# Topology induced by a norm

I came across the notion of a $$\textit{topology induced by a norm}$$.

If $$(X,\Vert\ . \Vert)$$ is a normed space w.r.t a norm $$\Vert\ . \Vert: X \to \mathbb{R}$$. Most sources define the topology $$\tau$$ on $$X$$ induced by $$\Vert\ . \Vert$$ as the sets $$U \subset X$$ open w.r.t. the metric $$d: X \times X \to \mathbb{R}$$ given by $$d(x,y) = \Vert x - y \Vert$$.

But would I be correct in assuming that an equivalent definition would be $$\tau = \{\Vert\ . \Vert^{-1}(U) \mid U \subset \mathbb{R}\ \textrm{open} \}$$?

• Note that the preimage of U would be a subset of $X\times X$, not $X$ – Ottavio Bartenor Feb 8 at 17:42
• Wouldnt you be applying the norm preimage to a real number (representing the norm) and trying to get the original set which has that value as a norm back out? Or am I misreading? – SquishyRhode Feb 8 at 17:46
• We know topologies are induced by metrics. But norms induce metrics too. So a norm should imply a metric, which then implies a topology. Assuming all other criteria are met. – SquishyRhode Feb 8 at 17:48
• @BrianMoehring My bad, didn't pay enough attention – Ottavio Bartenor Feb 8 at 17:48
• I think you only get a neighborhood basis at $0$ with this definition, but that uniquely determines the topology, so in a sense it is sufficient. – TSU Feb 8 at 17:48

With your version and $$\Bbb R$$ with norm $$|\cdot|$$, the set $$(-3,-2)\cup(2,3)$$ would be open, but $$(2,3)$$ not.
Note that your sets $$U'=\|\cdot\|^{-1}[U]$$ are always symmetrical in that $$x \in U$$ implies $$-x \in U'$$. Not all open sets will obey that. These sets will give the open balls around the origin for $$U$$ that are symmetric around $$0$$, but not all other translated balls.
In fact $$\tau$$ is the topology induced by $$\{ \lVert. \rVert^{-1}(U): U\subseteq \mathbb{R} \ \ \text{is Open} \ \}$$