# Geometrically independent set

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Demote once generalised, this is taken from an algebraic topology textbook. As such it deals with [ilmath]\mathbb{R}^N[/ilmath] 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

## Definition

Let [ilmath](V,\mathbb{F})[/ilmath] be a vector space over the field [ilmath]\mathbb{F} [/ilmath] and let [ilmath]\{a_0,\ldots,a_n\}\subseteq V[/ilmath] be a set of points. We say this set of points is "*geometrically independent*" if^{[1]}:

- [math]\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][/math]

## Equivalent statement

- [ilmath]\{a_0,\ldots,a_n\} [/ilmath] is
*geometrically independent**if and only if*the vectors [ilmath]\{a_1-a_0,\ldots,a_n-a_0\} [/ilmath] is a linearly independent set^{[1]}- We need not choose [ilmath]a_0[/ilmath] as our point to discard and subtract from everything else, we can choose any member. How to write this though...

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Where's the proof?

## See also

## References

- ↑
^{1.0}^{1.1}Elements of Algebraic Topology - James R. Munkres