View source for Geometrically independent set

{{Stub page|grade=A*|msg=Demote once generalised, this is taken from an algebraic topology textbook. As such it deals with {{M|\mathbb{R}^N}} not an arbitrary [[vector space]]
* Can we generalise further to a [[module]] rather than a vector space? Probably as this is a stone's throw from [[Linearly independent set]]
** Of course we can, see the equivalent statement}}
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==Definition==
Let {{M|(V,\mathbb{F})}} be a [[vector space]] over the [[field]] {{M|\mathbb{F} }} and let {{M|\{a_0,\ldots,a_n\}\subseteq V}} be a set of points. We say this set of points is "''geometrically independent''" if{{rEOATJRM}}:
* {{MM|\forall (t_i)_{i\eq 0}^n\subseteq\mathbb{F}\left[\left(\sum_{i\eq 0}^n t_ia_i\eq 0\wedge\sum_{i\eq 0}^n t_i\eq 0\right)\implies \forall i\in\{0,\ldots,n\}\subset\mathbb{N}[t_i\eq 0]\right]}}
==Equivalent statement==
* {{M|\{a_0,\ldots,a_n\} }} is ''geometrically independent'' {{iff}} the vectors {{M|\{a_1-a_0,\ldots,a_n-a_0\} }} is a [[linearly independent set]]<ref name="EOATJRM"/>
** We need not choose {{M|a_0}} as our point to discard and subtract from everything else, we can choose any member. How to write this though...
{{Requires work|grade=A*|msg=Where's the proof?}}
==See also==
* [[Simplex]]
* [[Affine transformation]]
* [[Linearly independent set]]
==References==
<references/>
{{Definition|Linear Algebra|Algebraic Topology|Topology|Abstract Algebra}}