Quotient Topological Vector Space is Hausdorff iff Linear Subspace is Closed/Proof 2
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 if and only if:
- $N$ is a closed linear subspace.
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 $\pi \sqbrk {N^-} = \set { {\mathbf 0}_{X/N} }$ if and only if $N^- = N$.
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 $\pi \sqbrk {N^-} = \set { {\mathbf 0}_{X/N} }$ if and only if $N^- = N$.
From Set is Closed iff Equals Topological Closure, we therefore have that $\pi \sqbrk {N^-} = \set { {\mathbf 0}_{X/N} }$ if and only if $N$ is closed.
That is, $N$ is closed if and only if:
- $\ds \bigcap_{U \in \VV_N} U = \set { {\mathbf 0}_{X/N} }$
From Characterization of Hausdorff Topological Vector Space, we conclude that $N$ is closed if and only if $X/N$ is Hausdorff.
$\blacksquare$