description/proof of that affine simplex on finite-dimensional real vectors space is closed and compact on canonical 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: Proof
Starting Context
- The reader knows a definition of affine simplex.
- 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 proposition that any affine simplex map into any finite-dimensional vectors space is continuous with respect to the canonical topologies of the domain and the codomain.
- The reader admits the proposition that the domain of any affine simplex map is closed and compact on the Euclidean topological superspace.
- The reader admits the proposition that for any continuous map between any topological spaces, the image of any compact subset of the domain is compact.
- The reader admits the proposition that any compact subset of any Hausdorff topological space is closed.
Target Context
- The reader will have a description and a proof of the proposition that any affine simplex on any finite-dimensional real vectors space is closed and compact on the canonical 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:
\(V\): \(\in \{\text{ the } d \text{ -dimensional real vectors spaces }\}\) with the canonical topology
\(\{p_0, ..., p_n\}\): \(\subseteq V\), \(\in \{\text{ the affine-independent sets of base points on } V\}\)
\([p_0, ..., p_n]\): \(= \text{ the affine simplex }\)
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Statements:
\([p_0, ..., p_n]\) is closed and compact on \(V\).
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\([p_0, ..., p_n]\) 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 any \(d\)-dimensional real vectors space, \(V\), with the canonical topology, and any affine-independent set of base points on \(V\), \(\{p_0, ..., p_n\} \subseteq V\), the affine simplex, \([p_0, ..., p_n]\), is closed and compact on \(V\).
3: Proof
For the affine simplex map, \(f: T \to V, t = (t^0, ..., t^n) \mapsto \sum_{j \in \{0, ..., n\}} t^j p_j\), where \(T := \{t = (t^0, ..., t^n) \in \mathbb{R}^{n + 1} \vert \sum_{j \in \{0, ..., n\}} t^j = 1 \land 0 \le t^j\} \subseteq \mathbb{R}^{n + 1}\) as the subspace of \(\mathbb{R}^{n + 1}\), \(f\) is continuous, by the proposition that any affine simplex map into any finite-dimensional vectors space is continuous with respect to the canonical topologies of the domain and the codomain, and \(T\) is compact, by the proposition that the domain of any affine simplex map is closed and compact on the Euclidean topological superspace.
So, \([p_0, ..., p_n]\) is compact on \(V\), by the proposition that for any continuous map between any topological spaces, the image of any compact subset of the domain is compact.
As \(V\) is a Hausdorff topological space (because \(V\) is homeomorphic to \(\mathbb{R}^d\) and \(\mathbb{R}^d\) is Hausdorff), \([p_0, ..., p_n]\) is closed on \(V\), by the proposition that any compact subset of any Hausdorff topological space is closed.
