Preimage of Open Sets forms Basis if Continuous Mappings Separate Points from Closed Sets

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Theorem

Let $X$ be a topological space.


Let $\family {Y_i}_{i \mathop \in I}$ be an indexed family of topological spaces for some indexing set $I$.

Let $\family {f_i : X \to Y_i}_{i \mathop \in I}$ be an indexed family of continuous mappings.

Let $\family {f_i}_{i \mathop \in I}$ separate points from closed sets.


Let $\BB = \set{f_i^{-1} \sqbrk V : i \in I, V \text{ is open in } Y_i}$.


Then:

$\BB$ is a basis for $X$

Proof

By definition of continuous:

$\forall B \in \BB : B$ is open in $X$


Let $U \subseteq X$ be open in $X$.

Let $x \in U$.


By definition of closed subset:

$X \setminus U$ is closed in $X$

By definition of mappings separating points from closed sets:

$\exists i \in I : \map {f_i} x \notin f_i \sqbrk {X \setminus U}^-$

Hence:

$\map {f_i} x \in Y_i \setminus f_i \sqbrk {X \setminus U}^-$

By definition of preimage of mapping:

$x \in f_i^{-1} \sqbrk {Y_i \setminus f_i \sqbrk {X \setminus U}^-}$


From Topological Closure is Closed:

$f_i \sqbrk {X \setminus U}^-$ is closed in $Y_i$

By definition of closed subset:

$Y_i \setminus f_i \sqbrk {X \setminus U}^-$ is open in $Y_i$

Hence:

$f_i^{-1} \sqbrk {Y_i \setminus f_i \sqbrk {X \setminus U}^-} \in \BB$


We have:

\(\ds y\) \(\in\) \(\ds f_i^{-1} \sqbrk {Y_i \setminus f_i \sqbrk {X \setminus U}^-}\)
\(\ds \leadsto \ \ \) \(\ds \map {f_i} y\) \(\in\) \(\ds Y_i \setminus f_i \sqbrk {X \setminus U}^-\) Definition of Preimage of Mapping
\(\ds \leadsto \ \ \) \(\ds \map {f_i} y\) \(\in\) \(\ds Y_i \setminus f_i \sqbrk {X \setminus U}\) Set is Subset of its Topological Closure and Set Difference with Subset is Superset of Set Difference
\(\ds \leadsto \ \ \) \(\ds \map {f_i} y\) \(\notin\) \(\ds f_i \sqbrk {X \setminus U}\) Definition of Set Difference
\(\ds \leadsto \ \ \) \(\ds y\) \(\notin\) \(\ds X \setminus U\) Definition of Image of Mapping
\(\ds \leadsto \ \ \) \(\ds y\) \(\in\) \(\ds U\) Definition of Set Difference

By definition of subset:

$f_i^{-1} \sqbrk {Y_i \setminus f_i \sqbrk {X \setminus U}^-} \subseteq U$


We have shown that:

$\forall$ open subsets $U \subseteq X, x \in U : \exists B \in \BB : x \in B \subseteq U$

Hence by definition $\BB$ is a basis for $X$.

$\blacksquare$

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