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

From ProofWiki
Jump to navigation Jump to search

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


Then:

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


Proof

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:

$\tau = \tau'$.

$\blacksquare$