Quotient Metric on Vector Space is Invariant Metric iff Vector Subspace is Closed

Theorem
Let $K$ be a field.

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

Let $d$ be an invariant metric on $X$.

Let $N$ be a vector subspace of $X$.

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

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

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

Then $d_N$ is an invariant metric $N$ is closed.

Proof
From Quotient Metric on Vector Space is Invariant Pseudometric, $d_N$ is an invariant pseudometric.

It remains to show that holds  $N$ is closed.

Note that holds  for $x, y \in X$:
 * $\map {d_N} {\map \pi x, \map \pi y} = 0$

implies that $\map \pi x = \map \pi y$.

That is, for $x, y \in X$:
 * $\ds \inf_{z \mathop \in N} \map d {x - y, z} = 0$

implies that $\map \pi x = \map \pi y$.

That is, for $x, y \in X$:
 * $\map d {x - y, N} = 0$

implies that $\map \pi x = \map \pi y$.

From Quotient Mapping is Linear Transformation and Kernel of Quotient Mapping, $\map \pi x = \map \pi y$ is equivalent to $x - y \in N$.

Setting $y = {\mathbf 0}_X$, we can see that:
 * for $x, y \in X$:
 * $\map d {x - y, N} = 0$
 * implies that $x - y \in N$

implies that:
 * for all $x \in X$:
 * $\map d {x, N} = 0$
 * implies that $x \in N$.

Clearly the converse implication also holds.

From Subset of Metric Space is Closed iff contains all Zero Distance Points, we have that:
 * for all $x \in X$:
 * $\map d {x, N} = 0$
 * implies that $x \in N$

holds $N$ is closed.

Hence holds  $N$ is closed.