# Final Topology is Topology

## Theorem

Let $X$ be a set.

Let $I$ be an indexing set.

Let $\left\langle {\left({Y_i, \tau_i}\right)} \right\rangle_{i \mathop \in I}$ be an $I$-indexed family of topological spaces.

Let $\left\langle {f_i: Y_i \to X} \right\rangle_{i \mathop \in I}$ be an $I$-indexed family of mappings.

Then $\tau$ is a topology on $X$.

## Proof

Define:

$\forall i \in I: \vartheta_i = \left\{{U \subseteq X: f_i^{-1} \left({U}\right) \in \tau_i}\right\} \subseteq \mathcal P \left({X}\right)$

Then, by the definition of intersection:

$\displaystyle \tau = \bigcap_{i \mathop \in I} \vartheta_i$

From the Intersection of Topologies is Topology, it suffices to show, for all $i \in I$, that $\vartheta_i$ is a topology on $X$.

We now verify the axioms for $\vartheta_i$ to be a topology on $X$.

### $({O1})$: Union of Open Sets

Let $\mathcal A \subseteq \vartheta_i$.

Then, by Preimage of Union under Mapping: General Result and by the definition of a topology, it follows that:

$\displaystyle f_i^{-1} \left({\bigcup \mathcal A}\right) = \bigcup_{U \mathop \in \mathcal A} f_i^{-1} \left({U}\right) \in \tau_i$

That is, $\displaystyle \bigcup \mathcal A \in \vartheta_i$.

$\Box$

### $({O2})$: Pairwise Intersection of Open Sets

Let $U, V \in \vartheta_i$.

Then, by Preimage of Intersection under Mapping and by the definition of a topology, it follows that:

$f_i^{-1} \left({U \cap V}\right) = f_i^{-1} \left({U}\right) \cap f_i^{-1} \left({V}\right) \in \tau_i$

That is, $U \cap V \in \vartheta_i$.

$\Box$

### $({O3})$: Set Itself

By the definition of a topology, it follows that:

$f_i^{-1} \left({X}\right) = Y_i \in \tau_i$

That is, $X \in \vartheta_i$.

$\blacksquare$