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

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:

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

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