# A question about Denseness in $\mathbb{P}^n$

$$\mathbb{P}^n=\bigcup_{j=1}^{n} U_j,$$ where $$U_j=\{[x_0:\dots,x_n]\in \mathbb{P}^n : x_j\neq 0\}=\{[x_0:\dots,x_n]\in \mathbb{P}^n : x_j=1\}$$ can be identified with $\mathbb{C}^n.$ In fact we need not use the complements of the hyperplanes defined by $x_i$ to define a cover $\{U_i\}$.Given any set of $n+1$ linearly independent hyperplanes not passing through the origin, the lines in $\mathbb{C}^{n+1}$ intersecting the $i'th$ hyperplane form a set $U_i$ that can be identified with $\mathbb{C}^n,$ and together these give a cover of $\mathbb{P}^n$ that differs from the one we first described only by a change of coordinates in $\mathbb{C}^{n+1}.$ Note that there is a natural Eucidean topology on $\mathbb{P}^n$ induced by virtue of the fact that $\mathbb{P}^n$ is a quotient of $\mathbb{C}^{n+1}\setminus\{0\}$. In particular, two points of $\mathbb{P}^n$ are close together if the corresponding lines in $\mathbb{C}^{n+1}$ have small angle between them. in this Eucledian topology on $\mathbb{P}^n$, each of the sets $U_i$ is open, and identification of $U_i$ with $\mathbb{C}^n$ descriebd above defines a homeomorphism of topological spaces when $\mathbb{C}^n$ considered with its Euclidean topology.

Why each $U_i$ is dense in $\mathbb{P}^n?$ and in fact, its complement is a lower dimensional space (namely, $\mathbb{P}^{n-1}$.)The intersection of $U_i$ with $U_j$ when $i\neq j$ is also dense.

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Where is the question? – martini Mar 13 '12 at 7:26
Why $ecah$ $U_i$ $is$ $dense$ $in$ $\mathbb{P}^n$ ?, and why its complement is a lower dimensional space (namely, $\mathbb{P}^{n-1}$.The intersection of $U_i$ with $U_j$ when $i\neq j$ is also dense? thank you dear sir – La Belle Noiseuse Mar 13 '12 at 7:29

## 1 Answer

We show first that $\mathbb P^n \setminus U_j$ can be identified with $\mathbb P^{n-1}$. By definition we have $$\mathbb P^n \setminus U_j = \{[x_0: \cdots : x_{j-1} : 0 : x_{j+1} : \cdots : x_n] \in \mathbb P^n \mid [x_0: \cdots : x_{j-1} : x_{j+1} : \cdots : x_n] \in \mathbb P^{n-1} \}.$$ So $\mathbb P^n \setminus U_j$ consists of the lines, which ly in the hyperplane $\{x_j = 0\}$ and can so be indentified with $\mathbb P^{n-1}$.

For denseness: Let $x \in \mathbb P^n \setminus U_j$ and $U$ an open subset contatining $x$, choose $i$ with $x \in U_i$. Write $x = [x_0 :\cdots: x_n]$. Then $x_j = 0$, $x_i = 1$. As $U \cap U_i$ is open in $U_i \cong \mathbb C^n$, there is some $y \in U\cap U_i$ with $y_j \ne 0$. Then $y \in U_j \cap U$, so each open subset containing $x$ intersects $U_j$.

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thank you sir ... – La Belle Noiseuse Mar 13 '12 at 9:02
It would be nice if you could accept the answer ... :) – martini Mar 13 '12 at 9:16
@martini, I also had that weird problem with the set braces {}. Using \\{ instead of \{ helped me :) – InvisiblePanda Mar 13 '12 at 9:47
I used \\{ first and the braces showed up ... now I reset it to \{ and it worked again :) – martini Mar 13 '12 at 9:48
Now that's strange oO – InvisiblePanda Mar 13 '12 at 10:23