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


Sources