Continuity Test using Sub-Basis/Proof 1

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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 $\mathcal S$ be an analytic sub-basis for $\tau_2$.

Suppose that:

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

where $f^{-1} \left({S}\right)$ denotes the preimage of $S$ under $f$.


Then $f$ is continuous.


Proof

Define:

$\displaystyle \mathcal B = \left\{{\bigcap \mathcal A: \mathcal A \subseteq \mathcal S, \, \mathcal A \text{ is finite}}\right\} \subseteq \mathcal P \left({X_2}\right)$

Let $B \in \mathcal B$.

Then there exists a finite subset $\mathcal A \subseteq \mathcal S$ such that:

$\displaystyle B = \bigcap \mathcal A$

Hence:

\(\displaystyle f^{-1} \left({B}\right)\) \(=\) \(\displaystyle f^{-1} \left({\bigcap \mathcal A}\right)\)
\(\displaystyle \) \(=\) \(\displaystyle \bigcap_{S \mathop \in \mathcal A} f^{-1} \left({S}\right)\) Preimage of Intersection under Mapping: General Result
\(\displaystyle \) \(\in\) \(\displaystyle \tau_1\) General Intersection Property of Topological Space


Define:

$\displaystyle \tau = \left\{{\bigcup \mathcal A: \mathcal A \subseteq \mathcal B}\right\} \subseteq \mathcal P \left({X_2}\right)$

By the definition of an analytic sub-basis, we have $\tau_2 \subseteq \tau$.

Let $U \in \tau_2$.

Then $U \in \tau$; therefore:

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

Hence:

\(\displaystyle f^{-1} \left({U}\right)\) \(=\) \(\displaystyle f^{-1} \left({\bigcup \mathcal A}\right)\)
\(\displaystyle \) \(=\) \(\displaystyle \bigcup_{S \mathop \in \mathcal A} f^{-1} \left({S}\right)\) Preimage of Union under Mapping: General Result
\(\displaystyle \) \(\in\) \(\displaystyle \tau_1\) because $\forall B \in \mathcal B: f^{-1} \left({B}\right) \in \tau_1$, and by open set axiom $({1})$ applied to $\tau_1$

That is, $f$ is continuous.

$\blacksquare$