description/proof of that linear map between Euclidean topological spaces is continuous
Topics
About: topological space
The table of contents of this article
Starting Context
- The reader knows a definition of Euclidean topological space.
- The reader knows a definition of Euclidean vectors space.
- The reader knows a definition of linear map.
- 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 any linear map between any Euclidean topological spaces 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:
\(d_1\): \(\in \mathbb{N} \setminus \{0\}\)
\(d_2\): \(\in \mathbb{N} \setminus \{0\}\)
\(\mathbb{R}^{d_1}\): \(= \text{ the Euclidean topological space }\), also as the Euclidean vectors space
\(\mathbb{R}^{d_2}\): \(= \text{ the Euclidean topological space }\), also as the Euclidean vectors space
\(f\): \(: \mathbb{R}^{d_1} \to \mathbb{R}^{d_2}\), \(\in \{\text{ the linear maps }\}\)
Statements:
\(f \in \{\text{ the continuous maps }\}\)
//
2: Proof
Whole Strategy: Step 1: for each \(r \in \mathbb{R}^{d_1}\), take any open ball around \(f (r)\), \(B_{f (r), \epsilon}\), and take an open ball around \(r\), \(B_{r, \delta}\), such that \(f (B_{r, \delta}) \subseteq B_{f (r), \epsilon}\): take any basis for \(\mathbb{R}^{d_1}\).
Step 1:
Let \(r \in \mathbb{R}^{d_1}\) be any.
Let us take any open ball around \(f (r)\), \(B_{f (r), \epsilon} \subseteq \mathbb{R}^{d_2}\).
Let us take an open ball around \(r\), \(B_{r, \delta} \subseteq \mathbb{R}^{d_1}\), such that \(f (B_{r, \delta}) \subseteq B_{f (r), \epsilon}\).
Let us take any basis for \(\mathbb{R}^{d_1}\), \(B = \{b_1, ..., b_{d_1}\}\).
Let \(r' \in B_{r, \delta}\) be any.
\(r' = r'^j b_j\).
As \(f\) is linear, \(f (r') = f (r'^j b_j) = r'^j f (b_j)\).
Let \(M := Max (\{\Vert f (b_1) \Vert, ..., \Vert f (b_{d_1}) \Vert\})\).
\(\Vert f (r') - f (r) \Vert = \Vert r'^j f (b_j) - r^j f (b_j) \Vert = \Vert (r'^j - r^j) f (b_j) \Vert \le \vert r'^j - r^j \vert \Vert f (b_j) \Vert \le \sum_{j \in \{1, ..., d_1\}} \vert r'^j - r^j \vert M\).
For each \(j \in \{1, ..., d_1\}\), \(\vert r'^j - r^j \vert \lt \delta\), because if \(\delta \le \vert r'^j - r^j \vert\) for a \(j\), \(\delta^2 \le \vert r'^j - r^j \vert^2\) and \(\delta^2 \le \sum_{j \in \{1, ..., d_1\}} \vert r'^j - r^j \vert^2 = \Vert r' - r \Vert^2\), a contradiction.
So, \(\Vert f (r') - f (r) \Vert \lt \sum_{j \in \{1, ..., d_1\}} \delta M = d_1 \delta M\).
So, let us take \(\delta := \epsilon / (d_1 M)\).
Then, \(\Vert f (r') - f (r) \Vert \lt \epsilon\).
That means that \(f (B_{r, \delta}) \subseteq B_{f (r), \epsilon}\).
So, \(f\) is continuous at \(r\).
As \(r\) is arbitrary, \(f\) is continuous.
