description/proof of that for finite-dimensional real vectors space with canonical topology and open neighborhood of \(0\), neighborhood multiplied by nonzero number is open neighborhood of \(0\)
Topics
About: vectors space
About: topological space
The table of contents of this article
Starting Context
- The reader knows a definition of canonical topology for finite-dimensional real vectors space.
- The reader admits the proposition that for any finite dimensional real vectors space, the topology defined by the Euclidean topology of the coordinates space based on any basis does not depend on the choice of 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 real vectors space with the canonical topology and any open neighborhood of \(0\), the neighborhood multiplied by any nonzero number is an open neighborhood of \(0\).
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:
\(d\): \(\in \mathbb{N} \setminus \{0\}\)
\(V\): \(\in \{\text{ the } d \text{ -dimensional real vectors spaces }\}\)
\(U_0\): \(\subseteq V\), \(\in \{\text{ the open neighborhoods of } 0\}\)
\(r\): \(\in \mathbb{R} \setminus \{0\}\)
\(r U_0\):
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Statements:
\(r U_0 \in \{\text{ the open neighborhoods of } 0\}\)
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2: Proof
Whole Strategy: Step 1: take any basis for \(V\) and the corresponding homeomorphism, \(f: V \to \mathbb{R}^d\); Step 2: see that \(0 \in r U_0\); Step 3: see that for each \(p \in r U_0\), there is an open neighborhood of \(p\) contained in \(r U_0\); Step 4: conclude the proposition.
Step 1:
Let us take any basis for \(V\), \(B\).
Let us take the corresponding homeomorphism, \(f: V \to \mathbb{R}^d\), by which the topology for \(V\) is defined.
The topology does not depend on the choice of the basis, by the proposition that for any finite dimensional real vectors space, the topology defined by the Euclidean topology of the coordinates space based on any basis does not depend on the choice of basis.
Step 2:
\(0 \in r U_0\), because \(0 \in U_0\) and \(0 = r 0\).
Step 3:
Let \(p \in r U_0\) be any.
\(p / r \in U_0\), because \(p = r u\) for a \(u \in U_0\) and \(p / r = u \in U_0\).
As \(U_0\) is an open neighborhood of \(0\), \(f (U_0)\) is an open neighborhood of \(0\) with \(f (p / r) \in f (U_0)\).
There is an open ball around \(f (p / r)\), \(B_{f (p / r), \epsilon} \subseteq \mathbb{R}^d\), such that \(B_{f (p / r), \epsilon} \subseteq f (U_0)\), by the definition of Euclidean topology.
\(f^{-1} (B_{f (p / r), \epsilon}) \subseteq U_0\).
\(f^{-1} (B_{f (p / r), \epsilon}) \subseteq V\) is an open subset of \(V\).
\(p / r \in f^{-1} (B_{f (p / r), \epsilon})\), because \(f (p / r) \in B_{f (p / r), \epsilon}\).
So, \(f^{-1} (B_{f (p / r), \epsilon})\) is an open neighborhood of \(p / r\).
\(r f^{-1} (B_{f (p / r), \epsilon}) \subseteq r U_0\).
\(p = r p / r \in r f^{-1} (B_{f (p / r), \epsilon})\).
\(f (r f^{-1} (B_{f (p / r), \epsilon})) = r f (f^{-1} (B_{f (p / r), \epsilon}))\), because \(f\) is obviously linear, \(= r B_{f (p / r), \epsilon}\).
But \(r B_{f (p / r), \epsilon} = B_{r f (p / r), \vert r \vert \epsilon}\), because for each \(p' \in r B_{f (p / r), \epsilon}\), \(p' / r \in B_{f (p / r), \epsilon}\), which means that \((p'^1 / r - f^1 (p / r))^2 + ... + (p'^d / r - f^d (p / r))^2 \lt \epsilon^2\), so, \(r^2 (p'^1 / r - f^1 (p / r))^2 + ... + r^2 (p'^d / r - f^d (p / r))^2 \lt r^2 \epsilon^2\), but the left hand side is \((p'^1 - r f^1 (p / r))^2 + ... + (p'^d - r f^d (p / r))^2\), which means that \(p' \in B_{r f (p / r), \vert r \vert \epsilon}\); for each \(p' \in B_{r f (p / r), \vert r \vert \epsilon}\), \((p'^1 - r f^1 (p / r))^2 + ... + (p'^d - r f^d (p / r))^2 \lt r^2 \epsilon^2\), so, \(1 / r^2 (p'^1 - r f^1 (p / r))^2 + ... + 1 / r^2 (p'^d - r f^d (p / r))^2 \lt \epsilon^2\), but the left hand side is \((p'^1 / r - f^1 (p / r))^2 + ... + (p'^d / r - f^d (p / r))^2\), which means that \(p' / r \in B_{f (p / r), \epsilon}\), so, \(p' \in r B_{f (p / r), \epsilon}\).
So, \(f (r f^{-1} (B_{f (p / r), \epsilon})) = B_{r f (p / r), \vert r \vert \epsilon}\), and \(r f^{-1} (B_{f (p / r), \epsilon}) = f^{-1} (f (r f^{-1} (B_{f (p / r), \epsilon}))) = f^{-1} (B_{r f (p / r), \vert r \vert \epsilon})\).
So, \(r f^{-1} (B_{f (p / r), \epsilon})\) is an open neighborhood of \(p\) contained in \(r U_0\).
Step 4:
By the local criterion for openness, \(r U_0\) is open.
So, \(r U_0\) is an open neighborhood of \(0\).
