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:

\(\ds f^{-1} \sqbrk U\)

\(=\)

\(\ds f^{-1} \sqbrk {\bigcup \AA}\)

\(\ds \)

\(=\)

\(\ds \bigcup_{B \mathop \in \AA} f^{-1} \sqbrk B\)

Preimage of Union under Mapping: General Result

\(\ds \)

\(\in\)

\(\ds \tau_1\)

by hypothesis, and by the definition of a topology

The result follows from the definition of continuity.

$\blacksquare$

Also see

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

Sources

• 1975: W.A. Sutherland: Introduction to Metric and Topological Spaces ... (previous) ... (next): $3$: Continuity generalized: topological spaces: $3.2$: Bases: Application $3.2.5$