description/proof of that for simplicial complex, intersection of 2 simplexes is simplex determined by intersection of sets of vertexes of simplexes
Topics
About: vectors space
The table of contents of this article
- Starting Context
- Target Context
- Orientation
- Main Body
- 1: Structured Description
- 2: Natural Language Description
- 3: Proof
Starting Context
- The reader knows a definition of simplicial complex.
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.
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:
\(V\): \(\in \{\text{ the real vectors spaces }\}\)
\(C\): \(\in \{\text{ the simplicial complexes on } V\}\)
\(S_1\): \(= [p_{1, 0}, ..., p_{1, n_1}]\), \(\in C\)
\(S_2\): \(= [p_{2, 0}, ..., p_{2, n_2}]\), \(\in C\)
\(\{p_0, ..., p_n\}\): \(= \{p_{1, 0}, ..., p_{1, n_1}\} \cap \{p_{2, 0}, ..., p_{2, n_2}\}\)
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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 \(\{p_0, ..., p_n\} = \{p_{1, 0}, ..., p_{1, n_1}\} \cap \{p_{2, 0}, ..., p_{2, n_2}\}\).
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, \(S_1 \cap S_2 = [p_0, ..., p_n]\).
