I have to prove the following: Let $S^k\neq\varnothing$ be a projective space. It follows that: $$X(x)\in\cap_{i\in I}S_i^{k_i}\Leftrightarrow [x]\subset (\cap_{i\in I}S_i^{k_i+1})\subset P^{n+1}.$$ It seems to be obvious but I can't prove it. Could someone explain it more intuitively?
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2Please add more context. What are $X,x,[x],P^{n+1}$? – Berci Oct 20 '21 at 21:36
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$X$ is a point in the projective space $P^n, [x]$ is the 1-dim subspace in the associate vector space $P^{n+1}$. – mathplayer Oct 21 '21 at 04:43
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Let me rephrase and improve notation.
Let $S_i$ be projective subspaces of the projective space $\Bbb RP^n$, and let $\pi:\Bbb R^{n+1}\setminus\{0\}\to\Bbb RP^n$ be the canonical projection, which satisfies
$$\pi(x)=\pi(y)\iff x=\lambda y\ \text{ for some }\,\lambda\in\Bbb R\,.$$
Then $\pi^{-1}(S_i)$ is a linear subspace (of dimension $\dim S_i+1$).
Further, instead of $[x]$ let's write $\Bbb Rx$ for the 1d linear subspace $\{\lambda x:\lambda\in\Bbb R\}$ of $\Bbb R^{n+1}$ generated by the given vector $x\in\Bbb R^{n+1}\setminus\{0\}$.
The statement is that for any $x\in\Bbb R^{n+1}$ we have $$\pi(x)\in\bigcap_iS_i\ \iff\ \Bbb R x\subseteq\bigcap_i \pi^{-1}(S_i)\,.$$ And to prove it just observe that for any $i$, we already have $\pi(x)\in S_i\iff\Bbb Rx\subseteq\pi^{-1}(S_i)$.
Berci
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