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Let $P_0,P_1,\ldots,P_r$ be distinct points in $\mathbb{P}^n$. Why there is a hyperplane $H$ in $\mathbb{P}^n$ passing through $P_0$ but not through any of $P_1,\ldots,P_r$?

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Maybe you want to say that $P_0 \neq \cdots$? – Dylan Moreland Jun 21 '12 at 16:11
Don't you need some restrictions on your points? – Derek Allums Jun 21 '12 at 16:11
The result is obviously false for projective space over a finite field $k$ : just take $P_0,P_1,...,P_n$ to be an enumeration of all the points of $\mathbb P^n(k)$ ! – Georges Elencwajg Jun 21 '12 at 17:44
up vote 4 down vote accepted

By projective duality, your question is equivalent to asking why, given a finite collection of distinct hyperplanes in $\mathbb P^n$, there is a point lying on exactly one of them. Does that make it any easier?

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Isn't this true in more general settings too? It seems to me that the essential situation here is that there is an embarrassment of riches: there are a lot of places to put the hyperplane, and only a very few of those will intersect one of the $P_i$. If the hyperplane accidentally does intersect a $P_i$, a small peturbation will move it aside so that it no longer does. – MJD Jun 21 '12 at 16:24
@Mark: Dear Mark, That's right. Regards, – Matt E Jun 21 '12 at 16:24
@Matt E: would it be valid to say that wlog the hyperplanes are $V(x_{i})$ for $i \in \{0,1,..r\}$ so take for example a point in which the first coordinate is zero and the rest are equal to $1$ this point would be in $V(x_{0})$ but not in the remaining ones. – user10 Jun 21 '12 at 16:43
@user10: Dear user, No, that wouldn't be valid, because not all hyperplane configurations are projectively equivalent to coordinate hyperplane configurations. (If $r$ is larger then $n+1$, then you can't find $r$ different coordinates in the first place.) Regards, – Matt E Jun 22 '12 at 6:51

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