# Equivalence of Definitions of Final Topology/Definition 1 Implies Definition 2

## Theorem

Let $X$ be a set.

Let $I$ be an indexing set.

Let $\family {\struct{Y_i, \tau_i}}_{i \mathop \in I}$ be an indexed family of topological spaces indexed by $I$.

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

Let:

$\tau = \set{U \subseteq X: \forall i \in I: \map {f_i^{-1}} U \in \tau_i} \subseteq \powerset X$

Then:

$\tau$ is the finest topology on $X$ such that each $f_i: Y_i \to X$ is $\tuple{\tau_i, \tau}$-continuous.

## Proof

From Final Topology is Topology, $\tau$ is a topology.

#### Mappings are continuous

Let $U \in \tau$.

Let $i \in I$.

Then $\map {f_i^{-1}} {U} \in \tau_i$ by definition of $\tau$.

It follows that for each $i \in I$, $f_i: Y_i \to X$ is $\tuple{\tau_i, \tau}$-continuous.

$\Box$

#### $\tau$ is the finest such topology

Let $\struct{X, \vartheta}$ be a topological space.

Let the mappings $\family {f_i: Y_i \to X}_{i \mathop \in I}$ be $\tuple{\tau_i, \vartheta}$-continuous.

Let $U \in \vartheta$.

By definition of continuity, for every $i \in I$:

$\map {f_i^{-1}} U \in \tau_i$

Then by definition of $\tau$:

$U \in \tau$

It follows that:

$\vartheta \subseteq \tau$

That is, $\tau$ is finer than $\vartheta$.

$\blacksquare$