Continuity Test using Basis

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Theorem

Let $\left({X_1, \tau_1}\right)$ and $\left({X_2, \tau_2}\right)$ be topological spaces.

Let $f: X_1 \to X_2$ be a mapping.

Let $\mathcal B$ be an analytic basis for $\tau_2$.

Suppose that:

$\forall B \in \mathcal B: f^{-1} \left({B}\right) \in \tau_1$

where $f^{-1} \left({B}\right)$ 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:

$\displaystyle \exists \mathcal A \subseteq \mathcal B: U = \bigcup \mathcal A$

Hence:

\(\displaystyle f^{-1} \left({U}\right)\) \(=\) \(\displaystyle f^{-1} \left({\bigcup \mathcal A}\right)\)
\(\displaystyle \) \(=\) \(\displaystyle \bigcup_{B \mathop \in \mathcal A} f^{-1} \left({B}\right)\) Preimage of Union under Mapping: General Result
\(\displaystyle \) \(\in\) \(\displaystyle \tau_1\) by hypothesis, and by the definition of a topology

The result follows from the definition of continuity.

$\blacksquare$


Also see


Sources