Topology induced by a norm

Question

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 $\tau = \{\Vert\ . \Vert^{-1}(U) \mid U \subset \mathbb{R}\ \textrm{open} \}$?

Answer 1

With your version and $\Bbb R$ with norm $|\cdot|$, the set $(-3,-2)\cup(2,3)$ would be open, but $(2,3)$ not.

Answer 2

In fact $\tau$ is the topology induced by $\{ \lVert. \rVert^{-1}(U): U\subseteq \mathbb{R} \ \ \text{is Open} \ \} $

Answer 3

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.