# The completeness of the quotient space (topological vector space)

## The Goal

We are going to show the completeness of \(X/N\) where \(X\) is a TVS and \(N\) a closed subspace. Alongside, a bunch of useful analysis tricks will be demonstrated (and that's why you may find this blog post a little tedious.). But what's more important, the theorem proved here will be used in the future.

## The main process

To make it clear, we should give a formal definition of \(F\)-space.

A topological space \(X\) is an \(F\)-space if its topology \(\tau\) is induced by a complete invariant metric \(d\).

A metric \(d\) on a vector space \(X\) will be called invariant if for all \(x,y,z \in X\), we have \[ d(x+z,y+z)=d(x,y). \] By complete we mean every Cauchy sequence of \((X,d)\) converges.

### Defining the quotient metric \(\rho\)

The metric can be inherited to the quotient space naturally (we will use this fact latter), that is

If \(X\) is a \(F\)-space, \(N\) is a closed subspace of a topological vector space \(X\), then \(X/N\) is still a \(F\)-space.

Suppose \(d\) is a complete invariant metric compatible with \(\tau_X\). The metric on \(X/N\) is defined by \[ \boxed{\rho(\pi(x),\pi(y))=\inf_{z \in N}d(x-y,z)} \] ### \(\rho\) is a metric

*Proof.* First, if \(\pi(x)=\pi(y)\), that is, \(x-y \in N\), we see \[
\rho(\pi(x),\pi(y))=\inf_{z \in N}d(x-y,z)=d(x-y,x-y)=0.
\] If \(\pi(x) \neq \pi(y)\) however, we shall show that \(\rho(\pi(x),\pi(y))>0\). In this case, we have \(x-y \notin N\). Since \(N\) is closed, \(N^c\) is open, and \(x-y\) is an interior point of \(X-N\). Therefore there exists an open ball \(B_r(x-y)\) centered at \(x-y\) with radius \(r>0\) such that \(B_r(x-y) \cap N = \varnothing\). Notice we have \(d(x-y,z)>r\) since otherwise \(z \in B_r(x-y)\). By putting \[
r_0=\sup\{r:B_r(x-y) \cap N = \varnothing\},
\] we see \(d(x-y,z) \geq r_0\) for all \(z \in N\) and indeed \(r_0=\inf_{z \in N}d(x-y,z)>0\) (the verification can be done by contradiction). In general, \(\inf_z d(x-y,z)=0\) if and only if \(x-y \in \overline{N}\).

Next, we shall show that \(\rho(\pi(x),\pi(y))=\rho(\pi(y),\pi(x))\), and it suffices to assume that \(\pi(x) \neq \pi(y)\). Sgince \(d\) is translate invariant, we get \[ \begin{aligned} d(x-y,z)&=d(x-y-z,0) \\ &=d(0,y-x+z) \\ &=d(-z,y-x) \\ &=d(y-x,-z). \end{aligned} \] Therefore the \(\inf\) of the left hand is equal to the one of the right hand. The identity is proved.

Finally, we need to verify the triangle inequality. Let \(r,s,t \in X\). For any \(\varepsilon>0\), there exist some \(z_\varepsilon\) and \(z_\varepsilon'\) such that \[
d(r-s,z_\varepsilon)<\rho(\pi(r),\pi(s))+\frac{\varepsilon}{2},\quad d(s-t,z'_\varepsilon)<\rho(\pi(s),\pi(t))+\frac{\varepsilon}{2}.
\] Since \(d\) is invariant, we see \[
\begin{aligned}
d(r-t,z_\varepsilon+z'_\varepsilon)&=d((r-s)+(s-t)-(z_\varepsilon+z'_\varepsilon),0) \\
&=d([(r-s)-z_\varepsilon]+[(s-t)-z'_\varepsilon],0) \\
&=d(r-s-z_\varepsilon,t-s+z'_\varepsilon) \\
&\leq d(r-s-z_\varepsilon,0)+d(t-s+z'_\varepsilon,0) \\
&=d(r-s,z_\varepsilon)+d(s-t,z'_\varepsilon)
\end{aligned}
\] *(I owe [@LeechLattice](https://onp4.com/@leechlattice) for the inequality above.)*

Therefore \[
\begin{aligned}
d(r-t,z_\varepsilon+z'_\varepsilon)&\leq d(r-s,z_\varepsilon)+d(s-t,z'_\varepsilon) \\
&<\rho(\pi(r),\pi(s))+\rho(\pi(s),\pi(t))+\varepsilon.
\end{aligned}
\] *(Warning: This does not imply that \(\rho(\pi(r),\pi(s))+\rho(\pi(s),\pi(t))=\inf_z d(r-t,z)\) since we don't know whether it is the lower bound or not.)*

If \(\rho(\pi(r),\pi(s))+\rho(\pi(s),\pi(t))<\rho(\pi(r),\pi(t))\) however, let \[ 0<\varepsilon<\rho(\pi(r),\pi(t))-(\rho(\pi(r),\pi(s))+\rho(\pi(s),\pi(t))) \] then there exists some \(z''_\varepsilon=z_\varepsilon+z'_\varepsilon\) such that \[ d(r-t,z''_\varepsilon)<\rho(\pi(r),\pi(t)) \] which is a contradiction since \(\rho(\pi(r),\pi(t)) \leq d(r-t,z)\) for all \(z \in N\).

*(We are using the \(\varepsilon\) definition of \(\inf\). See here.)*

### \(\rho\) is translate invariant

Since \(\pi\) is surjective, we see if \(u \in X/N\), there exists some \(a \in X\) such that \(\pi(a)=u\). Therefore \[ \begin{aligned} \rho(\pi(x)+u,\pi(y)+u) &=\rho(\pi(x)+\pi(a),\pi(y)+\pi(a)) \\ &=\rho(\pi(x+a),\pi(y+a)) \\ &=\inf_{z \in N}d(x+a-y-a,z) \\ &=\rho(\pi(x),\pi(y)). \end{aligned} \]

### \(\rho\) is well-defined

If \(\pi(x)=\pi(x')\) and \(\pi(y)=\pi(y')\), we have to show that \(\rho(\pi(x),\pi(y))=\rho(\pi(x'),\pi(y'))\). In fact, \[ \begin{aligned} \rho(\pi(x),\pi(y)) &\leq \rho(\pi(x),\pi(x'))+\rho(\pi(x'),\pi(y'))+\rho(\pi(y'),\pi(y)) \\ &=\rho(\pi(x'),\pi(y')) \end{aligned} \] since \(\rho(\pi(x),\pi(x'))=0\) as \(\pi(x)=\pi(x')\). Meanwhile \[ \begin{aligned} \rho(\pi(x'),\pi(y')) &\leq \rho(\pi(x'),\pi(x)) + \rho(\pi(x),\pi(y)) + \rho(\pi(y),\pi(y')) \\ &= \rho(\pi(x),\pi(y)). \end{aligned} \] therefore \(\rho(\pi(x),\pi(y))=\rho(\pi(x'),\pi(y'))\).

### \(\rho\) is compatible with \(\tau_N\)

By proving this, we need to show that a set \(E \subset X/N\) is open with respect to \(\tau_N\) if and only if \(E\) is a union of open balls. But we need to show a generalized version:

If \(\mathscr{B}\) is a local base for \(\tau\), then the collection \(\mathscr{B}_N\), which contains all sets \(\pi(V)\) where \(V \in \mathscr{B}\), forms a local base for \(\tau_N\).

*Proof.* We already know that \(\pi\) is continuous, linear and open. Therefore \(\pi(V)\) is open for all \(V \in \mathscr{B}\). For any open set around \(E \subset X/N\) containing \(\pi(0)\), we see \(\pi^{-1}(E)\) is open, and we have \[
\pi^{-1}(E)=\bigcup_{V\in\mathscr{B}}V
\] and therefore \[
E=\bigcup_{V \in \mathscr{B}}\pi(V).
\]

Now consider the local base \(\mathscr{B}\) containing all open balls around \(0 \in X\). Since \[ \pi(\{x:d(x,0)<r\})=\{u:\rho(u,\pi(0))<r\} \] we see \(\rho\) determines \(\mathscr{B}_N\). But we have already proved that \(\rho\) is invariant; hence \(\mathscr{B}_N\) determines \(\tau_N\).

### If \(d\) is complete, then \(\rho\) is complete.

Once this is proved, we are able to claim that, if \(X\) is a \(F\)-space, then \(X/N\) is still a \(F\)-space, since its topology is induced by a complete invariant metric \(\rho\).

*Proof.* Suppose \((x_n)\) is a Cauchy sequence in \(X/N\), relative to \(\rho\). There is a subsequence \((x_{n_k})\) with \(\rho(x_{n_k},x_{n_{k+1}})<2^{-k}\). Since \(\pi\) is surjective, we are able to pick some \(z_k \in X\) such that \(\pi(z_k) = x_{n_k}\) and such that \[
d(z_{k},z_{k+1})<2^{-k}.
\] (The existence can be verified by contradiction still.) By the inequality above, we see \((z_k)\) is Cauchy (can you see why?). Since \(X\) is complete, \(z_k \to z\) for some \(z \in X\). By the **continuity** of \(\pi\), we also see \(x_{n_k} \to \pi(z)\) as \(k \to \infty\). Therefore \((x_{n_k})\) converges. Hence \((x_n)\) converges since it has a convergent subsequence. \(\rho\) is complete.

## Remarks

This fact will be used to prove some corollaries in the open mapping theorem. For instance, for any continuous linear map \(\Lambda:X \to Y\), we see \(\ker(\Lambda)\) is closed, therefore if \(X\) is a \(F\)-space, then \(X/\ker(\Lambda)\) is a \(F\)-space as well. We will show in the future that \(X/\ker(\Lambda)\) and \(\Lambda(X)\) are homeomorphic if \(\Lambda(X)\) is of the second category.

There are more properties that can be inherited by \(X/N\) from \(X\). For example, normability, metrizability, local convexity. In particular, if \(X\) is Banach, then \(X/N\) is Banach as well. To do this, it suffices to define the quotient norm by \[ \lVert \pi(x) \rVert = \inf\{\lVert x-z \rVert:z \in N\}. \]