620: For Simplicial Complex, Intersection of 2 Simplexes Is Simplex Determined by Intersection of Sets of Vertexes of Simplexes

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description/proof of that for simplicial complex, intersection of 2 simplexes is simplex determined by intersection of sets of vertexes of simplexes

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Topics

About: vectors 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.

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Target Context

• The reader will have a description and a proof of the proposition that for any simplicial complex, the intersection of any 2 simplexes is the simplex determined by the intersection of the sets of the vertexes of the simplexes.

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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:
$V$: $\in \{\text{ the real vectors spaces }\}$
$C$: $\in \{\text{ the simplicial complexes on }V\}$
$S_1$: $=[p_{1,0},...,p_{1,n_1}]$, $\in C$
$S_2$: $=[p_{2,0},...,p_{2,n_2}]$, $\in C$
$\{p_0,...,p_n\}$: $=\{p_{1,0},...,p_{1,n_1}\}\cap \{p_{2,0},...,p_{2,n_2}\}$
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Statements:
$S_1\cap S_2=[p_0,...,p_n]$
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2: Natural Language Description

For any real vectors space, $V$, any simplicial complex, $C$, on $V$, and any simplexes, $S_1=[p_{1,0},...,p_{1,n_1}],S_2=[p_{2,0},...,p_{2,n_2}]\in C$, $S_1\cap S_2=[p_0,...,p_n]$, where $\{p_0,...,p_n\}=\{p_{1,0},...,p_{1,n_1}\}\cap \{p_{2,0},...,p_{2,n_2}\}$.

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3: Proof

$S_1\cap S_2$ is a face of $S_1$ and a face of $S_2$. As any face is a simplex in $C$, $S_1\cap S_2$ is a simplex in $C$.

Each vertex of $S_1\cap S_2$ is a vertex of $S_1$ and a vertex of $S_2$, by the definition of face. So, $Vert{S}_1\cap S_2\subseteq \{p_0,...,p_n\}$. That implies that $S_1\cap S_2\subseteq [p_0,...,p_n]$.

Let us prove that $[p_0,...,p_n]\subseteq S_1\cap S_2$. For each $p\in [p_0,...,p_n]$, $p=\sum _{j\in \{0,...,n\}}{t}^j{p}_j$, where $\sum _{j\in \{0,...,n\}}{t}^j=1$ and $0\le t^j$. $p\in S_1\cap S_2$, because $p$ is a special case of $\sum _{j\in \{0,...,n_1\}}{t}^j{p}_{1,j}$ and a special case of $\sum _{j\in \{0,...,n_2\}}{t}^j{p}_{2,j}$.

So, $S_1\cap S_2=[p_0,...,p_n]$.

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References

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