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Consider triples of points $u,v,w \in R^2$, which we may consider as single points $(u,v,w) \in R^6$. Show that for almost every $(u,v,w) \in R^6$, the points $u,v,w$ are not collinear.

I think I should use Sard's Theorem, simply because that is the only "almost every" statement in differential topology I've read so far. But I have no idea how to relate this to regular value etc, and to solve this problem.

Another Theorem related to this problem is Fubini Theorem (for measure zero):

Let $A$ be a closed subset of $R^n$ such that $A \cap V_c$ has measure zero in $V_c$ for all $c \in R^k$. Then $A$ has measure zero in $R^n$.

Thank you very much for your help!

1LiterTears
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1 Answers1

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$u,v,$ and $w$ are collinear if and only if there is some $\lambda\in\mathbb{R}$ with $w=v+\lambda(v-u)$. We can thus define a smooth function

$$\begin{array}{rcl}f:\mathbb{R}^5&\longrightarrow&\mathbb{R}^6\\(u,v,\lambda)&\longmapsto&(u,v,v+\lambda(v-u))\end{array}$$

By the equivalence mentioned in the first sentence, the image of $f$ is exactly the points $(u,v,w)$ in $\mathbb{R}^6$ with $u,v,$ and $w$ collinear. Now, because $5<6$, every point in $\mathbb{R}^5$ is a critical point, so that the entire image of $f$ has measure $0$, by Sard's theorem.

Jared
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