Initial Topology Generated by Countable Family of Functions Separating Points is Metrizable

== Theorem ==
Let $X$ be a set. 
For each $n \in \N$, let $\struct {Y_n, d_n}$ be a metric space. 
Let $\family {f_n}_{n \in \N}$ be a indexed family of functions that separates points.
Let $\tau$ be the initial topology on $X$ generated by $\family {f_n}_{n \in \N}$.
Then $\tau$ is metrizable. 
== Proof ==
For each $x, y \in X$, define: 
:$\map {\tilde {d_n} } {x, y} = \min \set {\map {d_n} {\map {f_n} x, \map {f_n} y}, 1}$
From Pointwise Minimum of Metric and Positive Real Number is Topologically Equivalent Metric, $\tilde {d_n}$ is a metric that is topologically equivalent to $d_n$. 
Now define, for each $x, y \in X$:
:$\ds \map d {x, y} = \sum_{n \mathop = 1}^\infty \frac {\map {\tilde {d_n} } {x, y} } {2^n}$
This is certainly finite, since: 
:$\ds \sum_{n \mathop = 1}^\infty \frac {\map {\tilde {d_n} } {x, y} } {2^n} \le 1$
from Sum of Infinite Geometric Sequence and since $\tilde {d_n} \le 1$. 
We show that $d$ is a metric on $X$.
=== Proof of ===
We have: 
=== Proof of ===
For each $x, y, z \in X$ we have:
=== Proof of ===
Now take $x, y \in X$.
We have, using for $\tilde {d_n}$ we have: 
=== Proof of ===
Suppose that $x, y \in X$ are such that: 
:$\ds \sum_{n \mathop = 1}^\infty \frac {\map {\tilde {d_n} } {x, y} } {2^n} = 0$
Then $\map {\tilde {d_n} } {x, y} = 0$.
Then: 
:$\map {d_n} {\map {f_n} x, \map {f_n} y} = 0$
Using for $d_n$, we have: 
:$\map {f_n} x = \map {f_n} y$ 
Since $\family {f_n}_{n \in \N}$ separates points, we have $x = y$. 
So $d$ is a metric.
We show that $d$ induces $\tau$. 
For this, we show that the identity mapping $\iota : \struct {X, d} \to \struct {X, \tau}$ is a homeomorphism.
From Continuity in Initial Topology, to show that $\iota : \struct {X, d} \to \struct {X, \tau}$ is continuous it is enough to show that $f_j \circ \iota : \struct {X, d} \to \struct {Y_j, d_j}$ is continuous for each $j \in \N$. 
Let $0 < \epsilon < 1$. 
Then for $x, y \in X$ with: 
:$\ds \map d {x, y} < \frac \epsilon {2^j}$
we have: 
:$\ds \sum_{n \mathop = 1}^\infty \frac {\map {\tilde {d_n} } {x, y} } {2^n} < \frac \epsilon {2^j}$ 
This gives that: 
:$\ds \map {\tilde {d_n} } {x, y} < \epsilon$
So we have that $f_j \circ \iota : \struct {X, d} \to \struct {Y_j, d_j}$ is continuous for each $j \in \N$.
To finish, we just need to show that $\iota : \struct {X, \tau} \to \struct {X, d}$ is continuous.
We show that each $d$-ball $\map {B_r} x$ is contained in $\tau$ for each $r > 0$ and $x \in X$.
By the definition of the initial topology we have that $f_n : \struct {X, \tau} \to \struct {Y_n, d_n}$ is continuous.
From Projection from Product Topology is Continuous, we have that the maps:
:$\struct {x, y} \mapsto x$
and:
:$\struct {x, y} \mapsto y$ 
are continuous mappings.
Hence the maps:
:$\struct {x, y} \mapsto \map {f_n} x$ 
and: 
:$\struct {x, y} \mapsto \map {f_n} y$ 
understood $\struct {X, \tau} \to \struct {X, d}$ are continuous, from Composite of Continuous Mappings is Continuous. 
From Continuous Mapping to Product Space, we have that the map:
:$\struct {x, y} \mapsto \struct {\map {f_n} x, \map {f_n} y}$ 
is continuous as a map $\struct {X, \tau} \times \struct {X, \tau} \to \struct {X, d}$.
Then, from Distance Function of Metric Space is Continuous:
:$\struct {x, y} \mapsto \map {\tilde {d_n} } {\map {f_n} x, \map {f_n} y}$ 
is continuous as a map $\struct {X, \tau} \times \struct {X, \tau} \to \R$.
Hence: 
:$\ds s_N = \sum_{n \mathop = 1}^N \frac {\map {\tilde {d_n} } {x, y} } {2^n}$ 
is continuous as a map $\struct {X, \tau} \times \struct {X, \tau} \to \R$.
To finish, from the Uniform Limit Theorem it is enough to show that $s_N \to d$ uniformly.
For $k, m \in \N$ with $m < k$ we have: 
:$\ds \size {s_n - s_k} = \sum_{n \mathop = m + 1}^k \frac {\map {\tilde {d_n} } {x, y} } {2^n} \le \sum_{n \mathop = m + 1}^k \frac 1 {2^n}$
From Sum of Infinite Geometric Sequence:
:$\ds \sum_{n \mathop = 1}^\infty \frac 1 {2^n}$ converges
We have therefore have that: 
:$\ds \sequence {\sum_{n \mathop = 1}^N \frac 1 {2^n} }_{N \in \N}$ is Cauchy. 
So for each $\epsilon > 0$ there exists $N \in \N$ such that: 
:$\ds \sum_{n \mathop = m + 1}^k \frac 1 {2^n} < \epsilon$ 
for each $m, k \ge N$. 
So we have: 
:$\ds \size {\map {s_n} {x, y} - \map {s_k} {x, y} } < \epsilon$
for each $x, y \in X$.
So $d : \struct {X, \tau} \times \struct {X, \tau} \to \R$ is continuous mapping.
From Vertical Section of Continuous Function is Continuous, it follows that for each $y \in X$: 
:$x \mapsto \map d {x, y}$ 
is continuous $\struct {X, \tau} \times \struct {X, d}$. 
Hence taking the preimage of the open set $\openint {-r} r$, we obtain that $\map {B_r} x$ is open in $\struct {X, \tau}$.
Since $\set {\map {B_r} x : r > 0, \, x \in X}$ is a basis for $\struct {X, d}$, it follows that $\iota : \struct {X, \tau} \to \struct {X, d}$ is continuous, completing the proof.
Category:Initial Topology
Category:Metrizable Topologies