A description/proof of that simplex is homeomorphic to same-dimensional closed ball
Topics
About: topological space
The table of contents of this article
Starting Context
- The reader knows a definition of topological space.
- 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 subspace topological space 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 topological space does not depend on the superspace of which the topological space is regarded to be a subspace.
- The reader admits the proposition that any map between any subspace topological spaces is continuous at any point if a map that is an extension of the original map to some open sets of the superspaces is continuous at the point.
Target Context
- The reader will have a description and a proof of the proposition that the n-dimensional simplex is homeomorphic to any n-dimensional closed ball.
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: Description
Between the n-dimensional simplex, \(\Delta^n\), and any n-dimensional close ball, \(\overline{B^n_{p-r}}\), there is a homeomorphism, \(f: \Delta^n \rightarrow \overline{B^n_{p-r}}\).
2: Proof
\(\Delta^n\) and \(\overline{B^n_{p-r}}\) are subspace topological spaces of the \(\mathbb{R}^n\) Euclidean topological space, but they can be regarded as subspace topological spaces of the \(\mathbb{R}^{n + 1}\) Euclidean topological space, because of the proposition that any Euclidean topological space nested in any Euclidean topological space is a subspace topological space of the nesting Euclidean topological space and the proposition that in any nest of topological subspaces, the openness of any subset on any subspace topological space does not depend on the superspace of which the topological space is regarded to be a subspace.
On \(\mathbb{R}^{n + 1}\), \(\sum_{i = 1, . . ., n + 1} x^i = 1\) is the \(\mathbb{R}^{n}\) space, and \(\Delta^n = \{x \in \mathbb{R}^{n + 1}| \sum_{i = 1, . . ., n + 1} x^i = 1, 0 \leq x^i \leq 1\}\); \(\overline{B^n_{p-r}} = \{x \in \mathbb{R}^{n + 1}| \sum_{i = 1, . . ., n + 1} x^i = 1, \sum_{i = 1, . . ., n + 1} (x^i - p^i)^2 \leq r^2\}\) where \(\sum_{i = 1, . . ., n + 1} p^i = 1\).
The barycenter of \(\Delta^n\) is \(c = (\frac{1}{n + 1}, \frac{1}{n + 1}, . . ., \frac{1}{n + 1})\). The function, \(g: \Delta^n \setminus \{c\} \rightarrow \Delta^n, x \mapsto g (x)\), is defined as the border point of \(\Delta^n\) 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, \(g (x) = c + (1 - {c^j}^{-1} x^j)^{-1} (x - c)\) where \(j\) is the index of the minimum component of \(x\) (if there are multiple minimums, choose any one of them), because \(g (x) = c + a (x - c)\) whose minimum component is 0, which is the condition of being on a border, so, \(c^j + a (x^j - c^j) = 0\), so, \(a = - c^j (x^j - c^j)^{-1} = (1 - {c^j}^{-1}x^j)^{-1}\).
\(f\) is obviously a bijection between \(\Delta^n\) and \(\overline{B^n_{p-r}}\), but in order to apply the proposition that any map between any subspace topological spaces is continuous at any point if a map that is an extension of the original map to some open sets of the superspaces is continuous at the point, the domain of \(f\) has to be extended to an open set of \(\mathbb{R}^{n + 1}\) with the range extended to an open set of \(\mathbb{R}^{n + 1}\). Let us define \(f': \mathbb{R}^{n + 1} \rightarrow \mathbb{R}^{n + 1}\) as follows: as an extension of \(g\), \(g': \mathbb{R}^{n} \setminus {c} \rightarrow \mathbb{R}^{n}, x \mapsto g' (x)\) is defined as outside \(\Delta^n\), the border point of \(\Delta^n\) that the radius from \(c\) to \(x\) passes, which just makes the maximum component to be chosen instead of minimum; as \(x'\) denotes the projection of \(x\) to the \(\mathbb{R}^{n}\) plane, which means that \(x - x'\) is the vector perpendicular to the \(\mathbb{R}^{n}\) plane, \(f'\) is defined as \(p + x - c\) when \(x' = c\) and \(p + r {|g' (x') - c|}^{-1} (x' - c) + x - x'\) otherwise, also which is obviously a bijection.
\(x \rightarrow x'\) is continuous (the \(\epsilon\)-'open ball' centered at \(x\) maps to the intersection of the \(\epsilon\)-'open ball' centered at \(x'\) and the \(\mathbb{R}^{n}\) plane); \(x \rightarrow g' (x' (x))\) is continuous (taking the minimum or maximum component is continuous, because the \(\epsilon\)-'open ball' centered at \(x'\) means that the variation of each component (including the minimum or maximum component) is less than \(\epsilon\), taking the vector length is continuous, taking the multiplicative inverse is continuous (\(x' = c\) is not inside the domain of \(g'\))); so, \(f'\) is continuous except when \(x' = c\) as a compound of continuous maps, and when \(x'\) is near \(c\), \(|{|g' (x') - c|}^{-1} (x' - c)| = {| {|(1 - {c^j}^{-1} x'^j)|}^{-1} |x' - c||}^{-1} |x' - c| = |(1 - {c^j}^{-1} x'^j)| \approx 0\), so, \(f' (x) = p + r {|g' (x') - c|}^{-1} (x' - c) + x - x' \approx p + x - c\), so, \(f'\) is continuous also when \(x' = c\).
On the other hand, \(f'^{-1} (y)\) is \(c + y - p\) when \(y' = p\) and \(c + r^{-1} |h' (y') - c| (y' - p) + y - y'\) otherwise, where \(y'\) denotes the projection of \(y\) to the \(\mathbb{R}^{n}\) plane and \(h' (y')\) denotes the border point of \(\Delta^n\) via which \(y'\) is mapped from \(x'\), which implies \(h' (y') = g' (x')\), because \(p + r {|g' (x') - c|}^{-1} (x' - c)\) is nothing but \(y'\), so, \(y' - p = r {|g' (x') - c|}^{-1} (x' - c)\), and \(f'^{-1} (y) = c + x' - c + y - y'\) as \(h' (y') = g' (x')\), but \(y - y' = x - x'\), as the vector perpendicular to the \(\mathbb{R}^{n}\) plane is not changed.
\(h' (y')\) is in fact \(g' (y' - p + c)\) because \(y'\) is mapped from \(x'\) such that the direction of \(y'\) from \(p\) is the same with the direction of \(x'\) from \(c\) and the border point depends on only the direction of the \(g'\) argument from \(c\). So, \(y \mapsto h' (y' (y))\) is continuous as \(y \mapsto g' (y' - p + c)\) is continuous. So, \(f'^{-1} (y)\) is continuous.
After all, by the proposition that any map between any subspace topological spaces is continuous at any point if a map that is an extension of the original map to some open sets of the superspaces is continuous at the point, \(f: \Delta^n \rightarrow \overline{B^n_{p-r}}\) is a homeomorphism.
3: Note
Any valid proof is not as simple as just bringing in an \(n + 1\) variables function \(f\) with the simplex domain, because we have to prove homeomorphism between \(\mathbb{R}^{n}\) subspace topological spaces, while the function is from a non-open subset of \(\mathbb{R}^{n + 1}\) to a non-open subset of \(\mathbb{R}^{n + 1}\), whose seeming-continuousness does not so self-evidently guarantee the desired homeomorphism.
