Quotient Topological Vector Space is Hausdorff iff Linear Subspace is Closed

Theorem
Let $\GF \in \set {\R, \C}$.

Let $\struct {X, \tau}$ be a topological vector space over $\GF$.

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

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

Let $\tau_N$ be the quotient topology on $X/N$.

Then $\struct {X/N, \tau_N}$ is Hausdorff :
 * $N$ is a closed linear subspace.

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

From the definition of the quotient topology, $\pi$ is continuous.

From Quotient Mapping on Quotient Topological Vector Space is Open Mapping, $\pi$ is an open mapping.

From Characterization of Hausdorff Topological Vector Space, we have that $\struct {X/N, \tau_N}$ is Hausdorff :
 * $\set { {\mathbf 0}_{X/N} }$ is closed in $\struct {X/N, \tau_N}$.

That is, :
 * $X/N \setminus \set { {\mathbf 0}_{X/N} }$ is open in $\struct {X/N, \tau_N}$.

That is, :
 * $\pi \sqbrk {X \setminus N}$ is open in $\struct {X/N, \tau_N}$

by Kernel of Quotient Mapping.

Since $\pi$ is continuous and surjective, if $\pi \sqbrk {X \setminus N}$ is open in $\struct {X/N, \tau_N}$, then $X \setminus N$ is open in $\struct {X, \tau}$.

Conversely, since $\pi$ is an open mapping, if $X \setminus N$ is open in $\struct {X, \tau}$, then $\pi \sqbrk {X \setminus N}$ is open in $\struct {X/N, \tau_N}$.

So we have:
 * $\pi \sqbrk {X \setminus N}$ is open in $\struct {X/N, \tau_N}$


 * $X \setminus N$ is open in $\struct {X, \tau}$.
 * $X \setminus N$ is open in $\struct {X, \tau}$.

Finally, we have:
 * $X \setminus N$ is open in $\struct {X, \tau}$


 * $N$ is closed in $\struct {X, \tau}$.
 * $N$ is closed in $\struct {X, \tau}$.

Hence we have shown that $\struct {X/N, \tau_N}$ is Hausdorff :
 * $N$ is a closed linear subspace.