Continuity Test using Sub-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 $\SS$ 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 1
Define:
- $\ds \BB = \set {\bigcap \AA: \AA \subseteq \SS, \AA \text{ is finite} } \subseteq \powerset {X_2}$
Let $B \in \BB$.
Then there exists a finite subset $\AA \subseteq \SS$ such that:
- $\ds B = \bigcap \AA$
Hence:
| \(\ds f^{-1} \sqbrk B\) | \(=\) | \(\ds f^{-1} \sqbrk {\bigcap \AA}\) | ||||||||||||
| \(\ds \) | \(=\) | \(\ds \bigcap_{S \mathop \in \AA} f^{-1} \sqbrk S\) | Preimage of Intersection under Mapping: General Result | |||||||||||
| \(\ds \) | \(\in\) | \(\ds \tau_1\) | General Intersection Property of Topological Space |
Define:
- $\ds \tau = \set {\bigcup \AA: \AA \subseteq \BB} \subseteq \powerset {X_2}$
By the definition of an analytic sub-basis, we have $\tau_2 \subseteq \tau$.
Let $U \in \tau_2$.
Then $U \in \tau$; therefore:
- $\ds \exists \AA \subseteq \BB: U = \bigcup \AA$
Hence:
| \(\ds f^{-1} \sqbrk U\) | \(=\) | \(\ds f^{-1} \sqbrk {\bigcup \AA}\) | ||||||||||||
| \(\ds \) | \(=\) | \(\ds \bigcup_{S \mathop \in \AA} f^{-1} \sqbrk S\) | Preimage of Union under Mapping: General Result | |||||||||||
| \(\ds \) | \(\in\) | \(\ds \tau_1\) | because $\forall B \in \BB: f^{-1} \sqbrk B \in \tau_1$, and by Open Set Axiom $\paren {\text O 1 }$: Union of Open Sets applied to $\tau_1$ |
That is, $f$ is continuous.
$\blacksquare$
Proof 2
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.
$\blacksquare$
Also see
- Analytic Basis is Analytic Sub-Basis, of which Continuity Test using Basis is seen as a special case.
Sources
- 1975: W.A. Sutherland: Introduction to Metric and Topological Spaces ... (previous) ... (next): $3$: Continuity generalized: topological spaces: $3.3$: Sub-bases and weak topologies