573: Domain of Affine Simplex Map Is Closed and Compact on Euclidean Topological Superspace

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description/proof of that domain of affine simplex map is closed and compact on Euclidean topological superspace

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Topics

About: vectors space
About: topological space

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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

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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.

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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.

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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.

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Main Body

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1: Structured Description

Here is the rules of Structured Description.

Entities:
${\mathbb{R}}^{n+1}$: $=\text{ the Euclidean topological space }$
$T$: $=\{t=(t^0,...,t^n)\in {\mathbb{R}}^{n+1}|\sum _{j\in \{0,...,n\}}{t}^j=1\wedge 0\le t^j\}\subseteq {\mathbb{R}}^{n+1}$, 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.

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2: Natural Language Description

For the Euclidean topological space, ${\mathbb{R}}^{n+1}$, $T:=\{t=(t^0,...,t^n)\in {\mathbb{R}}^{n+1}|\sum _{j\in \{0,...,n\}}{t}^j=1\wedge 0\le t^j\}\subseteq {\mathbb{R}}^{n+1}$, as the subspace of ${\mathbb{R}}^{n+1}$, is closed and compact on ${\mathbb{R}}^{n+1}$.

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3: Note

While any affine simplex does not explicitly appear in Description, $T$ is the domain of any affine simplex map, $f:T\to [p_0,...,p_n]=\{\sum _{j\in \{0,...,n\}}{t}^j{p}_j\in V|t^j\in \mathbb{R},\sum _{j\in \{0,...,n\}}{t}^j=1\wedge 0\le t^j\}$.

"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.

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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.

$T={f_1}^{-1}(\{1\})\cap {\cap }_j{f_{2,j}}^{-1}([0,\mathrm{\infty }))$, but ${f_1}^{-1}(\{1\})\subseteq {\mathbb{R}}^{n+1}$ and ${f_{2,j}}^{-1}([0,\mathrm{\infty }))\subseteq {\mathbb{R}}^{n+1}$ are closed on ${\mathbb{R}}^{n+1}$, because $\{1\},[0,\mathrm{\infty })\subseteq \mathbb{R}$ 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.

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References

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