description/proof of that for finite-product vectors space with finite-product inner product, topology induced by product inner product is product topology of topologies induced by inner products
Topics
About: vectors space
The table of contents of this article
Starting Context
- The reader knows a definition of finite-product inner product.
- 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 product topology.
- 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-product vectors space with any finite-product inner product, the topology induced by the product inner product is the product topology of the topologies induced by the inner products.
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
\(\{V_1, ..., V_n\}\): \(\subseteq \{\text{ the } F \text{ vectors spaces }\}\), with any inner products, \(\{\langle \bullet, \bullet \rangle_1, ..., \langle \bullet, \bullet \rangle_n\}\)
\(V_1 \times ... \times V_n\): \(= \text{ the product vectors space }\) with the product inner product, \(\langle \bullet, \bullet \rangle\)
\(O\): \(= \text{ the topology for } V_1 \times ... \times V_n \text{ induced by } \langle \bullet, \bullet \rangle\)
\(O'\): \(= \text{ the topology for } V_1 \times ... \times V_n \text{ as the product topology of the topologies induced by } \langle \bullet, \bullet \rangle_j \text{ s }\)
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Statements:
\(O = O'\)
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2: Proof
Whole Strategy: apply the local criterion for openness; Step 1: take any \(U \in O\) and for each \(u \in U\), see that there is an open neighborhood of \(u\) in \(O'\) contained in \(U\); Step 2: take any \(U \in O'\) and for each \(u \in U\), see that there is an open neighborhood of \(u\) in \(O\) contained in \(U\).
Step 1:
Let \(U \in O\) be any.
Let \(u = (u^1, ..., u^n) \in U\) be any.
There is an open ball around \(u\) in \(O\), \(B_{u, \epsilon} \in O\), such that \(B_{u, \epsilon} \subseteq U\).
Let us take \(\delta := \epsilon / \sqrt{n}\).
Let us take the product of the open balls around \(u\) in \(O'\), \(B_{u^1, \delta} \times ... \times B_{u^n, \delta} \in O'\).
Let us see that \(B_{u^1, \delta} \times ... \times B_{u^n, \delta} \subseteq B_{u, \epsilon}\).
Let \(b = (b^1, ..., b^n) \in B_{u^1, \delta} \times ... \times B_{u^n, \delta}\) be any.
\(\langle b^j - u^j, b^j - u^j \rangle_j \lt \delta^2\).
\(\langle b - u, b - u \rangle = \langle (b^1, ..., b^n) - (u^1, ..., u^n), (b^1, ..., b^n) - (u^1, ..., u^n) \rangle = \langle (b^1 - u^1, ..., b^n - u^n), (b^1 - u^1, ..., b^n - u^n) \rangle = \langle b^1 - u^1, b^1 - u^1 \rangle_1 + ... + \langle b^n - u^n, b^n - u^n \rangle_n \lt \delta^2 + ... + \delta^2 = n \delta^2 = \epsilon^2\).
That means that \(b \in B_{u, \epsilon}\).
So, \(B_{u^1, \delta} \times ... \times B_{u^n, \delta} \subseteq B_{u, \epsilon} \subseteq U\).
By the local criterion for openness, \(U \in O'\).
Step 2:
Let \(U \in O'\) be any.
Let \(u = (u^1, ..., u^n) \in U\) be any.
There is the product of some open balls around \(u\) in \(O'\), \(B_{u^1, \epsilon} \times ... \times B_{u^n, \epsilon} \in O'\), such that \(B_{u^1, \epsilon} \times ... \times B_{u^n, \epsilon} \subseteq U\).
Let us take \(\delta := \epsilon\).
Let us take the open ball around \(u\) in \(O\), \(B_{u, \delta} \in O\).
Let us see that \(B_{u, \delta} \subseteq B_{u^1, \epsilon} \times ... \times B_{u^n, \epsilon}\).
Let \(b \in B_{u, \delta}\) be any.
\(\langle b - u, b - u \rangle = \langle (b^1, ..., b^n) - (u^1, ..., u^n), (b^1, ..., b^n) - (u^1, ..., u^n) \rangle = \langle (b^1 - u^1, ..., b^n - u^n), (b^1 - u^1, ..., b^n - u^n) \rangle = \langle b^1 - u^1, b^1 - u^1 \rangle_1 + ... + \langle b^n - u^n, b^n - u^n \rangle_n \lt \delta^2\).
That means that \(\langle b^j - u^j, b^j - u^j \rangle_j \lt \delta^2 = \epsilon^2\).
That means that \(b^j \in B_{u^j, \epsilon}\), so, \(b \in B_{u^1, \epsilon} \times ... \times B_{u^n, \epsilon}\).
So, \(B_{u, \delta} \subseteq B_{u^1, \epsilon} \times ... \times B_{u^n, \epsilon} \subseteq U\).
So, by the local criterion for openness, \(U \in O\).
