# Hyperplane in projective space

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

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?
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
@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