633: For Finite Simplicial Complex, Stars of Vertexes of Simplexes Is Open Cover of Underlying Space

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description/proof of that for finite simplicial complex, stars of vertexes of simplexes is open cover of underlying space

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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: Proof

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Starting Context

• The reader knows a definition of simplicial complex.
• The reader knows a definition of star of vertex in simplicial complex.
• The reader admits the proposition that for any simplicial complex, any point on the underlying space is on the simplex interior of a unique simplex.
• The reader admits the proposition that for any simplicial complex, each vertex of each simplex that is on any another simplex is a vertex of the latter simplex.
• The reader admits 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.
• The reader admits the proposition that the simplex interior of any affine simplex is open on the affine simplex with the canonical topology.
• 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 local criterion for openness.

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Target Context

• The reader will have a description and a proof of the proposition that for any finite simplicial complex, the set of the stars of the vertexes of the simplexes is an open cover of the underlying space.

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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:
$V$: $\in \{\text{ the }{d}^{\prime }\text{ -dimensional real vectors spaces }\}$
$C$: $\in \{\text{ the finite simplicial complexes on }V\}$
$|C|$: $=\text{ the underlying space of }C$
$\{st(p)|p\in Vert(S),S\in C\}$:
//

Statements:
$\{st(p)\}\in \{\text{ the open covers of }|C|\}$.
//

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2: Natural Language Description

For any $d^{\prime }$-dimensional real vectors space, $V$, any finite simplicial complex, $C$, on $V$, and the underlying space, $|C|$, the set of the stars of the vertexes of the simplexes, $\{st(p)|p\in Vert(S),S\in C\}$, is an open cover of $|C|$.

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3: Proof

Let $C=\{S_j|j\in J^{\prime }\}$ where $J^{\prime }$ is a finite index set. Note that there is no duplication among the elements in $C$, because $C$ is defined as a set, and any set does not contain any duplication by definition.

Let us prove that $st(p)$ is open on $|C|$.

Let $st(p)={\cup }_{j\in J}{S}_j^{\circ }$ where $J\subseteq J^{\prime }$.

Let $p^{\prime }\in st(p)$ be any.

While $p^{\prime }\in S_k^{\circ }$ where $k\in J$, $S_k$ is the only such simplex in $C$, by the proposition that for any simplicial complex, any point on the underlying space is on the simplex interior of a unique simplex.

For each $S_j$ such that $j\in J^{\prime }\setminus J$, $p^{\prime }\notin S_j$, because otherwise $p^{\prime }\in S_k^{\circ }\cap S_j$ for a $k\in J$, and $p^{\prime }\in S_k\cap S_j$, but $S_k\cap S_j$ would be a face of $S_k$ but not any proper face of $S_k$, because $p^{\prime }$ would not be on any proper face of $S_k$, so, $S_k\cap S_j=S_k$, but as $p\in S_k$, $p\in S_j$, which would mean that $p$ was a vertex of $S_j$, by the proposition that for any simplicial complex, each vertex of each simplex that is on any another simplex is a vertex of the latter simplex, a contradiction against $j\notin J$.

As $S_j$ is closed on $|C|$, 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, $U_{p^{\prime },j}\subseteq |C|$, of $p^{\prime }$ on $|C|$ such that $U_{p^{\prime },j}\cap S_j=\mathrm{\varnothing }$, and $U_{p^{\prime }}:={\cap }_{j\in J^{\prime }\setminus J}{U}_{p^{\prime },j}\subseteq |C|$ is an open neighborhood of $p^{\prime }$. When we take an open neighborhood of $p^{\prime }$ on $|C|$ hereafter, we will always take it as contained in $U_{p^{\prime }}$, which is accomplished by taking the intersection of the neighborhood and $U_{p^{\prime }}$.

Let us suppose that $p^{\prime }=p$.

$\{p\}=S_k$ and $p^{\prime }\in S_k^{\circ }=S_k$, so, $p^{\prime }\notin S_l^{\circ }$ for each $l\in J$ such that $k\ne l$.

For each $S_l$ such that $l\in J$, let us take an open ball (so, an open neighborhood), $B_{p^{\prime },{ϵ}_l}\subseteq |C|$, of $p^{\prime }$ on $|C|$. Let us think of $B_{p^{\prime },{ϵ}_l}\cap S_l$. $S_l=[p^{\prime },p_1,...,p_n]$ and for any point, $p^{″}\in B_{p^{\prime },{ϵ}_l}\cap S_l$, $p^{″}=(1-\delta )p^{\prime }+\sum _{j\in \{1,...,n\}}{t}^j{p}_j$ where $1-\delta +\sum _{j\in \{1,...,n\}}{t}^j=1$. When ${ϵ}_l$ is small enough, $p^{″}$ is in the union of the interiors of the faces of $S_l$ that (the faces) belong to $st(p)$, because when $\delta =0$, $p^{″}=p^{\prime }$ while $[p^{\prime }]$ is the interior of the face, $[p^{\prime }]$, of $S_l$, and when $0<\delta <1$, if each of $t^j$ s is not $0$, $p^{″}$ is in the interior of $S_l$, and if some of $t^j$ s are $0$, $p^{″}$ is in the interior of the face of $S_l$ with the corresponding vertexes removed, while each of those faces belongs to $st(p)$ because as $0<1-\delta$, the face contains $p$ (for example, when only $t^1$ is $0$, $[p^{″},p_2,...,p_n]$ is a face of $S_l$ that (the face) belongs to $st(p)$, and $p^{″}=(1-\delta )p^{″}+\sum _{j\in \{2,...,n\}}{t}^j{p}_j$ is in the interior of $[p^{″},p_2,...,p_n]$). So, $B_{p^{\prime },{ϵ}_l}\cap S_l\subseteq st(p)$.

Let us take $B_{p^{\prime },ϵ}:={\cap }_{j\in J}{B}_{p^{\prime },{ϵ}_j}\subseteq |C|$, an open neighborhood of $p^{\prime }$ on $|C|$. $B_{p^{\prime },ϵ}=B_{p^{\prime },ϵ}\cap |C|=B_{p^{\prime },ϵ}\cap ({\cup }_{j\in J}{S}_j\cup {\cup }_{j\in J^{\prime }\setminus J}{S}_j)=(B_{p^{\prime },ϵ}\cap {\cup }_{j\in J}{S}_j)\cup (B_{p^{\prime },ϵ}\cap {\cup }_{j\in J^{\prime }\setminus J}{S}_j)=B_{p^{\prime },ϵ}\cap {\cup }_{j\in J}{S}_j={\cup }_{j\in J}(B_{p^{\prime },ϵ}\cap S_j)\subseteq {\cup }_{j\in J}(B_{p^{\prime },{ϵ}_j}\cap S_j)\subseteq {\cup }_{j\in J}st(p)=st(p)$.

Let us suppose that $p^{\prime }\ne p$.

$p^{\prime }\in S_k^{\circ }$ for a $k\in J$.

Let us take an open ball (so, an open neighborhood), $B_{p^{\prime },{ϵ}_k}\subseteq |C|$, of $p^{\prime }$ on $|C|$ such that $B_{p^{\prime },{ϵ}_k}\cap S_k\subseteq S_k^{\circ }$, which is possible by the proposition that the simplex interior of any affine simplex is open on the affine simplex with the canonical topology (while $S_k$ is the subspace of $V$, it is also the subspace of $|C|$, because $|C|$ is the subspace of $V$, 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). So, $B_{p^{\prime },{ϵ}_k}\cap S_k\subseteq S_k^{\circ }\subseteq st(p)$.

For each $S_l$ such that $l\in J$ and $k\ne l$, $p^{\prime }\notin S_l$ or $p^{\prime }\in S_l$.

When $p^{\prime }\notin S_l$, let us take an open ball (so, an open neighborhood), $B_{p^{\prime },{ϵ}_l}\subseteq |C|$, of $p^{\prime }$ on $|C|$ such that $B_{p^{\prime },{ϵ}_l}\cap S_l=\mathrm{\varnothing }$, which is possible because $S_l$ is closed on $|C|$, 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. So, $B_{p^{\prime },{ϵ}_l}\cap S_l\subseteq st(p)$.

When $p^{\prime }\in S_l$, let us take an open ball (so, an open neighborhood), $B_{p^{\prime },{ϵ}_l}\subseteq |C|$, of $p^{\prime }$ on $|C|$. Let us think of $B_{p^{\prime },{ϵ}_l}\cap S_l$. $S_l=[p,p_1,...,p_n]$ and $p^{\prime }$ is on a proper face of $S_l$, because $p^{\prime }\notin S_l^{\circ }$, which means that for $p^{\prime }=t^{\prime 0}p+\sum _{j\in \{1,...,n\}}{t}^{\prime j}{p}_j$ where $t^{\prime 0}+\sum _{j\in \{1,...,n\}}{t}^{\prime j}=1$, some of $t^{\prime j}$ s are $0$ but $0<t^{\prime 0}$, because $p^{\prime }\in S_k^{\circ }$ (with $S_k$ denoted as $[p,p_1^{\prime },...,p_m^{\prime }]$, $p^{\prime }\in S_k\cap S_l$, but $S_k\cap S_l$ is a face of $S_k$, which is not any proper face of $S_k$, so, $S_k\cap S_l=S_k$, which is a face of $S_l$, which means that while $p^{\prime }=t^{″0}p+\sum _{j\in \{1,...,m\}}{t}^{″j}{p}_m^{\prime }$, $p_j^{\prime }$ is in fact a vertex of $S_l$, and so, $p^{\prime }=t^{″0}p+\sum _{j\in \{1,...,m\}}{t}^{″j}{p}_m^{\prime }$ is a convex combination with respect to the vertexes of $S_l$, and as the convex combination is unique, $t^{\prime j}$ s equal the corresponding $t^{″j}$ s, especially, $t^{″0}=t^{\prime 0}$, but $0<t^{″0}$, because $p^{\prime }\in S_k^{\circ }$). For any point, $p^{″}\in B_{p^{\prime },{ϵ}_l}\cap S_l$, $p^{″}=(t^{\prime 0}+{\delta }_0)p+\sum _{j\in \{1,...,n\}}(t^{\prime j}+{\delta }_j)p_j$ where $t^{\prime 0}+{\delta }_0+\sum _{j\in \{1,...,n\}}(t^{\prime j}+{\delta }_j)=1$. When ${ϵ}_l$ is small enough, $0<t^{\prime 0}+{\delta }_0$, and if each of $t^{\prime j}+{\delta }_j$ s is not $0$, $p^{″}\in S_l^{\circ }$, and if some of $t^{\prime j}+{\delta }_j$ s are $0$, $p^{″}$ is on the interior of the face of $S_l$ with the corresponding vertexes removed that (the face) belongs to $st(p)$. So, $B_{p^{\prime },{ϵ}_l}\cap S_l\subseteq st(p)$.

Let us take $B_{p^{\prime },ϵ}:={\cap }_{j\in J}{B}_{p^{\prime },{ϵ}_j}\subseteq |C|$, an open neighborhood of $p^{\prime }$ on $|C|$. $B_{p^{\prime },ϵ}=B_{p^{\prime },ϵ}\cap |C|=B_{p^{\prime },ϵ}\cap ({\cup }_{j\in J}{S}_j\cup {\cup }_{j\in J^{\prime }\setminus J}{S}_j)=(B_{p^{\prime },ϵ}\cap {\cup }_{j\in J}{S}_j)\cup (B_{p^{\prime },ϵ}\cap {\cup }_{j\in J^{\prime }\setminus J}{S}_j)=B_{p^{\prime },ϵ}\cap {\cup }_{j\in J}{S}_j={\cup }_{j\in J}(B_{p^{\prime },ϵ}\cap S_j)\subseteq {\cup }_{j\in J}(B_{p^{\prime },{ϵ}_j}\cap S_j)\subseteq {\cup }_{j\in J}st(p)=st(p)$.

So, by the local criterion for openness, $st(p)$ is open on $|C|$.

Let us prove that $\{st(p)\}$ covers $|C|$.

For each $p^{\prime }\in |C|$, $p^{\prime }$ is on the interior of a simplex in $C$, by the proposition that for any simplicial complex, any point on the underlying space is on the simplex interior of a unique simplex. The simplex has at least 1 vertex, $p$, and $p^{\prime }\in st(p)$.

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References

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