Quotient Topological Vector Space is Hausdorff iff Linear Subspace is Closed/Proof 2

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

Let $\VV$ be the set of open neighborhoods of ${\mathbf 0}_X$.

Let $\VV_N$ be the set of open neighborhoods of ${\mathbf 0}_{X/N}$.

Then, we have, from Expression for Closure of Set in Topological Vector Space:
 * $\ds N = \bigcap_{V \mathop \in \VV} \paren {N + V}$

From Preimage of Image of Linear Transformation, we have $\pi^{-1} \sqbrk {\pi \sqbrk V} = \ker \pi + V$.

From Kernel of Quotient Mapping, we then obtain $\pi^{-1} \sqbrk {\pi \sqbrk V} = N + V$.

Hence, we have:
 * $\ds N^- = \bigcap_{V \mathop \in \VV} \pi^{-1} \sqbrk {\pi \sqbrk V}$

From Preimage of Intersection under Mapping: Family of Sets, we obtain:
 * $\ds N^- = \pi^{-1} \sqbrk {\bigcap_{V \mathop \in \VV} \pi \sqbrk V}$

Since $\pi$ is surjective, we obtain:
 * $\ds \pi \sqbrk {N^-} = \pi \sqbrk {\pi^{-1} \sqbrk {\bigcap_{V \mathop \in \VV} \pi \sqbrk V} } = \bigcap_{V \mathop \in \VV} \pi \sqbrk V$

From Open Set in Quotient Topological Vector Space, we have:
 * $\VV_N = \set {\pi \sqbrk U : U \in \VV}$

Hence, we have:
 * $\ds \pi \sqbrk {N^-} = \bigcap_{U \in \VV_N} U$

We argue that $\pi \sqbrk {N^-} = \set { {\mathbf 0}_{X/N} }$ $N^- = N$.

If $N^- = N$, then we immediately have $\pi \sqbrk {N^-} = \pi \sqbrk N = \set { {\mathbf 0}_{X/N} }$.

If $N^- \ne N$, then there exists $x_0 \in N^- \setminus N$.

From Kernel of Quotient Mapping, we then have $\map \pi {x_0} \ne {\mathbf 0}_{X/N}$ and hence $\pi \sqbrk {N^-} \ne \set { {\mathbf 0}_{X/N} }$.

Hence we have $\pi \sqbrk {N^-} = \set { {\mathbf 0}_{X/N} }$ $N^- = N$.

From Set is Closed iff Equals Topological Closure, we therefore have that $\pi \sqbrk {N^-} = \set { {\mathbf 0}_{X/N} }$ $N$ is closed.

That is, $N$ is closed :
 * $\ds \bigcap_{U \in \VV_N} U = \set { {\mathbf 0}_{X/N} }$

From Characterization of Hausdorff Topological Vector Space, we conclude that $N$ is closed $X/N$ is Hausdorff.