# Continuity Test using Sub-Basis/Proof 1

## 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

Define:

$\displaystyle \BB = \left\{{\bigcap \AA: \AA \subseteq \SS, \AA \text{ is finite}}\right\} \subseteq \powerset {X_2}$

Let $B \in \BB$.

Then there exists a finite subset $\AA \subseteq \SS$ such that:

$\displaystyle B = \bigcap \AA$

Hence:

 $\displaystyle f^{-1} \sqbrk B$ $=$ $\displaystyle f^{-1} \sqbrk {\bigcap \AA}$ $\displaystyle$ $=$ $\displaystyle \bigcap_{S \mathop \in \AA} f^{-1} \sqbrk S$ Preimage of Intersection under Mapping: General Result $\displaystyle$ $\in$ $\displaystyle \tau_1$ General Intersection Property of Topological Space

Define:

$\displaystyle \tau = \set {\bigcup \AA: \AA \subseteq \BB} \subseteq \powerset {X_2}$

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

Let $U \in \tau_2$.

Then $U \in \tau$; therefore:

$\displaystyle \exists \AA \subseteq \BB: U = \bigcup \AA$

Hence:

 $\displaystyle f^{-1} \sqbrk U$ $=$ $\displaystyle f^{-1} \sqbrk {\bigcup \AA}$ $\displaystyle$ $=$ $\displaystyle \bigcup_{S \mathop \in \AA} f^{-1} \sqbrk S$ Preimage of Union under Mapping: General Result $\displaystyle$ $\in$ $\displaystyle \tau_1$ because $\forall B \in \BB: f^{-1} \sqbrk B \in \tau_1$, and by open set axiom $({1})$ applied to $\tau_1$

That is, $f$ is continuous.

$\blacksquare$