description/proof of that domain of affine simplex map is closed and compact on Euclidean topological superspace
Topics
About: vectors space
About: topological space
The table of contents of this article
- Starting Context
- Target Context
- Orientation
- Main Body
- 1: Structured Description
- 2: Natural Language Description
- 3: Note
- 3: Proof
Starting Context
- The reader knows a definition of affine simplex.
- The reader knows a definition of topological subspace.
- The reader knows a definition of closed set.
- The reader knows a definition of compact subset of topological space.
- The reader knows a definition of continuous map.
- The reader admits the Heine-Borel theorem.
Target Context
- The reader will have a description and a proof of the proposition that the domain of any affine simplex map is closed and compact on the Euclidean topological superspace.
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:
\(\mathbb{R}^{n + 1}\): \(= \text{ the Euclidean topological space }\)
\(T\): \(= \{t = (t^0, ..., t^n) \in \mathbb{R}^{n + 1} \vert \sum_{j \in \{0, ..., n\}} t^j = 1 \land 0 \le t^j\} \subseteq \mathbb{R}^{n + 1}\), as the subspace of \(\mathbb{R}^{n + 1}\)
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Statements:
\(T \in \{\text{ the closed subsets of } \mathbb{R}^{n + 1}\} \cap \{\text{ the compact subsets of } \mathbb{R}^{n + 1}\}\).
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\(T\) is a compact topological space by itself, by the proposition that the compactness of any topological subset as a subset equals the compactness as a subspace.
2: Natural Language Description
For the Euclidean topological space, \(\mathbb{R}^{n + 1}\), \(T := \{t = (t^0, ..., t^n) \in \mathbb{R}^{n + 1} \vert \sum_{j \in \{0, ..., n\}} t^j = 1 \land 0 \le t^j\} \subseteq \mathbb{R}^{n + 1}\), as the subspace of \(\mathbb{R}^{n + 1}\), is closed and compact on \(\mathbb{R}^{n + 1}\).
3: Note
While any affine simplex does not explicitly appear in Description, \(T\) is the domain of any affine simplex map, \(f: T \to [p_0, ..., p_n] = \{\sum_{j \in \{0, ..., n\}} t^j p_j \in V \vert t^j \in \mathbb{R}, \sum_{j \in \{0, ..., n\}} t^j = 1 \land 0 \le t^j\}\).
"affine simplex map" may not be any established term, but means the map underlying any affine simplex, not any affine map from an affine simplex.
3: Proof
Let \(\mathbb{R}\) be the Euclidean topological space. Let us take the map, \(f_1: \mathbb{R}^{n + 1} \to \mathbb{R}, (t^1, ..., t^n) \mapsto \sum_{j \in \{0, ..., n\}} t^j\). \(f_1\) is obviously continuous. Let us take the map, \(f_{2, j}: \mathbb{R}^{n + 1} \to \mathbb{R}, (t^1, ..., t^n) \mapsto t^j\), for each \(j \in \{0, ..., n\}\). \(f_{2, j}\) is obviously continuous.
\(T = {f_1}^{-1} (\{1\}) \cap \cap_j {f_{2, j}}^{-1} ([0, \infty))\), but \({f_1}^{-1} (\{1\}) \subseteq \mathbb{R}^{n + 1}\) and \({f_{2, j}}^{-1} ([0, \infty)) \subseteq \mathbb{R}^{n + 1}\) are closed on \(\mathbb{R}^{n + 1}\), because \(\{1\}, [0, \infty) \subseteq \mathbb{R}\) are closed on \(\mathbb{R}\) and \(f_1\) and \(f_{2, j}\) are continuous.
So, \(T\) is closed on \(\mathbb{R}^{n + 1}\).
As \(T\) is bounded (inevitably \(t^j \le 1\)), \(T\) is compact on \(\mathbb{R}^{n + 1}\), by the Heine-Borel theorem.
