Showing posts with label Definitions and Propositions. Show all posts
Showing posts with label Definitions and Propositions. Show all posts

2024-05-19

586: For Topological Space, Subspace, and Subset of Superspace, Subspace Minus Subset as Subspace of Subspace Is Subspace of Superspace Minus Subset

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description/proof of that for topological space, subspace, and subset of superspace, subspace minus subset as subspace of subspace is subspace of superspace minus subset

Topics


About: topological space

The table of contents of this article

Starting Context

Target Context


  • The reader will have a description and a proof of the proposition that for any topological space, its any subspace, and any subset of the superspace, the subspace minus the subset as the subspace of the subspace is the subspace of the superspace minus the subset.

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:
\(T'\): \(\in \{\text{ the topological spaces }\}\)
\(T\): \(\subseteq T'\), with the subspace topology
\(S\): \(\subseteq T'\)
\(T' \setminus S \subseteq T'\): with the subspace topology
//

Statements:
\(T \setminus S\) as the subspace of \(T\) is \(T \setminus S\) as the subspace of \(T' \setminus S\).
//

2: Natural Language Description


For any topological space, \(T'\), any subspace, \(T \subseteq T'\), any \(S \subseteq T'\), and \(T' \setminus S \subseteq T'\) with the subspace topology, \(T \setminus S\) as the subspace of \(T\) is \(T \setminus S\) as the subspace of \(T' \setminus S\).

3: Proof


Let \(U \subseteq T \setminus S\) be any open subset of \(T \setminus S\) as the subspace of \(T\).

\(U = U' \cap (T \setminus S)\), where \(U' \subseteq T\) is an open subset of \(T\). \(U' = U'' \cap T\) where \(U'' \subseteq T'\) is an open subset of \(T'\). \(U = U'' \cap T \cap (T \setminus S) = U'' \cap (T \setminus S) = U'' \cap (T' \setminus S) \cap (T \setminus S)\), but \(U'' \cap (T' \setminus S) \subseteq T' \setminus S\) is an open subset of \(T' \setminus S\), and \(U'' \cap (T' \setminus S) \cap (T \setminus S)\) is an open subset of \(T \setminus S\) as the subspace of \(T' \setminus S\).

Let \(U \subseteq T \setminus S\) be any open subset of \(T \setminus S\) as the subspace of \(T' \setminus S\).

\(U = U' \cap (T \setminus S)\), where \(U' \subseteq T' \setminus S\) is an open subset of \(T' \setminus S\). \(U' = U'' \cap (T' \setminus S)\), where \(U'' \subseteq T'\) is an open subset of \(T'\). \(U = U'' \cap (T' \setminus S) \cap (T \setminus S) = U'' \cap (T \setminus S) = U'' \cap T \cap (T \setminus S)\), but \(U'' \cap T \subseteq T\) is an open subset of \(T\), and \(U'' \cap T \cap (T \setminus S) \subseteq T \setminus S\) is an open subset of \(T \setminus S\) as the subspace of \(T\).

So, \(T \setminus S\) as the subspace of \(T\) is \(T \setminus S\) as the subspace of \(T' \setminus S\).

References
<The previous article in this series | The table of contents of this series |

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585: Boundary of Subset of Topological Space Is Set of Points of Each of Which Each Neighborhood Intersects Both Subset and Complement of Subset

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description/proof of that boundary of subset of topological space is set of points of each of which each neighborhood intersects both subset and complement of subset

Topics


About: topological space

The table of contents of this article

Starting Context

Target Context


  • The reader will have a description and a proof of the proposition that the boundary of any subset of any topological space is the set of the points of each of which each neighborhood intersects both the subset and the complement of the subset.

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:
\(T\): \(\in \{\text{ the topological spaces }\}\)
\(S\): \(\subseteq T\)
\(\dot{S}\): \(= \text{ the boundary of } S\)
\(\tilde{S}\): \(= \{p \in T \vert \forall N_p \in \{\text{ the neighborhoods of } p \text{ on } T\} (N_p \cap S \neq \emptyset \land N_p \cap (T \setminus S) \neq \emptyset)\}\)
//

Statements:
\(\dot{S} = \tilde{S}\).
//

2: Natural Language Description


For any topological space, \(T\), and any subset, \(S \subseteq T\), the boundary of \(S\), \(\dot{S}\), equals \(\tilde{S} := \{p \in T \vert \forall N_p \in \{\text{ the neighborhoods of } p \text{ on } T\} (N_p \cap S \neq \emptyset \land N_p \cap (T \setminus S) \neq \emptyset)\}\).

3: Proof


For any \(p \in \dot{S}\), for any neighborhood, \(N_p \subseteq T\), of \(p\), \(N_p \cap S \neq \emptyset\) and \(N_p \cap (T \setminus S) \neq \emptyset\), so, \(p \in \tilde{S}\).

For any \(p \in \tilde{S}\), \(p \in \overline{S}\) and \(p \in \overline{T \setminus S}\), so, \(p \in \dot{S}\).

References
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584: For Simplicial Complex on Finite-Dimensional Real Vectors Space, Open Subset of Underlying Space That Intersects Star Intersects Simplex Interior of Maximal Simplex Involved in Star

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description/proof of that for simplicial complex on finite-dimensional real vectors space, open subset of underlying space that intersects star intersects simplex interior of maximal simplex involved in star

Topics


About: vectors space
About: topological space

The table of contents of this article

Starting Context

Target Context


  • The reader will have a description and a proof of the proposition that for any simplicial complex on any finite-dimensional real vectors space, any open subset of the underlying space that intersects any star intersects the simplex interior of a maximal simplex involved in the star.

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
\(C\): \(\in \{\text{ the simplicial complexes on } V\}\)
\(\vert C \vert\): \(= \text{ the underlying space of } C\)
\(U\): \(\subseteq \vert C \vert\), \(\in \{\text{ the open subsets of } \vert C \vert\}\)
\(st (p)\): \(= \text{ the star of } p\), where \(p \in Vert C\)
//

Statements:
\(U \cap st (p) \neq \emptyset\)
\(\implies\)
\(\exists S_k \in \{\text{ the maximal simplexes in } C\} (p \in Vert S_k \land S_k^\circ \cap U \neq \emptyset)\)
//

2: Natural Language Description


For any \(d\)-dimensional real vectors space, \(V\), any simplicial complex, \(C\), on \(V\), the underlying space, \(\vert C \vert\), any open subset, \(U \subset \vert C \vert\), and the star of any vertex, \(p \in Vert C\), \(st (p)\), if \(U \cap st (0) \neq \emptyset\), there is a maximal simplex, \(S_k \in C\), such that \(p \in Vert S_k\) and \(S_k^\circ \cap U \neq \emptyset\).

3: Proof


\(st (p)\) involves all the maximal simplexes that have \(p\) as a vertex (there are some positive number of such maximal simplexes), and any other simplex involved in \(st (p)\) is a proper face of at least 1 of the maximal simplexes, by the proposition that for any simplicial complex on any finite-dimensional real vectors space, each simplex in the complex is the faces of the elements of a subset of the maximal simplexes set.

For any \(p' \in U \cap st (p)\), \(p'\) is on the simplex interior of a simplex, \(S_l \in C\), involved in \(st (p)\). \(S_l\) is a (not necessarily proper) face of a maximal simplex, \(S_k \in C\), involved in \(st (p)\).

So, \(p' \in S_k = [p_0 = p, p_1 ..., p_n]\), and \(p' = \sum_{j \in \{0, ..., n\}} t^j p_j\).

If \(S_l\) is \(S_k\) itself, \(p' \in S_k^\circ \cap U \neq \emptyset\).

Let us suppose that \(S_l\) is a proper face of \(S_k\) hereafter.

\(p'\)'s being on a proper face of \(S_k\) means that some of \(t^j\) s are \(0\) but \(0 \lt t^0\), because the face is \([p, p'_1 ..., p'_m]\) where \(\{p, p'_1, ..., p'_m\} \subseteq \{p, p_1, ..., p_n\}\) (the missing vertexes correspond to the \(0\) \(t^j\) s), and if \(t^0 = 0\), \(p'\) would not be on the simplex interior of \([p, p'_1 ..., p'_m]\). That means that \(\sum_{j \in \{1, ..., m\}} t^j \lt 1\), because \(\sum_{j \in \{0, ..., m\}} t^j = 1\).

Let us take a basis of \(V\) as \(\{p_1 - p, ..., p_n - p, b_{n + 1}, ..., b_{d}\}\) and the canonical chart, \((V, \phi)\), \(\phi: s^1 (p_1 - p) + ... + s^n (p_n - p) + s^{n + 1} b_{n + 1} + ... + s^{d} b_{d} \mapsto (s^1, ..., s^{d})\).

As \(p' = t^0 p + \sum_{j \in \{1, ..., n\}} t^j p_j = \sum_{j \in \{1, ..., n\}} t^j (p_j - p) + t^0 p + \sum_{j \in \{1, ..., n\}} t^j p = \sum_{j \in \{1, ..., n\}} t^j (p_j - p) + \sum_{j \in \{0, ..., n\}} t^j p = \sum_{j \in \{1, ..., n\}} t^j (p_j - p) + p\), \(\phi (p') = (t^1 + p^1, ..., t^n + p^n, p^{n + 1}, ..., p^{d})\), where \(\phi (p) = (p^1, ..., p^{d})\).

As \(U\) is open on \(\vert C \vert\), there is an open ball, \(B_{p', \epsilon} \subseteq \vert C \vert\), around \(p'\) such that \(B_{p', \epsilon} \subseteq U\). For any point, \(p'' \in B_{p', \epsilon}\), \(\phi (p'') = (t^1 + \delta^1 + p^1, ..., t^n + \delta^n + p^n, p^{n + 1} + \delta^{n + 1}, ..., p^{d} + \delta^{d})\).

While some of \(t^j\) s are \(0\), the corresponding \(\delta^j\) s can be taken to be small enough positive and the other \(\delta^j\) s can be taken to be \(0\) such that \(t^1 + \delta^1 + ... + t^n + \delta^n \lt 1\) and \(\sqrt{\sum_{j \in \{1, ..., d\}} \delta_j^2} \lt \epsilon\). Then, \(p'' = \sum_{j \in \{1, ..., n\}} (t^j + \delta^j + p^j) (p_j - p) + \sum_{j \in \{n + 1, ..., d\}} p^{j} b_j = \sum_{j \in \{1, ..., n\}} (t^j + \delta^j) (p_j - p) + \sum_{j \in \{1, ..., n\}} p^j (p_j - p) + \sum_{j \in \{n + 1, ..., d\}} p^{j} b_j = \sum_{j \in \{1, ..., n\}} (t^j + \delta^j) (p_j - p) + p = \sum_{j \in \{1, ..., n\}} (t^j + \delta^j) (p_j - p) + \sum_{j \in \{0, ..., n\}} (t^j + \delta^j) p\), where \(\delta^0\) is defined such that \(\sum_{j \in \{0, ..., n\}} (t^j + \delta^j) = 1\), \( = (t^0 + \delta^0) p + \sum_{j \in \{1, ..., n\}} (t^j + \delta^j) p_j\). \(\sum_{j \in \{0, ..., n\}} (t^j + \delta^j) = 1\) and \(0 \lt t^j + \delta^j\). That means that \(p'' \in S_k^\circ\).

So, \(\emptyset \neq (B_{p', \epsilon} \cap S_k^\circ) \subseteq (U \cap S_k^\circ)\).

References
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583: Subset of Underlying Space of Finite Simplicial Complex on Finite-Dimensional Real Vectors Space Is Closed iff Its Intersection with Each Element of Complex Is Closed

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description/proof of that subset of underlying space of finite simplicial complex on finite-dimensional real vectors space is closed iff its intersection with each element of complex is closed

Topics


About: vectors space
About: topological space

The table of contents of this article

Starting Context

Target Context


  • The reader will have a description and a proof of the proposition that any subset of the underlying space of any finite simplicial complex on any finite-dimensional real vectors space is closed iff its intersection with each element of the complex is closed.

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
\(C\): \(\in \{\text{ the finite simplicial complexes on } V\}\)
\(\vert C \vert\): \(= \text{ the underlying space of } C\)
\(S\): \(\subseteq \vert C \vert\)
//

Statements:
\(S \in \{\text{ the closed subsets of } \vert C \vert\}\)
\(\iff\)
\(\forall s \in C (s \cap S \in \{\text{ the closed subsets of } \vert C \vert\})\)
//

2: Natural Language Description


For any \(d\)-dimensional real vectors space, \(V\), with the canonical topology, any finite simplicial complex, \(C\), on \(V\), and the underling space, \(\vert C \vert\), of \(C\), any subset, \(S \subseteq \vert C \vert\), is closed if and only if for each \(s \in C\), \(s \cap S\) is closed on \(\vert C \vert\).

3: Proof


Let us suppose that \(S\) is closed on \(\vert C \vert\).

\(s\) is closed on \(\vert C \vert\), by the proposition that each element of any simplicial complex on any finite-dimensional real vectors space is closed and compact on the underlying space of the complex. \(s \cap S\) is closed on \(\vert C \vert\) as the intersection of the closed subsets.

Let us suppose that \(s \cap S\) is closed on \(\vert C \vert\).

\(S = S \cap \vert C \vert = S \cap \cup_{s \in C} s = \cup_{s \in C} (s \cap S)\), by the proposition that for any set, the intersection of the union of any possibly uncountable number of subsets and any subset is the union of the intersections of each of the subsets and the latter subset. As \(s \cap S\) is closed on \(\vert C \vert\), the finite union, \(\cup_{s \in C} (s \cap S)\), is closed on \(\vert C \vert\).

References
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582: Intersection of Union of Subsets and Subset Is Union of Intersections of Each of Subsets and Latter Subset

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description/proof of that intersection of union of subsets and subset is union of intersections of each of subsets and latter subset

Topics


About: set

The table of contents of this article

Starting Context

Target Context


  • The reader will have a description and a proof of the proposition that for any set, the intersection of the union of any possibly uncountable number of subsets and any subset is the union of the intersections of each of the subsets and the latter subset.

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:
\(S'\): \(\in \{\text{ the sets }\}\)
\(A\): \(\in \{\text{ the possibly uncountable index sets }\}\)
\(\{S_\alpha\}\): \(\alpha \in A\), \(S_\alpha \subseteq S'\)
\(S\): \(S \subseteq S'\)
//

Statements:
\((\cup_{\alpha \in A} S_\alpha) \cap S = \cup_{\alpha \in A} (S_\alpha \cap S)\).
//

2: Natural Language Description


For any set, \(S'\), any possibly uncountable number of subsets, \(\{S_\alpha \subseteq S' \vert \alpha \in A\}\), where \(A\) is any possibly uncountable index set, and any subset, \(S \subseteq S'\), \((\cup_{\alpha \in A} S_\alpha) \cap S = \cup_{\alpha \in A} (S_\alpha \cap S)\).

3: Proof


For any \(p \in (\cup_{\alpha \in A} S_\alpha) \cap S\), \(p \in S_\alpha\) for an \(\alpha\) and \(p \in S\). \(p \in S_\alpha \cap S\) for an \(\alpha\). \(p \in \cup_{\alpha \in A} (S_\alpha \cap S)\).

For any \(p \in \cup_{\alpha \in A} (S_\alpha \cap S)\), \(p \in S_\alpha \cap S\) for an \(\alpha\). \(p \in S_\alpha\) for an \(\alpha\) and \(p \in S\). \(p \in \cup_{\alpha \in A} S_\alpha\) and \(p \in S\). \(p \in (\cup_{\alpha \in A} S_\alpha) \cap S\).

References
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2024-05-12

581: Star of Vertex in Simplicial Complex

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definition of star of vertex in simplicial complex

Topics


About: vectors space

The table of contents of this article

Starting Context

Target Context


  • The reader will have a definition of star of vertex in simplicial complex.

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:
\( V\): \(\in \{\text{ the real vectors spaces }\}\)
\( C\): \(\in \{\text{ the simplicial complexes on } V\}\)
\( p\): \(\in Vert (C)\)
\( Q_p\): \(= \{S_\alpha \in C \vert p \in Ver (S_\alpha)\}\)
\(*star (p)\): \(= \cup_{S_\alpha \in Q_p} S_\alpha^\circ\)
//

Conditions:
//

2: Natural Language Description


For any real vectors space, \(V\), any simplicial complex on \(V\), \(C\), any vertex in \(C\), \(p\), and \(Q_p := \{S_\alpha \in C \vert p \in Ver (S_\alpha)\}\), \(star (p) := \cup_{S_\alpha \in Q_p} S_\alpha^\circ\)

References
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580: Vertex in Simplicial Complex

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definition of vertex in simplicial complex

Topics


About: vectors space

The table of contents of this article

Starting Context

Target Context


  • The reader will have a definition of vertex in simplicial complex.

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:
\( V\): \(\in \{\text{ the real vectors spaces }\}\)
\( C\): \(\in \{\text{ the simplicial complexes on } V\}\)
\(*p\): \(\in Vert (S_\alpha)\) for any \(S_\alpha \in C\)
//

Conditions:
//

The set of the vertexes in \(C\) is denoted as \(Vert (C)\).

2: Natural Language Description


For any real vectors space, \(V\), and any simplicial complex on \(V\), \(C\), any vertex, \(p \in Vert (S_\alpha)\) for any \(S_\alpha \in C\)

The set of the vertexes in \(C\) is denoted as \(Vert (C)\).

References
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579: Vertex of Affine Simplex

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definition of vertex of affine simplex

Topics


About: vectors space

The table of contents of this article

Starting Context

Target Context


  • The reader will have a definition of vertex of affine simplex.

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:
\( V\): \(\in \{\text{ the real vectors spaces }\}\)
\( \{p_0, ..., p_n\}\): \(\subseteq V\), \(\in \{\text{ the affine-independent sets of base points on } V\}\)
\( [p_0, ..., p_n]\): \(= \text{ the affine simplex }\)
\(*p_j\): \(j \in \{0, ..., n\}\)
//

Conditions:
//

The set of the vertexes of \([p_0, ..., p_n]\) is denoted as \(Vert ([p_0, ..., p_n])\).

2: Natural Language Description


For any real vectors space, \(V\), any affine-independent set of base points, \(\{p_0, ..., p_n\} \subseteq V\), and the affine simplex, \([p_0, ..., p_n]\), any \(p_j\) where \(j \in \{0, ..., n\}\)

The set of the vertexes of \([p_0, ..., p_n]\) is denoted as \(Vert ([p_0, ..., p_n])\).

References
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578: For Finite Simplicial Complex on Finite-Dimensional Real Vectors Space, Simplex Interior of Maximal Simplex Is Open on Underlying Space of Complex

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description/proof of that for finite simplicial complex on finite-dimensional real vectors space, simplex interior of maximal simplex is open on underlying space of complex

Topics


About: vectors space

The table of contents of this article

Starting Context

Target Context


  • The reader will have a description and a proof of the proposition that for any finite simplicial complex on any finite-dimensional real vectors space, the simplex interior of each maximal simplex is open on the underlying space of the complex.

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
\(C\): \(\in \{\text{ the finite simplicial complexes on } V\}\)
\(\vert C \vert\): \(= \text{ the underlying space of } C\)
\(S_\alpha\): \(\in \{\text{ the maximal simplexes in } C\}\)
//

Statements:
\(S_\alpha^\circ \in \{\text{ the open subsets of } \vert C \vert\}\).
//

2: Natural Language Description


For any \(d\)-dimensional real vectors space, \(V\), with the canonical topology, any finite simplicial complex, \(C\), on \(V\), and any maximal simplex, \(S_\alpha \in C\), the simplex interior of \(S_\alpha\), \(S_\alpha^\circ\), is open on the underlying space of \(C\), \(\vert C \vert\).

3: Proof


For any \(p \in S_\beta^\circ\), \(p \notin S_\alpha\) for any \(S_\alpha \in C\) such that \(S_\alpha \neq S_\beta\), because \(S_\beta^\circ \cap S_\alpha = \emptyset\), by the proposition that for any simplicial complex, the simplex interior of any maximal simplex does not intersect any other simplex.

As \(S_\alpha\) is closed on \(\vert C \vert\), by the proposition that each element of any simplicial complex on any finite-dimensional real vectors space is closed and compact on the underlying space of the complex, there is an open neighborhood, \(U_{p, \alpha} \subseteq \vert C \vert\), of \(p\) on \(\vert C \vert\) such that \(U_{p, \alpha} \cap S_\alpha = \emptyset\).

As \(S_\beta^\circ\) is open on \(S_\beta\), by the proposition that the simplex interior of any affine simplex is open on the affine simplex with the canonical topology, there is an open neighborhood, \(U'_{p, \beta} \subseteq V\), of \(p\) on \(V\) such that \(U'_{p, \beta} \cap S_\beta \subseteq S_\beta^\circ\), while \(U_{p, \beta} := U'_{p, \beta} \cap \vert C \vert\) is open on \(\vert C \vert\) and \(U_{p, \beta} \cap S_\beta = U'_{p, \beta} \cap S_\beta \subseteq S_\beta^\circ\).

Let us take \(U_p := \cap_{S_\alpha \in C} U_{p, \alpha} \subseteq \vert C \vert\), which is an open neighborhood of \(p\) on \(\vert C \vert\), because \(C\) has only some finite elements. \(U_p \cap S_\alpha = \emptyset\) for \(\alpha \neq \beta\), because \(U_p \subseteq U_{p, \alpha}\) and \(U_{p, \alpha} \cap S_\alpha = \emptyset\). \(U_p = U_p \cap \vert C \vert = U_p \cap \cup_{S_\alpha \in C} S_\alpha = U_p \cap (S_\beta \cup \cup_{S_\alpha \in C \setminus \{S_\beta\}} S_\alpha) = (U_p \cap S_\beta) \cup (U_p \cap \cup_{S_\alpha \in C \setminus \{S_\beta\}} S_\alpha) = (U_p \cap S_\beta) \cup (\cup_{S_\alpha \in C \setminus \{S_\beta\}} (U_p \cap S_\alpha)) = U_p \cap S_\beta \subseteq U_{p, \beta} \cap S_\beta \subseteq S_\beta^\circ\).

By the local criterion for openness, \(S_\beta^\circ\) is open on \(\vert C \vert\).

4: Note


When \(C\) is not finite, \(S_\beta^\circ\) may not be open on \(\vert C \vert\). As a counterexample, let \(V = \mathbb{R}^2\), the Euclidean vectors space, and \(C\) consist of affine 1-simplexes (with their faces); \(S_0 = [(0, 0), (1, 0)]\); \(\{S_j = [(0, 0), (1, 1 / j)] \vert j \in \mathbb{N} \setminus \{0\}\}\); then, for \(p = (1 / 2, 0)\), whatever open ball, \(B_{p, \epsilon} \subseteq \mathbb{R}^2\), intersects an \(S_j\), because \(S_j\) nears \(S_0\) infinitely as \(j\) increases.

While we have proved another proposition that for any finite simplicial complex on any finite-dimensional real vectors space, the simplex interior of each complex-dimensional element is open on the underlying space of the complex, this proposition states that each maximal, not necessarily complex-dimensional, simplex is open: each complex-dimensional simplex is obviously a maximal simplex, but a maximal simplex is not necessarily complex-dimensional.

References
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577: Element of Simplicial Complex on Finite-Dimensional Real Vectors Space Is Closed and Compact on Underlying Space of Complex

<The previous article in this series | The table of contents of this series | The next article in this series>
description/proof of that element of simplicial complex on finite-dimensional real vectors space is closed and compact on underlying space of complex

Topics


About: vectors space
About: topological space

The table of contents of this article

Starting Context

Target Context


  • The reader will have a description and a proof of the proposition that each element of any simplicial complex on any finite-dimensional real vectors space is closed and compact on the underlying space of the complex.

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:
\(V\): \(\in \{\text{ the } d \text{ -dimensional real vectors spaces }\}\)
\(C\): \(\in \{\text{ the simplicial complexes on } V\}\)
\(\vert C \vert\): \(= \text{ the underlying space of } C\)
//

Statements:
\(\forall S_\alpha \in C (S_\alpha \in \{\text{ the closed subsets of } \vert C \vert\} \cap \{\text{ the compact subsets of } \vert C \vert\})\)
//

2: Natural Language Description


For any \(d\)-dimensional vectors space, \(V\), any simplicial complex, \(C\), on \(V\), and the underling space, \(\vert C \vert\), of \(C\), each \(S_\alpha \in C\) is a closed and compact subset of \(\vert C \vert\).

3: Proof


As \(S_\alpha\) is an affine simplex on \(V\), \(S_\alpha\) is closed and compact on \(V\), by the proposition that any affine simplex on any finite-dimensional real vectors space is closed and compact on the canonical topological superspace.

As \(S_\alpha = S_\alpha \cap \vert C \vert\), \(S_\alpha\) is a closed subset of \(\vert C \vert\).

\(S_\alpha\) is compact on \(\vert C \vert\), by the proposition that for any topological space, any subspace subset that is compact on the base space is compact on the subspace.

References
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576: Simplex Interior of Affine Simplex Is Open on Affine Simplex with Canonical Topology

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description/proof of that simplex interior of affine simplex is open on affine simplex with canonical topology

Topics


About: vectors space
About: topological space

The table of contents of this article

Starting Context

Target Context


  • The reader will have a description and a proof of the proposition that the simplex interior of any affine simplex is open on the affine simplex with the canonical topology.

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
\(\{p_0, ..., p_n\}\): \(\subseteq V\), \(\in \{\text{ the affine-independent sets of base points on } V\}\)
\([p_0, ..., p_n]\): \(= \text{ the affine simplex }\) as the topological subspace of \(V\)
//

Statements:
\([p_0, ..., p_n]^\circ \in \{\text{ the open subsets of } [p_0, ..., p_n]\}\).
//

2: Natural Language Description


For any \(d\)-dimensional real vectors space, \(V\), with the canonical topology, any affine-independent set of base points on \(V\), \(\{p_0, ..., p_n\} \subseteq V\), and the affine simplex, \([p_0, ..., p_n]\), as the topological subspace of \(V\), the simplex interior, \([p_0, ..., p_n]^\circ\) is an open subset of \([p_0, ..., p_n]\).

3: Proof


\([p_0, ..., p_n]^\circ = [p_0, ..., p_n] \setminus \cup_{j \in J} F_j\), where \(\{F_j \vert j \in J\}\) is the set of the proper faces of \([p_0, ..., p_n]\).

When \(n = 0\), \([p_0]^\circ = [p_0]\), which is open on \([p_0]\).

Let us suppose that \(1 \le n\).

Let \((V, \phi)\) be the canonical chart with respect to any basis of \(V\), \(\{p_1 - p_0, ..., p_n - p_0, b_{n + 1}, ..., b_d\} \subseteq V\).

For any \(p' \in [p_0, ..., p_n]\), \(p' = \sum_{j \in \{0, ..., n\}} t'^j p_j = \sum_{j \in \{0, ..., n\}} t'^j (p_j - p_0) + \sum_{j \in \{0, ..., n\}} t'^j p_0 = \sum_{j \in \{1, ..., n\}} t'^j (p_j - p_0) + p_0\), which means that \(\phi (p') = (t'^1 + p^1_0, ..., t'^n + p^n_0, p^{n + 1}_0, ..., p^d_0)\), where \(\phi (p_0) = (p^1_0, ..., p^d_0)\).

Let \(p \in [p_0, ..., p_n] \setminus \cup_{j \in J} F_j\) be any. \(\phi (p) = (t^1 + p^1_0, ..., t^n + p^n_0, p^{n + 1}_0, ..., p^d_0)\).

Let us take \(\delta\) such that \(0 \lt \delta \lt min (t^1, ..., t^n, (1 - \sum_{j \in \{1, ..., n\}} t^j) / n, 1 - t^1, ..., 1 - t^n, (\sum_{j \in \{1, ..., n\}} t^j) / n)\): \(0 \lt min (t^1, ..., t^n, (1 - (\sum_{j \in \{1, ..., n\}} t^j)) / n, 1 - t^1, ..., 1 - t^n, (\sum_{j \in \{1, ..., n\}} t^j) / n)\), because \(p\) is not on any proper face of \([p_0, ..., p_n]\) (the condition of being on a proper face is that \(t^j = 0\) for a \(j \in \{0, ..., n\}\), while \(t^0 = 1 - (\sum_{j \in \{1, ..., n\}} t^j)\), and \(t^j = 1\) implies that \(t^k = 0\) for \(k \neq j\)). Let us take \(U_{\phi (p)} = (t^1 + p^1_0 - \delta, t^1 + p^1_0 + \delta) \times ... \times (t^n + p^n_0 - \delta, t^n + p^n_0 + \delta) \times (p^{n + 1}_0 - \delta, p^{n + 1}_0 + \delta) \times ... \times (p^d_0 - \delta, p^d_0 + \delta) \subseteq \mathbb{R}^d\), an open neighborhood of \(\phi (p)\) on \(\mathbb{R}^d\).

\(U_{\phi (p)} \cap \phi ([p_0, ..., p_n])\), an open neighborhood of \(\phi (p)\) on \(\phi ([p_0, ..., p_n])\), is \((t^1 + p^1_0 - \delta, t^1 + p^1_0 + \delta) \times ... \times (t^n + p^n_0 - \delta, t^n + p^n_0 + \delta) \times \{p^{n + 1}_0\} \times ... \times \{p^d_0\}\), because \(\delta\) has been chosen such that \(0 \lt t^j - \delta \lt t^j + \delta \lt 1\) and \(0 \lt 1 - (\sum_{j \in \{1, ..., n\}} (t^j + \delta)) = 1 - (\sum_{j \in \{1, ..., n\}} t^j) - n \delta \lt 1 - (\sum_{j \in \{1, ..., n\}} (t^j - \delta)) = 1 - (\sum_{j \in \{1, ..., n\}} t^j) + n \delta \lt 1\): for any \(v = (t'^1 + p^1_0, ..., t'^n + p^n_0, p^{n + 1}_0, p^d_0) \in (t^1 + p^1_0 - \delta, t^1 + p^1_0 + \delta) \times ... \times (t^n + p^n_0 - \delta, t^n + p^n_0 + \delta) \times \{p^{n + 1}_0\} \times ... \times \{p^d_0\}\), \(t^j + p^j_0 - \delta \lt t'^j + p^j_0 \lt t^j + p^j_0 + \delta\), which implies that \(0 \lt t^j - \delta \lt t'^j \lt t^j + \delta \lt 1\) and \(\sum_{j \in \{1, ..., n\}} t^j - n \delta \lt \sum_{j \in \{1, ..., n\}} t'^j \lt \sum_{j \in \{1, ..., n\}} t^j + n \delta\), which implies that \(0 \lt 1 - \sum_{j \in \{1, ..., n\}} t^j - n \delta \lt 1 - \sum_{j \in \{1, ..., n\}} t'^j \lt 1 - \sum_{j \in \{1, ..., n\}} t^j + n \delta \lt 1\), and so, there is \(p' = \sum_{j \in \{1, ..., n\}} t'^j (p_j - p_0) + p_0 \in [p_0, ..., p_n]\) such that \(\phi (p') = v\). \(U_{\phi (p)} \cap \phi ([p_0, ..., p_n]) \subseteq \phi ([p_0, ..., p_n] \setminus \cup_{j \in J} F_j)\), because being on an \(F_j\) means that \(t'^k = 0\) for a \(k \in \{0, ..., n\}\), but we already know that \(0 \lt t'^j \lt 1\) for \(j \in \{0, ..., n\}\).

As \(\phi\) is a homeomorphism, \(\phi^{-1} (U_{\phi (p)} \cap \phi ([p_0, ..., p_n]))\) is an open neighborhood of \(p\) on \([p_0, ..., p_n]\) such that \(\phi^{-1} (U_{\phi (p)} \cap \phi ([p_0, ..., p_n])) \subseteq [p_0, ..., p_n] \setminus \cup_{j \in J} F_j\). So, \([p_0, ..., p_n] \setminus \cup_{j \in J} F_j\) is open on \([p_0, ..., p_n]\), by the local criterion for openness.

References
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575: Affine Simplex on Finite-Dimensional Real Vectors Space Is Closed and Compact on Canonical Topological Superspace

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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


  • 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
\(\{p_0, ..., p_n\}\): \(\subseteq V\), \(\in \{\text{ the affine-independent sets of base points on } V\}\)
\([p_0, ..., p_n]\): \(= \text{ the affine simplex }\)
//

Statements:
\([p_0, ..., p_n]\) is closed and compact on \(V\).
//

\([p_0, ..., p_n]\) 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.

References
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