description/proof of that for finite-'direct sum' of normed vectors spaces with topologies induced by norms, product norm induces product topology
Topics
About: vectors space
About: topological space
The table of contents of this article
Starting Context
- The reader knows a definition of normed vectors space.
- 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 product topology.
- The reader knows a definition of product norm on finite-'direct sum' of normed vectors spaces.
- The reader admits the local criterion for openness.
Target Context
- The reader will have a description and a proof of the proposition that for any finite-'direct sum' of normed vectors spaces with the topologies induced by the norms, the product norm induces the product topology.
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
\(J\): \(\in \{\text{ the finite index sets }\}\)
\(\{V_j \vert j \in J\}\): \(V_j \in \{\text{ the } F \text{ vectors spaces }\}\) with any norm, \(\Vert \bullet \Vert_j\)
\(\oplus_{j \in J} V_j\): \(= \text{ the direct sum }\)
\(\Vert \bullet \Vert\): \(= \text{ the product norm over } \oplus_{j \in J} V_j\)
\(O\): \(= \text{ the product topology for } \oplus_{j \in J} V_j \text{ with } V_J \text{ s given the topologies induced by } \Vert \bullet \Vert_j \text{ s }\)
\(O'\): \(= \text{ the topology for } \oplus_{j \in J} V_j \text{ induced by }\Vert \bullet \Vert\)
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Statements:
\(O = O'\)
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2: Proof
Whole Strategy: Step 1: let \(U \in O\) be any, and see that \(U \in O'\); Step 2: let \(U \in O'\) be any, and see that \(U \in O\).
Step 1:
Let \(U \in O\) be any.
Let us see that \(U \in O'\).
Let \(u \in U\) be any.
There is an open neighborhood of \(u\), \(U_u \in O\), such that \(U_u \subseteq U\), by the local criterion for openness.
There are some open neighborhoods of \(u_j\) s, \(\{U_{u_j} \subseteq V_j\}\), such that \(\times_{j \in J} U_{u_j} \subseteq U_u\), by the definition of product topology.
As \(U_{u_J}\) is open in the topology induced by \(\Vert \bullet \Vert_j\), there is an open ball around \(u_j\), \(B_{u_j, \epsilon_j} \subseteq V_j\), such that \(B_{u_j, \epsilon_j} \subseteq U_{u_J}\).
Let us take \(\epsilon = min \{\epsilon_j \vert j \in J\}\).
\(u \in \times_{j \in J} B_{u_j, \epsilon} \subseteq \times_{j \in J} U_{u_j} \subseteq U_u\).
Let us see that \(B_{u, \epsilon} \subseteq \times_{j \in J} B_{u_j, \epsilon}\).
Let \(u' \in B_{u, \epsilon}\) be any.
\(\Vert u' - u \Vert = \sqrt{\sum_{j \in J} {\Vert u'_j - u_j \Vert_j}^2} \lt \epsilon\).
That means that \(\Vert u'_j - u_j \Vert_j \lt \epsilon\).
That means that \(u' \in \times_{j \in J} B_{u_j, \epsilon}\).
So, \(u \in B_{u, \epsilon} \subseteq \times_{j \in J} B_{u_j, \epsilon} \subseteq U_u \subseteq U\).
So, \(U \in O'\).
Step 2:
Let \(U \in O'\) be any.
Let us see that \(U \in O\).
Let \(u \in U\) be any.
There is an open ball around \(u\), \(B_{u, \epsilon} \subseteq \oplus_{j \in J} V_j\), such that \(B_{u, \epsilon} \subseteq U\).
Let us see that for \(\delta := \epsilon / \sqrt{\vert J \vert}\), \(\times_{j \in J} B_{u_j, \delta} \subseteq B_{u, \epsilon}\).
Let \(u' \in \times_{j \in J} B_{u_j, \delta}\) be any.
\(\Vert u' - u \Vert = \sqrt{\sum_{j \in J} {\Vert u'_j - u_j \Vert_j}^2} \lt \sqrt{\sum_{j \in J} \delta^2} = \sqrt{\vert J \vert \delta^2} = \epsilon\).
So, \(u' \in B_{u, \epsilon}\).
So, \(u \in \times_{j \in J} B_{u_j, \delta} \subseteq B_{u, \epsilon} \subseteq U\).
But \(B_{u_j, \delta} \subseteq V_j\) is an open neighborhood of \(u_j\) in the topology induced by \(\Vert \bullet \Vert_j\), and \(\times_{j \in J} B_{u_j, \delta}\) is an open neighborhood of \(u\) in \(O\).
So, by the local criterion for openness, \(U \in O\).
