Quotient topology and normed vector spaces

This is a question on an exercise in topology, more precisely in quotient spaces.

Let $(E,||\cdot ||)$ be a normed vector space and $F\subseteq E$ a subspace. In the vector space $\widetilde{E}=E/F$ we define the map $p:\widetilde{E}\rightarrow\mathbb{R}$ with $p([x])=inf\{||x-y||:y\in F\}$.

I am wondering if the quotient topology $\mathcal{T}_\pi$ on $\widetilde{E}$ is the same as the smallest (weak) topology $\mathcal{T}_p$ on $\widetilde{E}$ such that $p$ is continuous (that is, the sets $p^{-1}(G)$ such that $G$ is open in $\mathbb{R}$).

In the notation, $\pi$ denotes the canonical projection $E\rightarrow\widetilde{E}$.

$p([x])$ is actually equal to $dist(x,F)$, the infimum of the distances of $x$ from the points of $F$.

Firstly, it is not hard to prove that $\mathcal{T}_p\subseteq \mathcal{T}_\pi$, using that $p\circ\pi$ is continuous. This is true because $E$ is also a metric space with the metric induced by the norm and in that case, $p\circ\pi$ is the usual $dist$ function on metric spaces and we know that it is continuous.

My question occurs when showing that $\mathcal{T}_\pi\subseteq \mathcal{T}_p$ holds. Let $U$ be open with the quotient topology, then we need to find an open subset $G$ of $\mathbb{R}$ such that $p^{-1}(G)=U$. This is the same as showing that $p$ is open under the quotient topology.

Under what conditions does this hold?

Finally, the textbook mentions that if $F$ is a closed subspace, then $p$ is a norm on $\widetilde{E}$. Is it true that the topology induced by a norm is the same as the smallest topology such that the norm is continuous as a map? If so, why? We know that in metric spaces, norms are continuous maps.

• On the last paragraph - I don't think so, for example in $\mathbb{R}^2$ the smallest topology making the norm continuous won't separate points at the same distance from the origin. – Daniel Schepler Aug 2 '17 at 22:47
• @DanielSchepler Indeed, I just saw that it is also stated here math.stackexchange.com/questions/2201878/… – iolig Aug 2 '17 at 22:49
• What you want to prove seems false to me. Maybe $\mathcal{T}_p$ is the topology induced by the norm $p$? – Idéophage Aug 2 '17 at 22:52
• @Idéophage $F$ is an arbitrary subspace, so $p$ might not be a norm. – iolig Aug 2 '17 at 22:56
• Concretely, it is $d(x,y) = d(\pi^{-1}(x), \pi^{-1}(y))$. But you should first do the theory of metric spaces I think. – Idéophage Aug 2 '17 at 23:26

1 Answer

Is it true that the topology induced by a norm is the same as the smallest topology such that the norm is continuous as a map?

No, this is not true. This is not even true in dimension $1$: the topology induced on $\mathbb{R}$ by the absolute value $\mathbb{R} \to \mathbb{R}_{\geq 0}$ does not distinguish a point and its opposite since they are sent on the same point (their neighborhoods are exactly the same).

The topologies induced by $\pi$ and $p$ never coincide (except when $F$ is dense), for the same reason. But the topology induced by the seminorm $p$ is always the same as the topology induced by $\pi$.

Given a metric space $E$ and a function $f : E \to F$, we define the metric induced by $f$ on $F$ as the “smallest metric such that $f$ does not increase the distances” (this can be made exactly the same definition as the quotient topology, in just a different setting). Concretely, this metric is defined as $d(x,y) := d(f^{-1}(x),f^{-1}(y))$ or alternatively $d(x,y) = \inf\,\{d(x',y') \,|\, f(x')=x, f(y')=y\}$. Note that $d(x,y)$ can be infinite or zero even when $x \neq y$ (I think a name for it would be an $\infty$-semimetric space).

With that construction, $F$ becomes a metric space $(F,d_f)$. So it induces a topology $\tau(d_f)$ on $F$. But $f$, as a function from a topological space to a set, also induces directly a topology $\tau_f$. Then $\tau_f$ is always finer than $\tau(d_f)$, since $f : E \to (F,\tau(d_f))$ is continuous because it is $1$-lipschitz. In the case of a quotient of a normed vector space, the converse is also true and these two topologies are equal, but not in general.

To prove that $\tau_\pi \subseteq \tau(d_\pi)$, take an open set $O \in \tau_\pi$ and a point $x \in O$. The goal is to prove that $d_\pi(x,\complement O) > 0$. Choose $x' \in \pi^{-1}(x)$. The distance we want to compute is then \begin{align*} d(\pi^{-1}(x),\pi^{-1}(\complement O)) &= d(x'+F,\pi^{-1}(\complement O))\\ &= d(x',\pi^{-1}(\complement O)-F)\\ &= d(x',\pi^{-1}(\complement O))\\ &= d(x',\complement \pi^{-1}(O))\text{.} \end{align*} But this is greater than $0$ since $\pi^{-1}(O)$ is open and that $x' \in \pi^{-1}(O)$.

Moreover, the metric induced by $p$ is $d_\pi$.