

By the definition of the final topology, it follows that $f$ is continuous.

$\blacksquare$