description/proof of that for affine simplex, simplex interior, and vertex, line segment from point on simplex interior to vertex is contained in union of simplex interior and vertex
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 affine simplex.
- The reader knows a definition of simplex interior of affine simplex.
Target Context
- The reader will have a description and a proof of the proposition that for any affine simplex, its simplex interior, and its any vertex, the line segment from any point on the simplex interior to the vertex is contained in the union of the simplex interior and the vertex.
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:
\(V\): \(\in \{\text{ the real vectors spaces }\}\)
\(\{p_0, ..., p_n\}\): \(\subseteq V\), \(\in \{\text{ the affine-independent sets of base points on } V\}\)
\([p_0, ..., p_n]\): \(= \text{ the affine simplex }\)
\([p_0, ..., p_n]^\circ\): \(= \text{ the simplex interior of } [p_0, ..., p_n]\)
\(p\): \(\in [p_0, ..., p_n]^\circ\)
\(\overline{p p_k}\): \(= \text{ the line segment from } p \text{ to } p_k\)
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Statements:
\(\overline{p p_k} \subseteq [p_0, ..., p_n]^\circ \cup \{p_k\}\)
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2: Natural Language Description
For any real vectors space, \(V\), any affine-independent set of base points on \(V\), \(\{p_0, ..., p_n\}\), the affine simplex, \([p_0, ..., p_n]\), the simplex interior of \([p_0, ..., p_n]\), \([p_0, ..., p_n]^\circ\), and any point, \(p \in [p_0, ..., p_n]^\circ\), the line segment from \(p\) to \(p_k\), \(\overline{p p_k}\), is contained in \([p_0, ..., p_n]^\circ \cup \{p_k\}\), which is \(\overline{p p_k} \subseteq [p_0, ..., p_n]^\circ \cup \{p_k\}\).
3: Proof
Whole Strategy: Step 1: express \(\overline{p p_k}\) as a set with \(p_k\) and \(p\); Step 2: express \(p\) as a linear combination of \(p_k\) and \(p_j - p_k\) s; Step 3: put the Step 2 expression into \(p\) of the Step 1 expression, and see that each element of \(\overline{p p_k}\) is on \([p_0, ..., p_n]^\circ\) or is \(p_k\).
Step 1:
\(\overline{p p_k} = \{p_k + t (p - p_k) \vert 0 \le t \le 1\}\).
Step 2:
\(p = \sum_{j \in \{0, 1, ..., n\} \setminus \{k\}} t^j (p_j - p_k) + p_k\), where \(0 \lt t^j\) and \(\sum_{j \in \{0, 1, ..., n\} \setminus \{k\}} t^j \lt 1\).
Step 3:
So, \(\overline{p p_k} = \{p_k + t (\sum_{j \in \{0, 1, ..., n\} \setminus \{k\}} t^j (p_j - p_k) + p_k - p_k) \vert 0 \le t \le 1\} = \{p_k + \sum_{j \in \{0, 1, ..., n\} \setminus \{k\}} t t^j (p_j - p_k) \vert 0 \le t \le 1\}\).
When \(0 \lt t\), \(0 \lt t t^j\) and \(\sum_{j \in \{0, 1, ..., n\} \setminus \{k\}} t t^j \lt 1\), which means that \(p_k + \sum_{j \in \{0, 1, ..., n\} \setminus \{k\}} t t^j (p_j - p_k) \in [p_0, ..., p_n]^\circ\). When \(0 = t\), \(p_k + \sum_{j \in \{0, 1, ..., n\} \setminus \{k\}} t t^j (p_j - p_k) = p_k\).
So, \(\overline{p p_k} \subseteq [p_0, ..., p_n]^\circ \cup \{p_k\}\).
