description/proof of that for finite-dimensional vectors space with inner product, topology induced by inner product is canonical topology
Topics
About: vectors space
The table of contents of this article
Starting Context
- The reader knows a definition of norm induced by inner product on real or complex 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 canonical topology for finite-dimensional real vectors space.
- The reader knows a definition of canonical topology for finite-dimensional complex vectors space.
- The reader admits the proposition that any countable-dimensional vectors space with any inner product has an orthonormal basis.
- 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-dimensional vectors space with any inner product, the topology induced by the inner product is the canonical 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
\(d\): \(\in \mathbb{N} \setminus \{0\}\)
\(V\): \(\in \{\text{ the } d \text{ -dimensional } F \text{ vectors spaces }\}\), with any inner product
\(O_i\): \(= \text{ the topology induced by the inner product for } V\)
\(O_c\): \(= \text{ the canonical topology for } V\)
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Statements:
\(O_i = O_c\)
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2: Proof
Whole Strategy: Step 1: take any orthonormal basis for \(V\), \(B = \{b_1, ..., b_d\}\); Step 2: see that each open ball with respect to \(O_i\) is an open ball with respect to \(O_c\) and vice versa; Step 3: see that for each \(U_i \in O_i\), \(U_i \in O_c\); Step 4: see that for each \(U_c \in O_c\), \(U_c \in O_i\); Step 5: conclude the proposition.
Step 1:
Let us take any orthonormal basis for \(V\), \(B = \{b_1, ..., b_d\}\), by the proposition that any countable-dimensional vectors space with any inner product has an orthonormal basis.
Note that \(O_c\) does not depend on the choice of basis, as is mentioned in the definition of canonical topology for finite-dimensional real vectors space or the definition of canonical topology for finite-dimensional complex vectors space.
Let \(f': V \to F^d\) be the canonical homeomorphism with respect to \(B\) by which \(O_c\) is defined.
When \(F = \mathbb{R}\), let \(f := f'\).
When \(F = \mathbb{C}\), let \(f'': \mathbb{R}^{2 d} \to \mathbb{C}^d\) be the canonical homeomorphism, and let \(f := f''^{-1} \circ f': V \to \mathbb{R}^{2 d}\), where \(f\) is a homeomorphism, obviously.
Step 2:
'the \(\epsilon\)-'open ball' around \(v \in V\) for \(O_c\)', \(B_{c, v, \epsilon} \subseteq V\), means \(f^{-1} (B_{f (v), \epsilon})\), where \(B_{f (v), \epsilon} \subseteq \mathbb{R}^d \text{ or } \mathbb{R}^{2 d}\) is the \(\epsilon\)-'open ball' around \(f (v) \in \mathbb{R}^d \text{ or } \mathbb{R}^{2 d}\) on the Euclidean topological space, \(\mathbb{R}^d \text{ or } \mathbb{R}^{2 d}\).
\(B_{c, v, \epsilon} \in O_c\), because \(f\) is a homeomorphism.
Let us see that each open ball with respect to \(O_i\) is an open ball with respect to \(O_c\).
Let \(v \in V\) and \(\epsilon \in \mathbb{R}\) such that \(0 \lt \epsilon\) be any.
Let \(B_{i, v, \epsilon} \subseteq V\) be the open ball around \(v\) for \(O_i\).
Let us see that \(B_{i, v, \epsilon} = B_{c, v, \epsilon}\).
Let \(p \in B_{i, v, \epsilon}\) be any.
\(\langle p - v, p - v \rangle \lt \epsilon^2\).
But \(p - v = p^j b_j - v^j b_j = (p^j - v^j) b_j\), so, \(\langle p - v, p - v \rangle = \langle (p^j - v^j) b_j, (p^j - v^j) b_j \rangle = \sum_{j \in \{1, ..., d\}} \vert p^j - v^j \vert^2\), because \(B\) is orthonormal, so, \(\sum_{j \in \{1, ..., d\}} \vert p^j - v^j \vert^2 \lt \epsilon^2\).
Let \(F = \mathbb{R}\).
\(\sum_{j \in \{1, ..., d\}} (p^j - v^j)^2 = \sum_{j \in \{1, ..., d\}} \vert p^j - v^j \vert^2 \lt \epsilon^2\).
\(f (p) - f (v) = (p^1, ..., p^d) - (v^1, ..., v^d) = (p^1 - v^1, ..., p^d - v^d)\), and \(\sum_{j \in \{1, ..., d\}} (p^j - v^j)^2 \lt \epsilon^2\) means that \(f (p) \in B_{f (v), \epsilon}\)
So, \(p \in f^{-1} (B_{f (v), \epsilon}) = B_{c, v, \epsilon}\).
So, \(B_{i, v, \epsilon} \subseteq B_{c, v, \epsilon}\).
Let \(p \in B_{c, v, \epsilon}\) be any.
\(f (p) \in f (B_{c, v, \epsilon}) = B_{f (v), \epsilon}\).
That means that \(\sum_{j \in \{1, ..., d\}} (p^j - v^j)^2 \lt \epsilon^2\).
\(\langle p - v, p - v \rangle = \sum_{j \in \{1, ..., d\}} (p^j - v^j)^2\), as before.
So, \(\langle p - v, p - v \rangle \lt \epsilon^2\).
So, \(p \in B_{i, v, \epsilon}\).
So, \(B_{c, v, \epsilon} \subseteq B_{i, v, \epsilon}\).
So, \(B_{i, v, \epsilon} = B_{c, v, \epsilon}\).
Let \(F = \mathbb{C}\).
\(\sum_{j \in \{1, ..., d\}} ((Re (p^j) - Re (v^j))^2 + (Im (p^j) - Im (v^j))^2) = \sum_{j \in \{1, ..., d\}} ((Re (p^j - v^j))^2 + (Im (p^j - v^j))^2) = \sum_{j \in \{1, ..., d\}} \vert p^j - v^j \vert^2 \lt \epsilon^2\).
\(f (p) - f (v) = (Re (p^1), Im (p^1)..., Re (p^d), Im (p^d)) - (Re (v^1), Im (v^1)..., Im (v^d), Im (v^d)) = (Re (p^1) - Re (v^1), Im (p^1) - Im (v^1), ..., Re (p^d) - Re (v^d), Im (p^d) - Im (v^d))\), and \(\sum_{j \in \{1, ..., d\}} ((Re (p^j) - Re (v^j))^2 + (Im (p^j) - Im (v^j))^2) \lt \epsilon^2\) means that \(f (p) \in B_{f (v), \epsilon}\).
So, \(p \in f^{-1} (B_{f (v), \epsilon}) = B_{c, v, \epsilon}\).
So, \(B_{i, v, \epsilon} \subseteq B_{c, v, \epsilon}\).
Let \(p \in B_{c, v, \epsilon}\) be any.
\(f (p) \in f (B_{c, v, \epsilon}) = B_{f (v), \epsilon}\).
That means that \(\sum_{j \in \{1, ..., d\}} ((Re (p^j) - Re (v^j))^2 + (Im (p^j) - Im (v^j))^2)\lt \epsilon^2\).
\(\langle p - v, p - v \rangle = \sum_{j \in \{1, ..., d\}} ((Re (p^j) - Re (v^j))^2 + (Im (p^j) - Im (v^j))^2)\), as before.
So, \(\langle p - v, p - v \rangle \lt \epsilon^2\).
So, \(p \in B_{i, v, \epsilon}\).
So, \(B_{c, v, \epsilon} \subseteq B_{i, v, \epsilon}\).
So, \(B_{i, v, \epsilon} = B_{c, v, \epsilon}\).
On the other hand, each open ball with respect to \(O_c\) is an open ball with respect to \(O_i\), because \(B_{c, v, \epsilon} = B_{i, v, \epsilon}\).
Step 3:
Let us see that for each \(U_i \in O_i\), \(U_i \in O_c\).
Let \(p \in U_i\) be any.
There is a \(B_{i, p, \epsilon} \subseteq V\) such that \(B_{i, p, \epsilon} \subseteq U_i\).
But as \(B_{c, p, \epsilon} = B_{i, p, \epsilon}\) by Step 2, \(B_{c, p, \epsilon} \subseteq U_i\).
By the local criterion for openness, \(U_i \in O_c\).
Step 4:
Let us see that for each \(U_c \in O_c\), \(U_c \in O_i\).
Let \(p \in U_c\) be any.
There is a \(B_{c, p, \epsilon} \subseteq V\) such that \(B_{c, p, \epsilon} \subseteq U_c\).
But as \(B_{i, p, \epsilon} = B_{c, p, \epsilon}\) by Step 2, \(B_{i, p, \epsilon} \subseteq U_c\).
By the local criterion for openness, \(U_c \in O_i\).
Step 5:
So, \(O_i = O_c\).
