description/proof of that for topological space induced by metric induced by norm on vectors space, norm map is continuous
Topics
About: vectors space
About: topological space
The table of contents of this article
Starting Context
- The reader knows a definition of metric induced by norm on real or complex vectors space.
- The reader knows a definition of topology induced by metric.
- The reader knows a definition of Euclidean topological space.
- The reader knows a definition of continuous, topological spaces map.
Target Context
- The reader will have a description and a proof of the proposition that for any vectors space with any norm and the topological space induced by the metric induced by the norm, the norm map is continuous.
Orientation
There is a list of definitions discussed so far in this site.
There is a list of propositions discussed so far in this site.
Main Body
1: Structured Description
Here is the rules of Structured Description.
Entities:
\(F\): \(\in \{\mathbb{R}, \mathbb{C}\}\), with the canonical field structure
\(V\): \(\in \{\text{ the } F \text{ vectors spaces }\}\), with any norm, \(\Vert \bullet \Vert: V \to \mathbb{R}\), with the topology induced by the metric induced by \(\Vert \bullet \Vert\)
\(\mathbb{R}\): \(= \text{ the Euclidean topological space }\)
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Statements:
\(\Vert \bullet \Vert \in \{\text{ the continuous maps }\}\)
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2: Proof
Whole Strategy: Step 1: for each \(v \in V\) and each open neighborhood of \(\Vert v \Vert\), \(U_{\Vert v \Vert}\), take an open ball around \(v\), \(B_{v, \epsilon}\), such that \(\Vert B_{v, \epsilon} \Vert \subseteq U_{\Vert v \Vert}\).
Let \(v \in V\) be any.
Let \(U_{\Vert v \Vert} \subseteq \mathbb{R}\) be any open neighborhood of \(\Vert v \Vert\).
There is an open ball around \(\Vert v \Vert\), \(B_{\Vert v \Vert, \epsilon} \subseteq \mathbb{R}\), such that \(B_{\Vert v \Vert, \epsilon} \subseteq U_{\Vert v \Vert}\), by the definition of Euclidean topological space.
Let us take the open ball around \(v\), \(B_{v, \epsilon} \subseteq V\).
Let \(v' \in B_{v, \epsilon}\) be any.
\(\Vert v' - v \Vert \lt \epsilon\).
\(\Vert v' \Vert = \Vert v' - v + v \Vert \leq \Vert v' - v \Vert + \Vert v \Vert \lt \epsilon + \Vert v \Vert\).
So, \(\Vert v' \Vert - \Vert v \Vert \lt \epsilon\).
\(\Vert v \Vert = \Vert v - v' + v' \Vert \leq \Vert v - v' \Vert + \Vert v' \Vert \lt \epsilon + \Vert v' \Vert\).
So, \(\Vert v \Vert - \Vert v' \Vert \lt \epsilon\).
So, \(\vert \Vert v' \Vert - \Vert v \Vert \vert \lt \epsilon\).
So, \(\Vert v' \Vert \in B_{\Vert v \Vert, \epsilon}\).
That means that \(\Vert B_{v, \epsilon} \Vert \subseteq B_{\Vert v \Vert, \epsilon} \subseteq U_{\Vert v \Vert}\).
So, \(\Vert \bullet \Vert\) is continuous at \(v\).
As \(v \in V\) is arbitrary, \(\Vert \bullet \Vert\) is continuous.
3: Note
While we have proved the proposition that for any finite-dimensional normed real vectors space with the canonical topology, the norm map is continuous and the proposition that for any finite-dimensional normed complex vectors space with the canonical topology, the norm map is continuous, the finite-dimensional requirements are really unnecessary.
