Ex: P^{n}(R) = RP^{n} = RP^{n} = (R^{n+1} − {0})/(x ∼ tx) = n-dimensional real projective space is a smooth manifold. Claim: RP^{n} is 2nd countable. We will use Lemma: If ∼ is open and if X has a countable basis, then X/ ∼ has a countable basis. [We will define a map which takes y ∈ [x] to ty ∈ [x] ] Let ϕ_{t} : R^{n+1} − {0} → R^{n+1} − {0}, ϕ_{t}(x) = tx. ϕ_{t} is invertible with inverse ϕ_{t}^{−1} = ϕ_{\frac{1}{t}}. Since ϕ_{t} and ϕ_{t}^{−1} are C^{1} (as well as C^{∞}), ϕ_{t} is a homeomorphism. Let U be open in R^{n+1} −{0}. Then ϕ_{t}(U) is open in R^{n+1} −{0}. Thus π^{−1}([U]) = ∪_{t∈R}ϕ_{t}(U) is open in R^{n+1} − {0}. Thus [U] is open in RP^{n}. Hence ∼ is open. Since R^{n} is 2nd countable, RP^{n} is 2nd countable. Claim RP^{n} is Hausdorff We will use Lemma: Let ∼ be open. Then {(x, y) | x ∼ y} is closed in X ×X iff X/ ∼ is Hausdorff. [We will show that {(x, y) | x ∼ y} = f^{−1}({0}) for some continuous function f. x ∼ y implies x_{i} = ty_{i}. Thus \frac{x_{i}}{y_{i}} = \frac{x_{j}}{y_{j}}. Hence x_{i}y_{j} − y_{i}x_{j} = 0 for all i, j.] Let f : R^{n+1} − {0} × R^{n+1} − {0} → R, f(x_{1}, ..., x_{n}, y_{1}, ..., y_{n}) = Σ_{i≠j}(x_{i}y_{j} − y_{i}x_{j})^{2}.

f is C^{1} (all partials of f exist and are continuous). Thus f is continuous.

Suppose y = tx, then f(x_{1}, ..., x_{n}, y_{1}, ..., y_{n}) = Σ_{i≠j}(x_{i}tx_{j} − tx_{i}x_{j})^{2} = 0.

Suppose f(x_{1}, ..., x_{n}, y_{1}, ..., y_{n}) = Σ_{i≠j}(x_{i}y_{j} − y_{i}x_{j})^{2} = 0. Then x_{i}y_{j} − y_{i}x_{j} = 0 for all i, j. Since x ≠ 0, there exists i_{0} such that x_{i_{0}} ≠ 0. Thus y_{j} = \frac{y_{i_{0}}}{x_{i_{0}}} x_{j} and y = \frac{y_{i_{0}}}{x_{i_{0}}} x. Hence x ∼ y.

Hence f^{−1}({0}) = {(x, y) | x ∼ y}. Since f is continuous and {0} is closed in R, {(x, y) | x ∼ y} is closed in R^{n+1} − {0} × R^{n+1} − {0}. Thus R^{n+1} − {0}/ ∼ is Hausdorff.

We will show that RP^{n} is locally Euclidean by finding a (pre) atlas:

Let V_{i} = {x ∈ R^{n+1} − {0} | x_{i} ≠ 0} ⊂ R^{n+1} − {0}

Let F_{i} : V_{i} → R^{n}, F_{i}(x_{1}, ..., x_{n+1}) = (\frac{x_{1}}{x_{i}} , \frac{x_{2}}{x_{i}} , ..., \frac{x_{i−1}}{x_{i}} , \frac{x_{i+1}}{x_{i}} , ..., \frac{x_{n+1}}{x_{i}}) = \frac{1}{x_{i}} (x_{1}, ..., x̂_{i}, ..., x_{n+1})

F_{i}(tx) = (\frac{tx_{1}}{tx_{i}} , \frac{tx_{2}}{tx_{i}} , ..., \frac{tx_{i−1}}{tx_{i}} , \frac{tx_{i+1}}{tx_{i}} , ..., \frac{tx_{n+1}}{tx_{i}})

= (\frac{x_{1}}{x_{i}} , \frac{x_{2}}{x_{i}} , ..., \frac{x_{i−1}}{x_{i}} , \frac{x_{i+1}}{x_{i}} , ..., \frac{x_{n+1}}{x_{i}}) = F_{i}(x).

Let U_{i} = π(V_{i}) Then ϕ_{i} : U_{i} → R^{n}, ϕ_{i}([x]) = F_{i}(x) is welldefined.

Claim: (ϕ_{i}, U_{i}) is a chart.

Subclaim 1: U_{i} is open in RP^{n}. π_{i}^{−1}(U_{i}) = π_{i}^{−1}(π(V_{i})) = V_{i} [by set theory]. Hence U_{i} is open in RP^{n}.

Subclaim 2: ϕ(U_{i}) is open in R^{n}.

Claim: ϕ_{i} is onto. Let (x_{1}, ..., x_{n}) ∈ R^{n}. ϕ_{i}([(x_{1}, ..., x_{i−1}, 1, x_{i}, .., x_{n}]) = (x_{1}, ..., x_{n}). Thus ϕ_{i} is onto. Since ϕ(U_{i}) = R^{n} , ϕ(U_{i}) is open in R^{n}.

Subclaim 3: ϕ_{i} is a homeomorphism. Claim: ϕ_{i} is continuous. Let V be open in R^{n}. Since F_{i} ∈ C^{1}, F_{i}^{−1}(V) is open in R^{n+1} − {0}. π^{−1} ∘ ϕ_{i}^{−1}(V) = F_{i}^{−1}(V). Thus ϕ_{i}^{−1}(V) is open in U_{i} and ϕ_{i} is continuous. Claim: ϕ_{i} is 1:1. If ϕ_{i}([x]) = ϕ_{i}([y]), then \frac{x_{j}}{x_{i}} = \frac{y_{j}}{y_{i}} for all j. Thus y_{i} = \frac{y_{i}}{x_{i}} x_{j} and thus y = \frac{y_{i}}{x_{i}} x. Thus ϕ_{i} is 1:1. Since ϕ_{i} is 1:1 and onto, ϕ_{i}^{−1} exists.

Claim: ϕ_{i}^{−1} : R^{n} → RP^{n} is continuous. ϕ_{i}^{−1}(x_{1}, ..., x_{n}) = [(x_{1}, ..., x_{i−1}, 1, x_{i}, .., x_{n})]. Let f_{i} : R^{n} → R^{n+1}, f_{i}(x) = (x_{1}, ..., x_{i−1}, 1, x_{i}, ..., x_{n}). f_{i} is C^{1} and hence continuous. π ∘ f(x) = [(x_{1}, ..., x_{i−1}, 1, x_{i}, .., x_{n})] = ϕ_{i}^{−1}(x_{1}, ..., x_{n}). Since π and f are continuous, ϕ_{i}^{−1} : R^{n} → RP^{n} is continuous. Thus (ϕ_{i}, U_{i}) is a chart. Claim: {(ϕ_{i}, U_{i}) | i = 1, ..., n + 1} is a (pre) atlas for RP^{n}. R^{n+1} − {0} = ∪_{i=1}^{n+1}V_{i}. Thus RP^{n} = ∪_{i=1}^{n+1}U_{i}. Claim: ϕ_{j} ∘ ϕ_{i}^{−1} : ϕ_{i}(U_{i} ∩ U_{j}) → ϕ_{i}(U_{i} ∩ U_{j}) is smooth. Suppose j < i. ϕ_{j}(ϕ_{i}^{−1}(x_{1}, ..., x_{n})) = ϕ_{j}([x_{1}, ..., x_{i−1}, 1, x_{i}, ..., x_{n}])

= (\frac{x_{1}}{x_{j}} , ..., \frac{x_{i−1}}{x_{j}} , \frac{1}{x_{j}} , \frac{x_{i}}{x_{j}} , ..., \frac{x_{n}}{x_{j}})

Since all the components of ϕ_{j} ∘ ϕ_{i}^{−1} are rational functions with non-vanishing denominators (0 is not in the domain of ϕ_{j} ∘ ϕ_{i}^{−1}), ϕ_{j} ∘ ϕ_{i}^{−1} is smooth. Similarly ϕ_{j} ∘ ϕ_{i}^{−1} is smooth when j > i. Thus {(ϕ_{i}, U_{i}) | i = 1, ..., n + 1} is a (pre) atlas for RP^{n}, and RP^{n} is a smooth manifold.