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 Metric Space Axiom $(\text M 1)$

We have:

\(\ds \map d {x, x}\)

\(=\)

\(\ds \sum_{n \mathop = 1}^\infty \frac {\map {\tilde {d_n} } {x, x} } {2^n}\)

\(\ds \)

\(=\)

\(\ds 0\)

Metric Space Axiom $(\text M 1)$ for each $\tilde {d_n}$

$\Box$

Proof of Metric Space Axiom $(\text M 2)$: Triangle Inequality

For each $x, y, z \in X$ we have:

\(\ds \map d {x, z}\)

\(=\)

\(\ds \sum_{n \mathop = 1}^\infty \frac {\map {\tilde {d_n} } {x, z} } {2^n}\)

\(\ds \)

\(\le\)

\(\ds \sum_{n \mathop = 1}^\infty \paren {\frac {\map {\tilde {d_n} } {x, y} } {2^n} + \frac {\map {\tilde {d_n} } {y, z} } {2^n} }\)

using Metric Space Axiom $(\text M 2)$: Triangle Inequality for $\tilde {d_n}$

\(\ds \)

\(=\)

\(\ds \sum_{n \mathop = 1}^\infty \frac {\map {\tilde {d_n} } {x, z} } {2^n} + \sum_{n \mathop = 1}^\infty \frac {\map {\tilde {d_n} } {y, z} } {2^n}\)

\(\ds \)

\(=\)

\(\ds \map d {x, y} + \map d {y, z}\)

$\Box$

Proof of Metric Space Axiom $(\text M 3)$

Now take $x, y \in X$.

We have, using Metric Space Axiom $(\text M 3)$ for $\tilde {d_n}$ we have:

\(\ds \map d {y, x}\)

\(=\)

\(\ds \sum_{n \mathop = 1}^\infty \frac {\map {\tilde {d_n} } {y, x} } {2^n}\)

\(\ds \)

\(=\)

\(\ds \sum_{n \mathop = 1}^\infty \frac {\map {\tilde {d_n} } {x, y} } {2^n}\)

using Metric Space Axiom $(\text M 3)$ for each $\tilde {d_n}$

\(\ds \)

\(=\)

\(\ds \map d {x, y}\)

$\Box$

Proof of Metric Space Axiom $(\text M 4)$

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 Metric Space Axiom $(\text M 4)$ for $d_n$, we have:

$\map {f_n} x = \map {f_n} y$

Since $\family {f_n}_{n \in \N}$ separates points, we have $x = y$.

$\Box$

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.

$\blacksquare$