95: Standard Simplex Is Homeomorphic to Same-Dimensional Closed Ball

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description/proof of that standard simplex is homeomorphic to same-dimensional closed ball

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Topics

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: Proof
• 4: Note

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Starting Context

• The reader knows a definition of homeomorphism.
• The reader admits the proposition that any Euclidean topological space nested in any Euclidean topological space is a topological subspace of the nesting Euclidean topological space.
• The reader admits the proposition that in any nest of topological subspaces, the openness of any subset on any subspace does not depend on the superspace of which the subspace is regarded to be a subspace.
• The reader admits the proposition that any topological spaces map is continuous at any point if the spaces are the subspaces of some $C^{\mathrm{\infty }}$ manifolds and there are some charts of the manifolds around the point and the point image and a map between the chart open subsets which (the map) is restricted to the original map whose (the chart open subsets map's) coordinates function's restriction is continuous.
• The reader admits the proposition that any restriction of any continuous map on the domain and the codomain is continuous.

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Target Context

• The reader will have a description and a proof of the proposition that the n-dimensional standard simplex is homeomorphic to any n-dimensional closed ball.

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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:
${\mathbb{R}}^{n+1}$: $=\text{ the Euclidean vectors space }$ with the Euclidean topology
${\mathrm{\Delta }}^n$: $=\text{ the standard n-simplex }$, $\subseteq {\mathbb{R}}^{n+1}$, with the subspace topology of ${\mathbb{R}}^{n+1}$
${\mathbb{R}}^n$: $=\text{ the Euclidean topological space }$
$\overline{B_{p,r}^n}$: $=\text{ the closed ball }$, $\subseteq {\mathbb{R}}^n$, with the subspace topology of ${\mathbb{R}}^n$
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Statements:
$\mathrm{\exists }f:{\mathrm{\Delta }}^n\to \overline{B_{p,r}^n}(f\in \{\text{ the homeomorphisms }\})$
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2: Natural Language Description

For the Euclidean vectors space, ${\mathbb{R}}^{n+1}$, with the Euclidean topology, the standard n-simplex, ${\mathrm{\Delta }}^n\subseteq {\mathbb{R}}^{n+1}$, with the subspace topology of ${\mathbb{R}}^{n+1}$, the Euclidean topological space, ${\mathbb{R}}^n$, and the closed ball, $\overline{B_{p,r}^n}\subseteq {\mathbb{R}}^n$, with the subspace topology of ${\mathbb{R}}^n$, there is a homeomorphism, $f:{\mathrm{\Delta }}^n\to \overline{B_{p,r}^n}$.

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3: Proof

${\mathbb{R}}^n$ can be nested in ${\mathbb{R}}^{n+1}$ as ${\mathbb{R}}^n\times \{0\}$ being rotated and translated such that ${\mathbb{R}}^n=\{(x^1,...,x^{n+1})\in {\mathbb{R}}^{n+1}|\sum _{j\in \{1,...,n+1\}}{x}^j=1\}$: think of the hyperplane that contains the origin perpendicular to the vector, $(1/(n+1),...,1/(n+1))\in {\mathbb{R}}^{n+1}$, which implies that $(x^1,...,x^{n+1})(1/(n+1),...,1/(n+1))=x^{1}1/(n+1)+...+x^{n+1}1/(n+1)=1/(n+1)(x^1+...+x^{n+1})=0$, which implies that $x^1+...+x^{n+1}=0$, and translate it by the vector, $(1/(n+1),...,1/(n+1))\in {\mathbb{R}}^{n+1}$, which implies that $x^1+...+x^{n+1}=1$. Then, ${\mathbb{R}}^n$ is the topological subspace of ${\mathbb{R}}^{n+1}$, by the proposition that any Euclidean topological space nested in any Euclidean topological space is a topological subspace of the nesting Euclidean topological space.

$\overline{B_{p,r}^n}\subseteq {\mathbb{R}}^n\subseteq {\mathbb{R}}^{n+1}$ is the topological subspace of ${\mathbb{R}}^{n+1}$, by the proposition that in any nest of topological subspaces, the openness of any subset on any subspace does not depend on the superspace of which the subspace is regarded to be a subspace.

${\mathrm{\Delta }}^n=\{x\in {\mathbb{R}}^{n+1}|\sum _{j\in \{1,...,n+1\}}{x}^j=1\wedge 0\le x^j\}$. $\overline{B_{p,r}^n}=\{x\in {\mathbb{R}}^{n+1}|\sum _{j\in \{1,...,n+1\}}{x}^j=1\wedge \sum _{j\in \{1,...,n+1\}}(x^j-p^j{)}^2\le r^2\}$, where $\sum _{j\in \{1,...,n+1\}}{p}^j=1$.

The barycenter of ${\mathrm{\Delta }}^n$ is $c=(1/(n+1),...,1/(n+1))$. The map, $g:{\mathrm{\Delta }}^n\setminus \{c\}\to {\mathrm{\Delta }}^n$, is defined as the boundary point of ${\mathrm{\Delta }}^n$ that the radius from $c$ to $x$ passes. $f$ is defined to be $p$ for $c$ and $p+r{|g(x)-c|}^{-1}(x-c)$ otherwise.

In fact, $g(x)=c+(1-{c^k}^{-1}{x}^k{)}^{-1}(x-c)$, where $k$ is the index of the minimum component of $x$ (if there are some multiple minimums, choose any one of them), because $g(x)=c+a(x-c)$ whose minimum component is 0 ($0<a$ and obviously, the minimum component of x corresponds to the minimum component of $g(x)$), which is the condition of being on a boundary, so, $c^k+a(x^k-c^k)=0$, so, $a=-c^k(x^k-c^k{)}^{-1}=(1-{c^k}^{-1}{x}^k{)}^{-1}$.

$f$ is obviously a bijection between ${\mathrm{\Delta }}^n$ and $\overline{B_{p,r}^n}$.

Let us apply the proposition that any topological spaces map is continuous at any point if the spaces are the subspaces of some $C^{\mathrm{\infty }}$ manifolds and there are some charts of the manifolds around the point and the point image and a map between the chart open subsets which (the map) is restricted to the original map whose (the chart open subsets map's) coordinates function's restriction is continuous.

${\mathbb{R}}^{n+1}$ is canonically a $C^{\mathrm{\infty }}$ manifold. Let us take the standard chart for ${\mathbb{R}}^{n+1}$, $({\mathbb{R}}^{n+1}\subseteq {\mathbb{R}}^{n+1},id)$. Let us define the extension of $g$, $g^{\prime }:{\mathbb{R}}^n\setminus c\to {\mathbb{R}}^n$, such that outside ${\mathrm{\Delta }}^n$, the value is the boundary point of ${\mathrm{\Delta }}^n$ that the radius from $c$ to $x$ passes: as before, $g^{\prime }(x)=c+(1-{c^k}^{-1}{x}^k{)}^{-1}(x-c)$ where $k$ is the index of the minimum component of $x$. Let us define the extension of $f$, $f^{\prime }:{\mathbb{R}}^{n+1}\to {\mathbb{R}}^{n+1}$, such that denoting the projection of $x-c$ to the ${\mathbb{R}}^n$ hyperplane as $x^{\prime }$, which means that $x-c-x^{\prime }$ is the vector perpendicular to the ${\mathbb{R}}^n$ hyperplane, $p+x-c$ when $x^{\prime }=0$ and $p+r{|g^{\prime }(c+x^{\prime })-c|}^{-1}{x}^{\prime }+x-c-x^{\prime }$ otherwise.

Obviously, $f^{\prime }$ is indeed restricted to $f$ on ${\mathrm{\Delta }}^n$.

$f^{\prime }$ is obviously a bijection.

$x\mapsto x^{\prime }$ is continuous (the $ϵ$-'open ball' centered at $x$ maps to the intersection of the $ϵ$-'open ball' centered at $c+x^{\prime }$ and the ${\mathbb{R}}^n$ hyperplane, and $c+x^{\prime }\mapsto x^{\prime }$ is continuous); $x\mapsto g^{\prime }(c+x^{\prime }(x))$ is continuous (taking the minimum component is continuous, because the $ϵ$-'open ball' centered at $c+x^{\prime }$ means that the variation of each component (including the minimum component) is less than $ϵ$, taking the multiplicative inverse is continuous); taking the vector length is continuous and taking the multiplicative inverse is continuous; so, $f^{\prime }$ is continuous except at $x^{\prime }=0$ as a compound of continuous maps, and when $x^{\prime }$ is near $0$, $|{|g^{\prime }(c+x^{\prime })-c|}^{-1}{x}^{\prime }|={|{|(1-{c^j}^{-1}(c^j+x^{\prime j}))|}^{-1}|x^{\prime }||}^{-1}|x^{\prime }|=|(1-{c^j}^{-1}(c^j+x^{\prime j}))|\approx 0$, so, $f^{\prime }(x)=p+r{|g^{\prime }(c+x^{\prime })-c|}^{-1}{x}^{\prime }+x-c-x^{\prime }\approx p+x-c$, so, $f^{\prime }$ is continuous also at $x^{\prime }=0$.

On the other hand, denoting the projection of $y-p$ to the ${\mathbb{R}}^n$ hyperplane as $y^{\prime }$ and denoting the boundary point of ${\mathrm{\Delta }}^n$ via which $p+y^{\prime }$ is mapped from $c+x^{\prime }$ as $h^{\prime }(p+y^{\prime })$, which implies that $h^{\prime }(p+y^{\prime })=g^{\prime }(c+x^{\prime })$, $f^{\prime -1}(y)$ is $c+y-p$ when $y^{\prime }=0$ and $c+r^{-1}|h^{\prime }(p+y^{\prime })-c|y^{\prime }+y-p-y^{\prime }$ otherwise, because $p+r{|g^{\prime }(c+x^{\prime })-c|}^{-1}{x}^{\prime }$ is nothing but $p+y^{\prime }$, so, $y^{\prime }=r{|g^{\prime }(c+x^{\prime })-c|}^{-1}{x}^{\prime }$, and $f^{\prime -1}(y)=x=c+x^{\prime }+x-c-x^{\prime }=c+x^{\prime }+y-p-y^{\prime }$, because $y-p-y^{\prime }=x-c-x^{\prime }$, as the vector perpendicular to the ${\mathbb{R}}^n$ hyperplane is not changed.

$h^{\prime }(p+y^{\prime })$ is in fact $g^{\prime }(c+y^{\prime })$ because $p+y^{\prime }$ is mapped from $c+x^{\prime }$ such that the direction of $p+y^{\prime }$ from $p$ is the same with the direction of $c+x^{\prime }$ from $c$ and the boundary point depends on only the direction of the $g^{\prime }$ argument from $c$. So, $y\mapsto h^{\prime }(p+y^{\prime }(y))$ is continuous as $y\mapsto g^{\prime }(c+y^{\prime })$ is continuous; when $y^{\prime }$ is near 0, $c+r^{-1}|h^{\prime }(p+y^{\prime })-c|y^{\prime }+y-p-y^{\prime }=c+r^{-1}|g^{\prime }(c+y^{\prime })-c|y^{\prime }+y-p-y^{\prime }\approx c+y-p$. So, $f^{\prime -1}(y)$ is continuous.

As $f^{\prime }$ is a homeomorphism, the restriction, $f^{\prime }{|}_{{\mathrm{\Delta }}^n}:{\mathrm{\Delta }}^n\to \overline{B_{p,r}^n}$ is a homeomorphism, by the proposition that any restriction of any continuous map on the domain and the codomain is continuous.

After all, by the proposition that any topological spaces map is continuous at any point if the spaces are the subspaces of some $C^{\mathrm{\infty }}$ manifolds and there are some charts of the manifolds around the point and the point image and a map between the chart open subsets which (the map) is restricted to the original map whose (the chart open subsets map's) coordinates function's restriction is continuous, $f:{\mathrm{\Delta }}^n\to \overline{B_{p,r}^n}$ is a homeomorphism: as ${\varphi }_p^{\prime }=id$ and ${\varphi }_{f(p)}^{\prime }=id$, ${\varphi }_p^{\prime }$ and ${\varphi }_{f(p)}^{\prime }$ are not conspicuous, but $f^{\prime }$ is indeed a coordinates function.

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4: Note

We could not directly claim the homeomorphism of $f$ by its expression with the coordinates for ${\mathbb{R}}^{n+1}$, because the coordinates are not for ${\mathrm{\Delta }}^n$ or $\overline{B_{p,r}^n}$. So, we employed the proposition that any topological spaces map is continuous at any point if the spaces are the subspaces of some $C^{\mathrm{\infty }}$ manifolds and there are some charts of the manifolds around the point and the point image and a map between the chart open subsets which (the map) is restricted to the original map whose (the chart open subsets map's) coordinates function's restriction is continuous, which required having $f^{\prime }$.

$f^{\prime }$ did not particularly have to be wholly homeomorphic: just the restriction of $f^{\prime }$ had to be homeomorphic, but we established the wholly homeomorphic $f^{\prime }$ anyway because we could.

The boundary of ${\mathrm{\Delta }}^n$ as the subspace of ${\mathrm{\Delta }}^n$, which (the boundary) is also the subspace of ${\mathbb{R}}^{n+1}$, is homeomorphic to the boundary of $\overline{B_{p,r}^n}$ as the subspace of $\overline{B_{p,r}^n}$, which (the boundary) is also the subspace of ${\mathbb{R}}^n$, which (the boundary) is homeomorphic to $S^n$, because $f$ maps the boundary onto the boundary, and the restriction of $f$ on the domain and the codomain is homeomorphic.

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References

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