Continuity Test using Basis
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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 $3.2.5$