> [!definition] > > Let > $ > \mathbb{P}^n = \bracs{x\real: x \ne 0, x \in \real^{n + 1}} > $ > be the space of all one-dimensional linear subspaces of $\real^{n + 1}$, and let $\pi: \real^{n + 1} \setminus \bracs{0} \to \mathbb{P}^n$ with $x \mapsto x\real$ be the canonical projection. > > For each $i \in [n + 1]$, let $\widetilde{U}_i \in \mathbb{R}^{n+1}$ be the set where $x^i \ne 0$, and let $U_i = \pi(\widetilde{U_i})$. Since $\widetilde U_i$ is saturated, $U_i$ is open, and $\pi|_{\widetilde U_i}: \tilde U_i \to U_i$ is a [[Quotient Topology|quotient map]]. Define a map $\varphi_i: U_i \to \real^n$ by > $ > \pi(x^1, \cdots, x^{n + 1}) \mapsto \paren{\frac{x^1}{x^i}, \cdots, \frac{x^{i - 1}}{x^i}, \frac{x^{i + 1}}{x^i}, \cdots, \frac{x^{n + 1}}{x^i}} > $ > which is well-defined as it is independent of scalar multiplications, continuous because $\varphi_i \circ \pi$ is, and has an inverse > $ > (u^1, \cdots, u^n) \mapsto \pi(u^1, \cdots, u^{i - 1}, 1, u^i, \cdots, u^n) > $ > Since the charts $\bracs{(\widetilde U_i, \varphi_i)}_1^n$ cover the space, $\mathbb P^n$ is a $\real^n$-[[Manifold|manifold]], known as the **projective space**[^1]. > [!theorem] > > $\mathbb{P}^n$ is Hausdorff. > > *Proof*. Let $x\real, y\real \in \mathbb{P}^n$. Assume without of generality that $\norm{x} = \norm{y} = 1$ ($x, y \in \mathbb{S}^n$). Let $U_x, U_y$ be separating neighbourhoods of $x$ and $y$ on $\mathbb{S}^n$ such that $\pm U_x \cap \pm U_y = \emptyset$. Since $\mathbb{S}^n \times \real^+ \iso \real^{n + 1} \setminus \bracs{0}$, the sets > $ > \bigcup_{\alpha > 0}\alpha( U_x \cup -U_x) \quad \bigcup_{\alpha > 0}\alpha (U_y \cup -U_y) > $ > are saturated open sets that are disjoint. Therefore their images under the projection are disjoint open sets that separate $\pi(x)$ and $\pi(y)$. > [!theorem] > > The canonical projection $\pi: U \to V$ is open. > > *Proof*. Let $U \subset \real^{n + 1} \setminus \bracs{0}$ be an open set, and $x \in U$, then there exists $\varepsilon > 0$ such that $B(x, \varepsilon) \subset U$. > > Hence the open cone > $ > E = \bigcup_{x' \in x\real \setminus \bracs{0}}B\paren{x', \varepsilon \cdot \frac{\norm{x'}}{\norm{x}}} > $ > is contained in $\pi^{-1}(\pi(U))$. Applying this to every point in $U$ yields that $\pi^{-1}(\pi(U))$ is a union of open cones, and thus a saturated open set. Therefore $\pi(U)$ is open. > [!theorem] > > $\mathbb{P}^n$ is second countable. > > *Proof*. Since the canonical projection is open, a countable base from $\real^{n + 1}$ can be transported to $\mathbb{P}^n$. > [!theorem] > > $\mathbb{P}^n$ is [[Compactness|compact]]. > > *Proof*. $\pi|_{\mathbb{S}^n}$ is surjective. Since $\mathbb{S}^n$ is compact, $\mathbb{P}^n$ is as well. [^1]: Lee *Introduction to Smooth Manifolds*, Chapter 1. [^2]: [[Projection From Sphere Shenanigans]]