Equivalence of Definitions of Final Topology

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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 $I$.


The following definitions of the concept of Final Topology are equivalent:

Definition 1

The final topology on $X$ with respect to $\family {f_i}_{i \mathop \in I}$ is defined as:

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

Definition 2

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

Then $\tau$ is known as the final topology on $X$ with respect to $\family{f_i}_{i \mathop \in I}$.



Proof

Definition 1 Implies Definition 2

Let:

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


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 $\vartheta$.

$\Box$

Definition 2 Implies Definition 1

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


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

$\tau'$ contains $\tau$

Let $U \in \tau$.

By definition of $\tuple{\tau_i, \tau}$-continuity for each $i \in I$:

$\forall i \in I : \map {f_i^{-1}} U \in \tau_i$

So:

$U \in \tau'$.

Since $U$ was arbitrary:

$\tau \subseteq \tau'$.

$\Box$

$\tau$ contains $\tau'$

From Final Topology is Topology then $\tau'$ is a topology.

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.

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

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

$\tau' \subseteq \tau$

By definition of set equality:

$\tau = \tau'$.

$\blacksquare$