Quotient Metric on Vector Space induces Quotient Topology

Theorem
Let $K$ be a topological field.

Let $X$ be a vector space over $K$.

Let $d$ be an invariant metric such that the induced topology $\tau$ makes $\struct {X, \tau}$ a topological vector space.

Let $N$ be a closed linear subspace of $X$.

Let $X/N$ be the quotient vector space of $X$ modulo $N$.

Let $\struct {X/N, \tau_N}$ be the quotient topological vector space of $X$ modulo $N$.

Let $d_N$ be the quotient metric on $X/N$ induced by $d$.

Then $d_N$ induces $\tau_N$.

Proof
From Quotient Metric on Vector Space is Invariant Metric iff Vector Subspace is Closed, $d_N$ is a metric and hence:
 * $\struct {X/N, d_N}$ is a metric space.

Let $\pi : X \to X/N$ be the quotient mapping.

We first show that:
 * $\pi \sqbrk {\set {x \in X : \map d {x, {\mathbf 0}_X} < r} } = \set {\map \pi x \in X/N : \map {d_N} {\map \pi x, {\mathbf 0}_{X/N} } < r}$

Let $x \in \set {x \in X : \map d {x, {\mathbf 0}_X} < r}$.

We want to show that $\map \pi x \in \set {\map \pi x \in X/N : \map {d_N} {\map \pi x, {\mathbf 0}_{X/N} } < r}$.

Since $N$ is a linear subspace, we have ${\mathbf 0}_X \in N$.

Hence, we have:
 * $\ds \inf_{z \mathop \in N} \map d {x - {\mathbf 0}_X, z} \le \map d {x, {\mathbf 0}_X} < r$

Hence, we have:
 * $\map {d_N} {\map \pi x, {\mathbf 0}_{X/N} } < r$

So, we have:
 * $\pi \sqbrk {\set {x \in X : \map d {x, {\mathbf 0}_X} < r} } \subseteq \set {\map \pi x \in X/N : \map {d_N} {\map \pi x, {\mathbf 0}_{X/N} } < r}$

Now let $\map \pi x \in \set {x \in X : \map {d_N} {\map \pi x, {\mathbf 0}_{X/N} } < r}$.

Then, we have:
 * $\ds \inf_{z \mathop \in N} \map d {x, z} < r$

Since $d$ is invariant metric, we have that:
 * $\ds \inf_{z \mathop \in N} \map d {x - z, {\mathbf 0}_X} < r$

Hence by the definition of infimum there exists $z' \in N$ such that:
 * $\map d {x - z', {\mathbf 0}_X} < r$

By Kernel of Quotient Mapping, we have that $\map \pi x = \map \pi {x - z'}$.

We have $x - z' \in \set {x \in X : \map d {x, {\mathbf 0}_X} < r}$, so we obtain:
 * $\map \pi x \in \pi \sqbrk {\set {x \in X : \map d {x, {\mathbf 0}_X} < r} }$

Hence we have:
 * $\set {\map \pi x \in X/N : \map {d_N} {\map \pi x, {\mathbf 0}_{X/N} } < r} \subseteq \pi \sqbrk {\set {x \in X : \map d {x, {\mathbf 0}_X} < r} }$

So we can conclude that:
 * $\pi \sqbrk {\set {x \in X : \map d {x, {\mathbf 0}_X} < r} } = \set {\map \pi x \in X/N : \map {d_N} {\map \pi x, {\mathbf 0}_{X/N} } < r}$

For $\epsilon > 0$ and $x \in X$, let $\map {B_\epsilon^d} x$ be the open ball in $\struct {X, d}$ of radius $\epsilon$ and center $x$.

For $\epsilon > 0$ and $\map \pi x \in X/N$, let $\map {B_\epsilon^{d_N} } x$ be the open ball in $\struct {X/N, d_N}$ of radius $\epsilon$ and center $\map \pi x$.

We have already shown that:
 * $\pi \sqbrk {\map {B_\epsilon^d} { {\mathbf 0}_X} } = \map {B_\epsilon^{d_N} } { {\mathbf 0}_{X/N} }$

Let $x \in X$.

Then, we have:

We now aim to show that the open sets of $\struct {X/N, d_N}$ are precisely the sets of the form $\pi \sqbrk V$ for $V \in \tau$.

From Open Set in Quotient Topological Vector Space, we will then have that the topology induced by $d_N$ is precisely $\tau_N$.

First let $U$ be open in $\struct {X/N, d_N}$.

From the definition of an open set in a metric space, for each $\mathbf x \in U$ there exists a real number $\map \epsilon {\map \pi x} > 0$ such that together we have:
 * $\ds U = \bigcup_{\map \pi x \in U} \map {B_{\map \epsilon {\map \pi x} }^{d_N} } {\map \pi x}$

We now have:

As the union of open sets:
 * $\ds \bigcup_{x \in \pi^{-1} \sqbrk U} \map {B_{\map \epsilon {\map \pi x} }^d} x$ is open in $\struct {X, d}$.

Hence every open set has the form $\pi \sqbrk V$ for some $V$ open in $\struct {X, d}$.

Conversely, let $V$ be an open set in $\struct {X, d}$.

Again, from the definition of an open set in a metric space, for each $x \in V$ there exists $\epsilon_x > 0$ such that:
 * $\ds V = \bigcup_{x \in V} \map {B_{\epsilon_x}^d} x$

We then have, from Image of Union under Mapping: General Result:
 * $\ds \pi \sqbrk V = \bigcup_{x \in V} \pi \sqbrk {\map {B_{\epsilon_x}^d} x} = \bigcup_{x \in V} \map {B_\epsilon^{d_N} } {\map \pi x}$

So $\pi \sqbrk V$ is the union of open sets in $\struct {X/N, d_N}$.

Hence every set of the form $\pi \sqbrk V$ with $V$ open in $\struct {X, d}$, is open in $\struct {X/N, d_N}$.

So the open sets of $\struct {X/N, d_N}$ are precisely those sets of the form $\pi \sqbrk V$ for $V$ open in $\struct {X, d}$.

By Open Set in Quotient Topological Vector Space, we obtain that the topology induced by $d_N$ is $\tau_N$.