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

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  • $\begingroup$ Note that the preimage of U would be a subset of $X\times X$, not $X$ $\endgroup$ – Ottavio Bartenor Feb 8 at 17:42
  • $\begingroup$ 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? $\endgroup$ – SquishyRhode Feb 8 at 17:46
  • $\begingroup$ 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. $\endgroup$ – SquishyRhode Feb 8 at 17:48
  • $\begingroup$ @BrianMoehring My bad, didn't pay enough attention $\endgroup$ – Ottavio Bartenor Feb 8 at 17:48
  • $\begingroup$ 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. $\endgroup$ – TSU Feb 8 at 17:48
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With your version and $\Bbb R$ with norm $|\cdot|$, the set $(-3,-2)\cup(2,3)$ would be open, but $(2,3)$ not.

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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.

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

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  • $\begingroup$ @Brian Moehring Thankyou. I edited it $\endgroup$ – Jo Jomax Feb 9 at 6:22

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