543: Convex Set Spanned by Possibly-Non-Affine-Independent Set of Base Points on Real Vectors Space Is Convex

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description/proof of that convex set spanned by possibly-non-affine-independent set of base points on real vectors space is convex

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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 convex set spanned by possibly-non-affine-independent set of base points on real vectors space.
• The reader knows a definition of convex subset of real vectors space.

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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.

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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:
$V$: $\in \{\text{ the real vectors spaces }\}$
$\{p_0,...,p_n\}$: $\subseteq V$, $\in \{\text{ the possibly-non-affine-independent sets of base points on }V\}$
$S$: $=\{\sum _{j=0\sim n}{t}^j{p}_j\in V|t^j\in \mathbb{R},\sum _{j=0\sim n}{t}^j=1\wedge 0\le t^j\}$
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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, $S:=\{\sum _{j=0\sim n}{t}^j{p}_j\in V|t^j\in \mathbb{R},\sum _{j=0\sim n}{t}^j=1\wedge 0\le t^j\}$, is convex.

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3: Proof

Let $\sum _{j=0\sim n}{t}_1^j{p}_j,\sum _{j=0\sim n}{t}_2^j{p}_j\in S$ be any points. S's being convex is about that $\sum _{j=0\sim n}{t}_1^j{p}_j+t(\sum _{j=0\sim n}{t}_2^j{p}_j-\sum _{j=0\sim n}{t}_1^j{p}_j)$ is on $S$ whenever $0\le t\le 1$.

$\sum _{j=0\sim n}{t}_1^j{p}_j+t(\sum _{j=0\sim n}{t}_2^j{p}_j-\sum _{j=0\sim n}{t}_1^j{p}_j)=\sum _{j=0\sim n}(t_1^j(1-t)+t{t}_2^j)p_j$. $\sum _{j=0\sim n}(t_1^j(1-t)+t{t}_2^j)=\sum _{j=0\sim n}(t_1^j(1-t))+\sum _{j=0\sim n}(t{t}_2^j)=(1-t)\sum _{j=0\sim n}{t}_1^j+t\sum _{j=0\sim n}{t}_2^j=1-t+t=1$. $0\le t_1^j(1-t)+t{t}_2^j$, because $0\le t_1^j,1-t,t,t_2^j$.

So, $\sum _{j=0\sim n}{t}_1^j{p}_j+t(\sum _{j=0\sim n}{t}_2^j{p}_j-\sum _{j=0\sim n}{t}_1^j{p}_j)\in S$ whenever $0\le t\le 1$.

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References

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