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How many regions does n non-concurrent lines divide a Projective Plane into? (probably a standard problem, but I am having conflicting answers)

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    I would say $1+\frac{n(n-1)}{2}$. What are your answers?2012-11-08
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    Mine is $2^{n-1}$. How do you arrive at that?2012-11-08
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    I have posted an answer. This is actually my second solution. The first one was to use the similar problem for a usual Euclidean plane. For the usual plane $\mathbb{R}^2$, the answer is $1+n(n+1)/2$, and you you can turn $\mathbb{R}P^1$ into $\mathbb{R}^2$ by removing one of the lines, so the answer is $1+n(n-1)/2$.2012-11-08
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    @ Dan: thanks for the answer. I forgot to clarify, but by projective plane I wanted to mean $\mathbb{R}P^{2}$. But I guess we are talking of the same thing under different notations, right?2012-11-08
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    Also, it would be helpful if you find me the error in my argument. I was trying to go directly through dividing $\mathbb{R}{}^{3}$ by planes passing through origin. Each such plane $P$ creates regions $P^{+}$ and $P^{-}$. n such planes, no two of which intersect in a line, creates $2^{n}$ regions. (for 2 planes, they would be like $P_{1}^{+}P_{2}^{+},P_{1}^{+}P_{2}^{-},P_{1}^{-}P_{2}^{+},P_{1}^{-}P_{2}^{-}$).2012-11-08
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    Now because of projectivization, $P_{1}^{+}P_{2}^{+}=P_{1}^{-}P_{2}^{-}$ and $P_{1}^{+}P_{2}^{-}=P_{1}^{-}P_{2}^{+}$(since $P_{1}(a,b,c)\lesseqqgtr0\implies P_{1}(-a,-b,-c)\gtreqqless0$).Hence the feeling $2^{n-1}$. Do you see where am I going wrong?2012-11-08
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    Yep, this is my mistake. I wrote $\mathbb{R}P^1$ instead $\mathbb{R}P^2$ for some reason. Don't know what came onto me.2012-11-08
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    And yes, I see where you're going wrong. Even before the projectivization, when you're in \mathbb{R}^3. If $n$ is large, then most of the $2^n$ regions that you're talking about will be simply empty. $3$ planes will divide $\mathbb{R}^3$ into $8$ regions. But $4$ planes will only divide it into $14$ regions, not $16$.2012-11-08
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    I guess it will become clearer if you lower the dimension and look at lines passing through the origin in $\mathbb{R}^2$. $2$ lines will divide it inso $4$ regions, but $3$ lines produce only $6$ regions instead of $8$.2012-11-08

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