description/proof of that convex set spanned by possibly-non-affine-independent set of base points on real vectors space is convex
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
Target Context
- The reader will have a description and a proof of the proposition that the convex set spanned by any possibly-non-affine-independent set of base points on any real vectors space is convex.
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:
\(V\): \(\in \{\text{ the real vectors spaces }\}\)
\(\{p_0, ..., p_n\}\): \(\subseteq V\), \(\in \{\text{ the possibly-non-affine-independent sets of base points on } V\}\)
\(S\): \(= \{\sum_{j = 0 \sim n} t^j p_j \in V \vert t^j \in \mathbb{R}, \sum_{j = 0 \sim n} t^j = 1 \land 0 \le t^j\}\)
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\(S \in \{\text{ the convex sets }\}\)
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2: Natural Language Description
For any real vectors space, \(V\), and any possibly-non-affine-independent set of base points, \(\{p_0, ..., p_n\} \subseteq V\), the convex set spanned by the set of the base points, \(S := \{\sum_{j = 0 \sim n} t^j p_j \in V \vert t^j \in \mathbb{R}, \sum_{j = 0 \sim n} t^j = 1 \land 0 \le t^j\}\), is convex.
3: Proof
Let \(\sum_{j = 0 \sim n} t^j_1 p_j, \sum_{j = 0 \sim n} t^j_2 p_j \in S\) be any points. S's being convex is about that \(\sum_{j = 0 \sim n} t^j_1 p_j + t (\sum_{j = 0 \sim n} t^j_2 p_j - \sum_{j = 0 \sim n} t^j_1 p_j)\) is on \(S\) whenever \(0 \le t \le 1\).
\(\sum_{j = 0 \sim n} t^j_1 p_j + t (\sum_{j = 0 \sim n} t^j_2 p_j - \sum_{j = 0 \sim n} t^j_1 p_j) = \sum_{j = 0 \sim n} (t^j_1 (1 - t) + t t^j_2) p_j\). \(\sum_{j = 0 \sim n} (t^j_1 (1 - t) + t t^j_2) = \sum_{j = 0 \sim n} (t^j_1 (1 - t)) + \sum_{j = 0 \sim n} (t t^j_2) = (1 - t) \sum_{j = 0 \sim n} t^j_1 + t \sum_{j = 0 \sim n} t^j_2 = 1 - t + t = 1\). \(0 \le t^j_1 (1 - t) + t t^j_2\), because \(0 \le t^j_1, 1 - t, t, t^j_2\).
So, \(\sum_{j = 0 \sim n} t^j_1 p_j + t (\sum_{j = 0 \sim n} t^j_2 p_j - \sum_{j = 0 \sim n} t^j_1 p_j) \in S\) whenever \(0 \le t \le 1\).
